CLP-4 Vector Calculus

\begin{align*} &\big(x(\phi),y(\phi)\big) =\big(a\cos(\phi-\tfrac{\pi}{2}),a\sin(\phi-\tfrac{\pi}{2})\big) =\big(a\sin \phi ,-a\cos \phi \big),\\\ &\tfrac{\pi}{2}\le\phi\le\pi \end{align*}

\begin{equation*} \big(x(s),y(s)\big) =\left(a\cos\left(\frac{\pi}{2}-\frac{s}{a}\right)\,,\, a\sin\left(\frac{\pi}{2}-\frac{s}{a}\right)\right) ,\ 0\le s\le\frac{\pi}{2}a \end{equation*}

\begin{align*} 1&=x^2-\frac{1}{4}y^2+3z^2=(-y-z)^2-\frac14y^2+3z^2\\ &=\frac{3}{4}y^2+4z^2+2yz\\ \end{align*}

Completing the square,

\begin{align*} 1&=\frac{1}{2}y^2+\left(2z+\frac{y}{2}\right)^2\\ 1-\frac{y^2}{2}&=\left(2z+\frac{y}{2}\right)^2\\ \end{align*}

Since \(y\) is small, the left hand is close to \(1\) and the right hand side is close to \((2z)^2\text{.}\) So \((2z)^2\approx 1\text{.}\) Since \(z\) is negative, \(z\approx -\frac{1}{2}\) and \(2z+\frac{y}{2} \lt 0\text{.}\) Also, \(1-\frac{y^2}{2}\) is positive, so it has a real square root.

\begin{align*} -\sqrt{1-\frac{y^2}{2}}&=2z+\frac{y}{2}\\ -\frac12\sqrt{1-\frac{y^2}{2}}-\frac{y}{4}&=z \end{align*}

\begin{equation*} \vr'(t)=-e^{-t}\,\hi-\frac{1}{t^2}\,\hj+4(t-1)(t-2)(t-3)\hk \end{equation*}

\begin{align*} \diff{ }{t}\big[ (\vr \times \vr')\cdot\vr'' \big] &= (\vr' \times \vr')\cdot\vr'' +(\vr \times \vr'')\cdot\vr'' +(\vr \times \vr')\cdot\vr''' \end{align*}

\begin{equation*} \diff{ }{t}\big[ (\vr \times \vr')\cdot\vr'' \big] = (\vr \times \vr')\cdot\vr''' \end{equation*}

\begin{equation*} \diff{}{t}|\vr(t)|^2 =\diff{}{t}\big[\vr(t)\cdot\vr(t)\big] =2\vr(t)\cdot\vr'(t)=0 \end{equation*}

\begin{equation*} \vv(t) = \vr'(t) = 5 \sqrt{2}\,\hi + 5e^{5t}\,\hj +5 e^{-5t}\,\hk \end{equation*}

\begin{equation*} |\vv(t)| = |\vr'(t)| = 5 \big|\sqrt{2}\,\hi + e^{5t}\,\hj + e^{-5t}\,\hk\big| = 5\sqrt{2+ e^{10t}+ e^{-10t}} \end{equation*}

\begin{equation*} \vr(t)= a \cos t\,\hi+a\sin t\,\hj+ct\,\hk \end{equation*}

\begin{align*} \text{velocity}&= \vv(t)= \vr'(t)=-a \sin t\,\hi+a\cos t\,\hj+c\,\hk\\ \text{speed}&=\diff{s}{t}(t) = |\vr'(t)| = \sqrt{a^2+c^2}\\ \text{acceleration}&=\va(t)= \vr''(t)=-a \cos t\,\hi-a\sin t\,\hj \end{align*}

\begin{equation*} \hat\vT(1) = \frac{(2,0,1)}{\sqrt{5}} \end{equation*}

\begin{align*} \text{arc length} &= \int_{-1}^0 \sqrt{4t^2+t^4}\ \dee{t} \end{align*}

\begin{align*} \text{arc length} &= \int_0^1 \sqrt{4t^2+t^4}\ \dee{t} =\int_0^1 t\sqrt{4+t^2}\ \dee{t} =\Big[\tfrac{1}{3}{(4+t^2)}^{3/2}\Big]_0^1\\ &=\tfrac{1}{3}\big[5^{3/2}-8\big] \end{align*}

\begin{align*} \vr(t)&=\left(t,\sqrt{\frac{3}{2}}t^2,t^3\right)\\ \diff{\vr}{t}(t)&=\left(1,\sqrt{6}t,3t^2\right)\\ \left|\diff{\vr}{t}(t)\right|&=\sqrt{1^2+(\sqrt{6}t)^2+(3t^2)^2}=\sqrt{1+6t^2+9t^4}\\ &=\sqrt{(3t^2+1)^2}=3t^2+1\\ \int_{0}^1\left| \diff{\vr}{t}(t)\right|\dee t&=\int_0^1\left(3t^2+1 \right)\dee t=2 \end{align*}

\begin{align*} x'(t)&=a\big[\cos^2 t-\sin^2 t\big]=a\cos 2t\\ y'(t)&=2a\sin t\cos t=a\sin 2t\\ z'(t)&=b \end{align*}

\begin{equation*} \diff{s}{t}(t) =\sqrt{x'(t)^2+y'(t)^2+z'(t)^2}=\sqrt{a^2+b^2} \end{equation*}

\begin{align*} \vr(t)&=(t+\sin t, \cos t)\\ \vr'(t)&=(1+\cos t,-\sin t)\\ \vr''(t)&=(-\sin t,-\cos t)\\ |\vr''(t)|&=\sqrt{\sin^2t+\cos^2t}=1 \end{align*}

\begin{align*} \diff{s}{t}(t) =\big|\vr'(t)\big| & = \left|\left(2\cos t - 2t\sin t\,,\, 2\sin t + 2t \cos t\,,\, t^2\right)\right|\\ &=\sqrt{\big(2\cos t - 2t\sin t\big)^2 +\big(2\sin t + 2t \cos t\big)^2 +t^4}\\ &= \sqrt{4+ 4t^2 +t^4}\\ &= 2+t^2 \end{align*}

\begin{align*} \text{length } &=\int_0^2 \diff{s}{t}\,\dee{t} =\int_0^2 (2+t^2)\,\dee{t} = \left[2t +\frac{t^3}{3}\right]_0^2 =\frac{20}{3} \end{align*}

\begin{equation*} \vr'(\pi) = \left(2\cos\pi - 2\pi\sin\pi\,,\, 2\sin\pi + 2\pi \cos\pi\,,\, \pi^2\right) = (-2\,,\,-2\pi\,,\,\pi^2) \end{equation*}

\begin{align*} x(t) &= -2\pi -2t\\ y(t) &= -2\pi t\\ z(t) &= \frac{\pi^3}{3} + \pi^2 t \end{align*}

\begin{equation*} \vr'(t) = \big(-3 \sin t, 3 \cos t, 4\big) \end{equation*}

\begin{equation*} \diff{s}{t}(t) = |\vr'(t)| = \big|\big(-3 \sin t, 3 \cos t, 4\big)\big| =5 \end{equation*}

\begin{equation*} \int_1^2\dee{t}\ \diff{s}{t}(t) =\int_1^2\dee{t}\ 5 =5 \end{equation*}

\begin{align*} &x(\theta)=3\cos\theta,\ y(\theta)=3\sin\theta,\ z(\theta)=2x(\theta)+3 y(\theta) =6\cos\theta+9\sin\theta\\ & 0\le\theta\le 2\pi \end{align*}

\begin{align*} \diff{s}{\theta} &=\sqrt{x'(\theta)^2+y'(\theta)^2+z'(\theta)^2}\\ &=\sqrt{9\sin^2\theta+9\cos^2\theta+36\sin^2\theta +81\cos^2\theta-108\sin\theta\cos\theta}\\ &=\sqrt{45+45\cos^2\theta-108\sin\theta\cos\theta} \end{align*}

\begin{equation*} s=\int_0^{2\pi} \sqrt{45+45\cos^2\theta-108\sin\theta\cos\theta}\,\dee{\theta} \end{equation*}

\begin{align*} \vr'(t) & = -\sin t\cos^2 t\,\hi + \sin^2 t\cos t\,\hi + 3\sin^2 t\cos t\,\hk\\ &= \sin t\cos t\big(-\cos t\,\hi +\sin t\,\hj +3\sin t\,\hk\big)\\ \diff{s}{t}(t) & = |\sin t\cos t|\sqrt{\cos^2 t + \sin ^2 t + 9\sin^2 t} = |\sin t\cos t|\sqrt{1+ 9\sin^2 t} \end{align*}

\begin{align*} &\int_0^{\pi/2} \diff{s}{t}(t)\,\dee{t} =\int_0^{\pi/2} \sin t\cos t \sqrt{1+ 9\sin^2 t}\,\dee{t}\\ &\hskip0.5in=\frac{1}{18}\int_1^{10} \sqrt{u}\ \dee{u} \qquad \text{with } u = 1+ 9\sin^2 t,\ \dee{u} = 18\sin t\cos t\,\dee{t}\\ &\hskip0.5in=\frac{1}{18}\Big[\frac{2}{3}u^{3/2}\Big]_1^{10}\\ &\hskip0.5in=\frac{1}{27}\big(10\sqrt{10}-1\big) \end{align*}

\begin{align*} \int_0^{\pi} \diff{s}{t}(t)\,\dee{t} &=\int_0^{\pi} |\sin t\cos t| \sqrt{1+ 9\sin^2 t}\,\dee{t}\\ &\hskip1in\text{Don't forget the absolute value signs!}\\ &=2\int_0^{\pi/2} |\sin t\cos t| \sqrt{1+ 9\sin^2 t}\,\dee{t}\\ &= 2\int_0^{\pi/2} \sin t\cos t \sqrt{1+ 9\sin^2 t}\,\dee{t} \end{align*}

\begin{align*} \vr(t)&= \frac{t^3}{3}\,\hi + \frac{t^2}{2}\,\hj + \frac{t}{2}\,\hk\\ \vr'(t)&= t^2\,\hi + t\,\hj + \frac{1}{2}\,\hk\\ \diff{s}{t}(t)=|\vr'(t)|&=\sqrt{t^4+t^2+\frac{1}{4}} =\sqrt{\Big(t^2+\frac{1}{2}\Big)^2}=t^2+\frac{1}{2} \end{align*}

\begin{gather*} s(t)=\int_0^t \diff{s}{t}(u)\,\dee{u} =\int_0^t \Big(u^2+\frac{1}{2}\Big)\,\dee{u} =\frac{t^3}{3} +\frac{t}{2} \end{gather*}

\begin{align*} \vr(t) & = t^m\,\hi + t^m\,\hj + t^{3m/2}\,\hk\\ \vr'(t) &= mt^{m-1}\,\hi + mt^{m-1}\,\hj +\frac{3m}{2}t^{3m/2-1}\,\hk\\ \diff{s}{t} = |\vr'(t)| & = \sqrt{ 2m^2 t^{2m-2} +\frac{9m^2}{4} t^{3m-2} } = mt^{m-1}\sqrt{2 + \frac{9}{4}t^m } \end{align*}

\begin{align*} \int_a^b \diff{s}{t}(t)\,\dee{t} &=\int_a^b mt^{m-1}\sqrt{2 + \frac{9}{4}t^m }\ \dee{t}\\ &=\frac{4}{9}\int_{2 + \frac{9}{4}a^m}^{2 + \frac{9}{4}b^m}\sqrt{u}\,\dee{u} \qquad\text{with } u = 2 + \frac{9}{4}t^m,\ \dee{u} = \frac{9m}{4}t^{m-1}\\ &=\frac{4}{9}\Big[\frac{2}{3}u^{3/2}\Big]_{2 + \frac{9}{4}a^m} ^{2 + \frac{9}{4}b^m}\\ &=\frac{8}{27}\Big[\Big(2 + \frac{9}{4}b^m\Big)^{3/2} -\Big(2 + \frac{9}{4}a^m\Big)^{3/2}\Big] \end{align*}

\begin{align*} \vr(x)&=x\,\hi+\sqrt{x}\,\hj+\frac{2}{3}x^{3/2}\,\hk \end{align*}

\begin{align*} \vr'(x)&=\hi+\frac{1}{2\sqrt{x}}\,\hj+\sqrt{x}\,\hk\\ \vr''(x)&=-\frac{1}{4x^{3/2}}\,\hj+\frac{1}{2\sqrt{x}}\,\hk\\ \diff{s}{x}=|\vr'(x)| &=\sqrt{1+\frac{1}{4x}+x} =\sqrt{\left(\frac{1}{2\sqrt{x}}+\sqrt{x}\right)^2} =\frac{1}{2\sqrt{x}}+\sqrt{x}\\ \diff{s}{x}(1)&=\frac{3}{2} \end{align*}

\begin{align*} \int_C \ ds &=\int_0^9 \diff{s}{x}\ \dee{x} =\int_0^9 \left(\frac{1}{2\sqrt{x}}+\sqrt{x}\right)\ \dee{x} =\left[\sqrt{x}+\frac{2}{3}x^{3/2}\right]_0^9\\ &=3+18 =21 \end{align*}

\begin{align*} &\vR(t)=\vr\big(x(t)\big)\implies \vR'(t)=\vr'\big(x(t)\big)\diff{x}{t}(t)\\ &\hskip0.25in\implies 9=|\vR'(t)|=\diff{s}{x}\big(x(t)\big) \diff{x}{t}(t) =\left(\frac{1}{2\sqrt{x(t)}}+\sqrt{x(t)}\right)\diff{x}{t}(t) \end{align*}

\begin{equation*} 9=\left(\frac{1}{2\sqrt{1}}+\sqrt{1}\right)\diff{x}{t}(0) \implies \diff{x}{t}(0)=6 \end{equation*}

\begin{equation*} \vR'(0)=\vr'(1)\diff{x}{t}(0) =\left(\hi+\frac{1}{2}\,\hj+\hk\right)6 =\ 6\,\hi+3\,\hj+6\,\hk \end{equation*}

\begin{equation*} \vR'(t)=\vr'\big(x(t)\big)\diff{x}{t}(t)\implies \vR''(t)=\vr''\big(x(t)\big)\left(\diff{x}{t}(t)\right)^2 +\vr'\big(x(t)\big)\difftwo{x}{t}(t) \end{equation*}

\begin{equation*} \diff{x}{t}(t)=9\left(\frac{1}{2\sqrt{x(t)}}+\sqrt{x(t)}\right)^{-1} \end{equation*}

\begin{equation*} \frac{d^2x}{dt^2}(t)=-9\left(\frac{1}{2\sqrt{x(t)}}+\sqrt{x(t)}\right)^{-2} \left(-\frac{1}{4x(t)^{3/2}}+\frac{1}{2\sqrt{x(t)}}\right)\diff{x}{t}(t) \end{equation*}

\begin{equation*} \difftwo{x}{t}(0)=-9\left(\frac{3}{2}\right)^{-2} \left(\frac{1}{4}\right)6=-6 \end{equation*}

\begin{align*} \vR''(0)&=\vr''(1)\ \big(6\big)^2+\vr'(1)\big(-6\big) =36\left(-\frac{1}{4}\,\hj+\frac{1}{2}\,\hk\right) -6\left(\hi+\frac{1}{2}\,\hj+\hk\right)\\ &=-6\,\hi-12\,\hj+12\,\hk \end{align*}

\begin{equation*} \vv(t)=\vr'(t)=(\cos t, -\sin t , 1) \end{equation*}

\begin{equation*} \vL(t)=\vr \times \vv=(\sin t , \cos t , t) \times (\cos t, -\sin t , 1). \end{equation*}

\begin{align*} \vr(u)&=u^3\,\hi+3u^2\,\hj+6u\,\hk \end{align*}

\begin{align*} \vr'(u)&=3u^2\,\hi+6u\,\hj+6\,\hk\\ \vr''(u)&=6u\,\hi+6\,\hj\\ \diff{s}{u}(u)=|\vr'(u)| &=\sqrt{9u^4+36u^2+36}=3\big(u^2+2\big) \end{align*}

\begin{equation*} \int_\cC \ ds =\int_0^1 \diff{s}{u}\ \dee{u} =\int_0^1 3\big(u^2+2\big)\ \dee{u} =\big[u^3+6u\big]_0^1 =7 \end{equation*}

\begin{equation*} \vR(t)=\vr\big(u(t)\big)\implies \vR'(t)=\vr'\big(u(t)\big)\diff{u}{t} \end{equation*}

\begin{equation*} 6\,\hi+12\,\hj+12\,\hk=\vR'(t_1)=\vr'(1)\diff{u}{t}(t_1) =\big(3\,\hi+6\,\hj+6\,\hk\big)\diff{u}{t}(t_1) \end{equation*}

\begin{equation*} \vR'(t)=\vr'\big(u(t)\big)\diff{u}{t}\implies \vR''(t)=\vr''\big(u(t)\big)\Big(\diff{u}{t}\Big)^2 +\vr'\big(u(t)\big)\difftwo{u}{t} \end{equation*}

\begin{align*} 27\,\hi+30\,\hj+6\,\hk &=\vR''(t_1)=\vr''(1)\Big(\diff{u}{t}(t_1)\Big)^2 +\vr'\big(1\big)\difftwo{u}{t}(t_1)\\ &=\big(6\,\hi+6\,\hj\big)2^2+\big(3\,\hi+6\,\hj+6\,\hk\big)\difftwo{u}{t}(t_1) \end{align*}

\begin{equation*} 3\,\hi+6\,\hj+6\,\hk =\big(3\,\hi+6\,\hj+6\,\hk\big)\difftwo{u}{t}(t_1) \implies \difftwo{u}{t}(t_1)=1 \end{equation*}

\begin{equation*} m\vr''(t) = \vF(t)\qquad\text{so that}\qquad \vr''(t) = -3t\,\hi + \sin t\,\hj + 2e^{2t}\,\hk \end{equation*}

\begin{gather*} \vr'(t) = -3\frac{t^2}{2}\,\hi - \cos t\,\hj + e^{2t}\,\hk +\vc \end{gather*}

\begin{gather*} \vr'(t) = \left(\frac{\pi^2}{2}-\frac{3t^2}{2}\right)\,\hi +(1- \cos t)\,\hj + \big(e^{2t}-1\big)\,\hk \end{gather*}

\begin{gather*} \vr(t) = \left(\frac{\pi^2 t}{2}-\frac{t^3}{2}\right)\,\hi +(t- \sin t)\,\hj + \left(\frac{1}{2}e^{2t}-t\right)\,\hk + \vc \end{gather*}

\begin{gather*} \vr(t) = \left(\frac{\pi^2 t}{2}-\frac{t^3}{2}\right)\,\hi +(t- \sin t)\,\hj + \left(\frac{1}{2}e^{2t}-t\right)\,\hk \end{gather*}

\begin{gather*} 0=\left(\frac{\pi^2 t}{2}-\frac{t^3}{2}\right) =\frac{t}{2}(\pi^2-t^2) \iff t=0, \pm\pi \end{gather*}

\begin{align*} \vr'(\pi) &= \left(\frac{\pi^2}{2}-\frac{3\pi^2}{2}\right)\,\hi +(1- \cos\pi)\,\hj + \big(e^{2\pi}-1\big)\,\hk\\ &= -\pi^2\,\hi +2\,\hj + \big(e^{2\pi}-1\big)\,\hk \end{align*}

\begin{align*} \vr(x)&=x\,\hi+x^2\,\hj+\frac{2}{3}x^3\,\hk\\ \vr'(x)&=\hi+2x\,\hj+2x^2\,\hk\\ \vr''(x)&=2\,\hj+4x\,\hk\\ \diff{s}{x} &=|\vr'(x)| =\sqrt{1+4x^2+4x^4}=1+2x^2 \end{align*}

\begin{equation*} \int_C \ \dee{s} =\int_0^3 \diff{s}{x}\ \dee{x} =\int_0^3 \big(1+2x^2\big)\ \dee{x} ={\Big[x+\frac{2}{3}x^3\Big]}_0^3 =21 \end{equation*}

\begin{equation*} \vR(t)=\vr\big(x(t)\big)\implies \vR'(t)=\vr'\big(x(t)\big)\diff{x}{t} \end{equation*}

\begin{equation*} 6=\diff{s}{t}=\diff{s}{x}\diff{x}{t}=(1+2x^2)\diff{x}{t} \implies \diff{x}{t}=\frac{6}{1+2x^2} \end{equation*}

\begin{equation*} \vR'=\vr'\big(1\big)\diff{x}{t} =\big(\hi+2\,\hj+2\,\hk\big)2 =2\hi+4\,\hj+4\,\hk \end{equation*}

\begin{equation*} \vR'(t)=\vr'\big(x(t)\big)\diff{x}{t}\implies \vR''(t)=\vr''\big(x(t)\big){\Big(\diff{x}{t}\Big)}^2 +\vr'\big(x(t)\big)\difftwo{x}{t} \end{equation*}

\begin{equation*} 0=4x{\Big(\diff{x}{t}\Big)}^2+(1+2x^2)\difftwo{x}{t} \end{equation*}

\begin{equation*} \vR''=\big(2\,\hj+4\,\hk\big)\big(2\big)^2 -\big(\hi+2\,\hj+2\,\hk\big)\frac{16}{3} = -\frac{8}{3}\big(2\hi+\,\hj-2\,\hk\big) \end{equation*}

\begin{align*} \tan\theta&=\frac{y}{x}\\ \end{align*}

Differentiating implicitly with respect to \(t\text{:}\)

\begin{align*} \sec^2\theta\,\diff{\theta}{t}&=\frac{xy'-yx'}{x^2}\\ \diff{\theta}{t}&=\cos^2\theta\left(\frac{xy'-yx'}{x^2}\right)=\left(\frac{x}{\sqrt{x^2+y^2}}\right)^2\left(\frac{xy'-yx'}{x^2}\right)\\ &=\frac{xy'-yx'}{{x^2+y^2}} \end{align*}

\begin{align*} \diff{\vr}{t}&=(\sqrt{2t},t,1)\\ \left| \diff{\vr}{t}\right|&=\sqrt{2t+t^2+1}=|t+1|\\ L&=\int_0^{10}(t+1)\dee{t}=60 \end{align*}

\begin{equation*} \vr(\theta)=a\cos\theta\,\hi+a\sin\theta\,\hj+b\theta\,\hk \end{equation*}

\begin{align*} \int_0^{2\pi}\sqrt{9+\frac{1}{400\pi^2}}\ \dee{\theta} &=2\pi\sqrt{9+\frac{1}{400\pi^2}} \end{align*}

\begin{equation*} \frac{1000}{2\pi\sqrt{9+\frac{1}{400\pi^2}}} =\frac{500}{\pi\sqrt{9+\frac{1}{400\pi^2}}} \end{equation*}

\begin{equation*} \frac{500}{\pi\sqrt{9+\frac{1}{400\pi^2}}}\cdot \frac{1}{10} =\frac{50}{\pi\sqrt{9+\frac{1}{400\pi^2}}} \approx 5.3\textrm{ cm} \end{equation*}

\begin{align*} \diff{\vu}{t}(t)&=\alpha e^{\alpha t}\diff{\vr}{t}(t) +e^{\alpha t}\frac{d^2\vr}{dt^2}(t)\\ &=\alpha e^{\alpha t}\diff{\vr}{t}(t) -ge^{\alpha t}\hk -\alpha e^{\alpha t}\diff{\vr}{t}(t)\\ &=-ge^{\alpha t}\hk \end{align*}

\begin{equation*} \vu(T)-\vu(0)=-g\frac{e^{\alpha T}-1}{\alpha}\hk \end{equation*}

\begin{equation*} \vu(T)=\vu(0)-g\frac{e^{\alpha T}-1}{\alpha}\hk =\diff{\vr}{t}(0)-g\frac{e^{\alpha T}-1}{\alpha}\hk =\vv_0-g\frac{e^{\alpha T}-1}{\alpha}\hk \end{equation*}

\begin{equation*} \diff{\vr}{t}(T) =e^{-\alpha T}\vv_0-g\frac{1-e^{-\alpha T}}{\alpha}\hk \end{equation*}

\begin{gather*} \vr(t)-\vr(0) =\frac{e^{-\alpha t}-1}{-\alpha}\vv_0-g\frac{ t}{\alpha}\hk +g\frac{e^{-\alpha t}-1}{-\alpha^2}\hk \end{gather*}

\begin{gather*} \vr(t)=\vr_0-\frac{e^{-\alpha t}-1}{\alpha}\vv_0 +g\frac{1-\alpha t-e^{-\alpha t}}{\alpha^2}\hk \end{gather*}

\begin{align*} \diff{\vb}{s}&=\diff{}{s}[\va(t(s))]=\diff{\va}{t}\,\diff{t}{s} \end{align*}

\begin{align*} \vr'(t) &= (6\sin^2(t)\cos t\,,\, -6\sin t\cos^2(t)\,,\, 3\cos^2t-3\sin^2t)\\ &= 3\big(\sin t \sin(2t)\,,\,-\cos t\sin(2t)\,,\,\cos(2t)\big) \end{align*}

\begin{align*} \vr'(\pi/3) &= 3\big(\frac{3}{4}\,,\, -\frac{\sqrt{3}}{4}\,,\,-\frac{1}{2}\big) \end{align*}

\begin{align*} \hat\vT &= \frac{\big(\frac{3}{4}\,,\, -\frac{\sqrt{3}}{4}\,,\,-\frac{1}{2}\big)} {\big|\big(\frac{3}{4}\,,\, -\frac{\sqrt{3}}{4}\,,\,-\frac{1}{2}\big)\big|} =\big(\frac{3}{4}\,,\, -\frac{\sqrt{3}}{4}\,,\,-\frac{1}{2}\big) \end{align*}

\begin{align*} \diff{s}{t} &= |\vr'(t)| = 3\sqrt{\sin^2t\sin^2(2t)+\cos^2t\sin^2(2t)+\cos^2(2t)}\\ &= 3\sqrt{\sin^2(2t)+\cos^2(2t)}\\ &=3 \end{align*}

\begin{gather*} \vR(s) = \big(2 \sin^3(\frac{s}{3}) , 2\cos^3(\frac{s}{3}), 3 \sin(\frac{s}{3}) \cos(\frac{s}{3})\big) \end{gather*}

\begin{gather*} \vr'(t) = e^t (\cos t, \sin t) + e^t (-\sin t, \cos t) =e^t\big(\cos t -\sin t\,,\,\sin t + \cos t\big) \end{gather*}

\begin{gather*} \diff{s}{t} = \big|\vr'(t)\big| = e^t \sqrt{(\cos t-\sin t)^2+(\sin t+\cos t)^2} = \sqrt{2}e^t \end{gather*}

\begin{equation*} s(t)=\int_{-\infty}^t \diff{s}{t}(\tilde t)\ \dee{\tilde t} =\int_{-\infty}^t \sqrt{2} e^{\tilde t}\ \dee{\tilde t} =\sqrt{2}e^t \end{equation*}

\begin{align*} &\vR(s) = e^t (\cos t, \sin t)\Big|_{t=\ln\left(\frac{s}{\sqrt{2}}\right)} =\frac{s}{\sqrt{2}}\left(\cos\Big(\ln\Big(\frac{s}{\sqrt{2}}\Big)\Big)\,,\, \sin\Big(\ln\Big(\frac{s}{\sqrt{2}}\Big)\Big)\right)\\ &\quad\text{with } s \gt 0 \end{align*}

\begin{align*} \vr(t)&=\left(\frac{1}{\sqrt{1+t^2}}, \frac{\arctan t}{\sqrt{1+t^{-2}}}, \arctan t\right)\\ &=\left(\frac{1}{\sqrt{1+\tan^2z}}, \frac{z}{\sqrt{1+\cot^{2}z}}, z\right)\\ &=\left(\frac{1}{|\sec z|}, \frac{z}{|\csc z|}, z\right)\\ &=\left(|\cos z|, z|\sin z|, z\right)\\ \end{align*}

Since \(0 \le t \text{,}\) and \(\arctan t \lt \pi/2\) we have \(0 \le z \lt \pi/2\text{,}\) so \(\cos z\) and \(\sin z\) are both nonnegative.

\begin{align*} &=\left(\cos z, z\sin z, z\right) \end{align*}

\begin{align*} \vr(t)&=(\tfrac12 t^2 , \tfrac13 t^3)\\ \vr'(t)&=(t,t^2)\\ |\vr'(t)|&=\sqrt{t^2+t^4}=|t|\sqrt{1+t^2}\\ s(t)&=\int_{-1}^t|x|\sqrt{1+x^2}\dee{x}\\ &=\begin{cases} \int_{-1}^t -x\sqrt{1+x^2}\dee{x} & \text{when } t \le 0 \\ \int_{-1}^0 -x\sqrt{1+x^2}\dee{x} +\int_{0}^t x\sqrt{1+x^2}\dee{x} & \text{when } t \gt 0 \end{cases}\\ \end{align*}

Let \(u=1+x^2, \frac12\dee{u}=x\dee{x}\)

\begin{align*} &=\begin{cases} -\int_{2}^{1+t^2} \frac{1}{2}\sqrt{u}\dee{u} & \text{when } t \le 0 \\ -\int_{2}^1 \frac12\sqrt{u}\dee{u}+ \int_{1}^{1+t^2} \frac12\sqrt{u}\dee{u} & \text{when } t \gt 0 \end{cases}\\ &=\begin{cases} -\frac{1}{3}u^{3/2}|_{2}^{1+t^2} & \text{when } t \le 0 \\ -\frac{1}{3}u^{3/2}|_{2}^{1}+ \frac13u^{3/2}|_{1}^{1+t^2} & \text{when } t \gt 0 \end{cases}\\ &=\begin{cases} \frac{2^{3/2}}{3}-\frac{1}{3}(1+t^2)^{3/2} & \text{when } t \le 0 \\ -\frac23+\frac{2^{3/2}}{3}+\frac13(1+t^2)^{3/2} & \text{when } t \gt 0 \end{cases}\\ \end{align*}

Solving for \(t\) in terms of \(s\text{:}\)

\begin{align*} 1+t^2&=\begin{cases}(2\sqrt2-3s)^{2/3} &\text{when } t \le 0 \\ (3s+2-2\sqrt{2})^{2/3} &\text{when }t \gt 0\end{cases}\\ t^2&=\begin{cases}(2\sqrt2-3s)^{2/3}-1 &\text{when } t \le 0 \\ (3s+2-2\sqrt{2})^{2/3}-1 &\text{when }t \gt 0 \end{cases}\\ \end{align*}

Remembering that \(\sqrt{t^2}=|t|\text{:}\)

\begin{align*} t&=\begin{cases}-\sqrt{(2\sqrt2-3s)^{2/3}-1} &\text{when } t \le 0 \\ \sqrt{(3s+2-2\sqrt{2})^{2/3}-1} &\text{when }t \gt 0 \end{cases} \end{align*}

\begin{equation*} \vR(s)= \Big(\frac12\left[ (2\sqrt2-3s)^{2/3}-1\right], -\frac13\left[(2\sqrt2-3s)^{2/3}-1\right]^{3/2}\Big) \end{equation*}

\begin{equation*} \vR(s)= \Big(\frac12\left[(3s+2-2\sqrt2)^{2/3}-1 \right], \frac13\left[(3s+2-2\sqrt2)^{2/3}-1\right]^{3/2}\Big) \end{equation*}

\begin{align*} \vv(t)&=\vr'(t)=(3\cos t, -3\sin t) \\ \hT(t)&=\frac{\vr'(t)}{|\vr'(t)|}=\frac{(3\cos t, -3\sin t)}{3}=(\cos t, -\sin t)\\ \hT(s)&=\vR'(s)=(\cos(s/3),-\sin(s/3)) \end{align*}

\begin{align*} \diff{\hT}{s}(s) &= \ka(s)\,\hN(s)\\ \left(-\frac13\sin(s/3) , -\frac13\cos(s/3) \right)&=\frac13\hN(s)\\ \left(-\sin(s/3) , -\cos(s/3) \right)&=\hN(s) \end{align*}

\begin{align*} \diff{\hT}{t} &= \ka \diff{s}{t} \hN(t)\\ (-\sin t ,-\cos t) &= \frac13 (3) \hN(t)\\ (-\sin t ,-\cos t) &= \hN(t) \end{align*}

\begin{align*} \hT(t)&=\frac{\vv(t)}{|\vv(t)|}=\frac{\vr'(t)}{|\vr'(t)|}\\ &=\frac{\big(e^t(\cos t - \sin t) , e^t(\cos t + \sin t)\big)}{\sqrt{e^{2t}(\cos t - \sin t)^2+e^{2t}(\cos t + \sin t)^2}}\\ &=\frac1{\sqrt2}\big(\cos t - \sin t , \cos t + \sin t\big)\\ \diff{\hT}{t}&=\frac1{\sqrt2}\big(-\sin t - \cos t , -\sin t + \cos t\big)\\ \end{align*}

Since \(\vR(s)\) is parametrized with respect to arclength, \(|\vR'(s)|=1\text{.}\)

\begin{align*} \hT(s)&=\vR'(s)\\ \end{align*}

Making ample use of the chain rule, and setting \(U(s)=\left(\ln\left( s/\sqrt{2}\right)\right)\text{,}\) we have \(U'(s)=\frac{1}{s}\text{:}\)

\begin{align*} \hT(s)&=\frac{1}{\sqrt2}\left( \cos U(s)-\sin U(s) , \cos U(s) + \sin U(s)\right)\\ \diff{\hT}{s}&=\frac{1}{\sqrt{2}\,s}\left( -\sin U(s)-\cos U(s) , -\sin U(s)+\cos U(s)\right) \end{align*}

\begin{equation*} y = g(x) = r - \sqrt{r^2-x^2} \end{equation*}

\begin{alignat*}{2} g'(x) &=\frac{x}{\sqrt{r^2-x^2}} & g'(0) &=0\\ g''(x)&=\frac{1}{\sqrt{r^2-x^2}} +\frac{x^2}{[{r^2-x^2]}^{3/2}}\qquad & g''(0)&=\frac{1}{r} \end{alignat*}

\begin{alignat*}{2} \vr'(x)& = \hi +f'(x)\,\hj\qquad & \vr'(0)& = \hi +f'(0)\,\hj = \hi\\ \vr''(x)& = f''(x)\,\hj & \vr''(0)& = f''(0)\,\hj \end{alignat*}

\begin{align*} \ka(0)& = \frac{|\vr'(0)\times\vr''(0)|}{|\vr'(0)|^3} = \frac{|f''(0)\,\hi\times\hj|}{|\hi|^3} =f''(0) \end{align*}

\begin{align*} x(t)&= \cos t+\sin t =\sqrt{2}\big[\cos t \cos\tfrac{\pi}{4} +\sin t \sin\tfrac{\pi}{4}\big]\\ &=\sqrt{2}\cos\big(t-\tfrac{\pi}{4}\big)\\ y(t)&= \sin t-\cos t =\sqrt{2}\big[\sin t \cos\tfrac{\pi}{4}-\cos t\sin \tfrac{\pi}{4}\big]\\ &=\sqrt{2}\sin\big(t-\tfrac{\pi}{4}\big) \end{align*}

\begin{align*} \vv(t)&=\vr'(t)=(-\sin t +\cos t, \cos t + \sin t)\\ \va(t)&=\vr''(t)=(-\cos t -\sin t, -\sin t + \cos t)\\ \vv(t) \times \va(t)&=\big[(-\sin t + \cos t)^2+(\cos t + \sin t)^2\big]\hk=2\hk\\ \diff{s}{t}&=\left|\diff{\vv}{t} \right|=\sqrt{(-\sin t + \cos t)^2+(\cos t + \sin t )^2}=\sqrt2\\ \ka&=\left|\frac{\vv(t) \times \va(t)}{\left(\diff{s}{t}\right)^3}\right|=\left|\frac{2\hk}{\sqrt{2}^3}\right|=\frac{1}{\sqrt 2} \end{align*}

\begin{align*} \vr(t)&= a\cos t\ \hi +b\sin t\ \hj\\ \vv(t)&= -a\sin t\ \hi +b\cos t\ \hj\\ |\vv(t)|& = \sqrt{a^2\sin^2t+b^2\cos^2 t}\\ \va(t)&= -a\cos t\ \hi -b\sin t\ \hj\\ \vv(t)\times\va(t) &=\det\left[\begin{matrix} \hi & \hj & \hk\\ -a\sin t & b\cos t & 0\\ -a\cos t & -b\sin t & 0 \end{matrix}\right] = ab\,\hk\\ \ka(t)&=\frac{|\vv(t)\times\va(t)|}{|\vv(t)|^3} =\frac{ab}{{[a^2\sin^2t+b^2\cos^2 t]}^{3/2}} \end{align*}

\begin{align*} \vv(t) & = \hi + e^t\,\hj & \vv(0) & = \hi +\hj \\ \frac{ds}{dt} &=|\vv(t)|=\sqrt{1+e^{2t}} & \frac{ds}{dt}(0) &=\sqrt{2} \\ \hT(t) & = \frac{\vv(t)}{|\vv(t)|} = \frac{\hi+e^t\hj}{\sqrt{1+e^{2t}}} & \hT(0) & = \frac{\vv(0)}{|\vv(0)|} = \frac{\hi+\hj}{\sqrt{2}}\\ \va(t) & = e^t\,\hj & \va(0) & = \hj \end{align*}

\begin{align*} \ka &=\frac{\big|\difftwo{y}{x}\big|} { {\big[1+\big(\diff{y}{x}\big)^2\big]}^{3/2} } = \frac{e^x} { {\big[1+\big(e^x\big)^2\big]}^{3/2} }\\ \ka(0)&=\frac{1}{[1+1]^{3/2}}=2^{-3/2} \end{align*}

\begin{equation*} \vr(t) = (t,1) - (\sin t,\cos t) \end{equation*}

\begin{align*} \vr(t) &= \big(t - \sin t\,,\, 1 - \cos t\big)\\ \vv(t) =\vr'(t)&= \big(1-\cos t\,,\, \sin t\big)\\ \diff{s}{t}(t)=|\vv(t)|&=\sqrt{2-2\cos t}\\ \va(t) = \vv'(t)&=\big(\sin t\,,\, \cos t\big)\\ \vv(t)\times \va(t)&=\det\left[\begin{matrix}\hi&\hj&\hk \\ 1-\cos t & \sin t & 0 \\ \sin t & \cos t & 0\end{matrix}\right] =\big(\cos t-1\big)\,\hk \end{align*}

\begin{gather*} \ka(t) =\frac{|\vv(t)\times\va(t)|}{|\vv(t)|^3} =\frac{|\cos t-1|}{(2-2\cos t)^{3/2}} =\frac{1}{2^{3/2}\sqrt{1-\cos t}} \end{gather*}

\begin{equation*} \rho(\pi) = \frac{1}{\ka(\pi)} = \frac{1}{1/2^{3/2}\sqrt{2}} =4 \end{equation*}

\begin{equation*} \vr(\pi) + \rho(\pi) \hN(\pi) =(\pi,2) + 4(0,-1) =(\pi,-2) \end{equation*}

\begin{equation*} (x-\pi)^2 +(y+2)^2 = 16 \end{equation*}

\begin{equation*} \vv(\theta)=x'(\theta)\,\hi+y'(\theta)\,\hj = \cos\big(\half\pi\theta^2\big)\,\hi + \sin\big(\half\pi\theta^2\big)\,\hj \end{equation*}

\begin{equation*} \diff{s}{\theta}(\theta)= |\vv(\theta)|=1 \implies s(\theta)=\theta+s(0) \end{equation*}

\begin{equation*} \hT(s)=\vv(s) = \cos\big(\half\pi s^2\big)\,\hi + \sin\big(\half\pi s^2\big)\,\hj \end{equation*}

\begin{equation*} \ka(s)=\left|\diff{\hT}{s}(s)\right| =\big|-\pi s\sin\big(\half\pi s^2\big)\,\hi +\pi s \cos\big(\half\pi s^2\big)\,\hj\big| =\pi s \end{equation*}

\begin{align*} \ka(x) &=\frac{\big|\difftwo{y}{x}(x)\big|} {\Big[1+\big(\diff{y}{x}(x)\big)^2\Big]^{3/2}} =\frac{\big|2x\big|} {\big[1+x^4\big]^{3/2}} \end{align*}

\begin{align*} 0&=\diff{ }{x}\ka(x)^2 =\diff{ }{x}\frac{4x^2}{{(1+x^4)}^3} =\frac{8x}{{(1+x^4)}^3} - 3\frac{16x^5}{{(1+x^4)}^4}\\ &=\frac{8x(1+x^4)-3\times 16x^5}{{(1+x^4)}^4} =\frac{8x(1-5x^4)}{{(1+x^4)}^4} \end{align*}

\begin{align*} \vr(t)&=\big(\cos t\,,\,\mp 2\sin t \,,\,\pm t/2\big)\\ \vv(t)&=\big(-\sin t\,,\,\mp 2\cos t \,,\,\pm 1/2\big)\\ \va(t)&=\big(-\cos t\,,\,\pm 2\sin t \,,\,0\big)\\ \vv(t)\times\va(t)&=\big(-\sin t\,,\,\mp\cos t/2\,,\,\mp 2\big)\\ \diff{\va}{t}(t)&=\big(\sin t\,,\,\pm 2\cos t\,,\,0\big)\\ \vv(t)\times\va(t)\cdot \diff{\va}{t}(t) &= -1\\ \tau(t)=\frac{\vv(t)\times\va(t)\cdot \diff{\va}{t}(t)}{|\vv(t)\times\va(t)|^2} &=-\frac{1}{\sin^2 t+\frac{1}{4}\cos^2 t+4} \lt 0 \end{align*}

\begin{equation*} \diff{}{s} (\vr(s)-\vr(0))\cdot\hB =\hT(s)\cdot\hB=0 \end{equation*}

\begin{equation*} \vr_c(s)=\vr(s)+\frac{1}{\ka(s)}\hN(s) \end{equation*}

\begin{equation*} \diff{}{s}\vr_c(s) =\hT(s)+\frac{1}{\ka_0}\big[\tau(s)\hB-\ka(s)\hT\big] =\hT(s)+\frac{1}{\ka_0}\big[0\hB-\ka_0\hT\big] =0 \end{equation*}

\begin{align*} \vr'(t)&=\big(e^t + e^{-t}\big)\,\hi + \big(e^t - e^{-t}\big)\,\hj +2\,\hk\\ \diff{s}{t}(t) &=|\vr'(t)|= \sqrt{4+2e^{2t}+2e^{-2t}} =\sqrt{2}\big(e^t+e^{-t}\big)\\ \vr''(t)&=\big(e^t - e^{-t}\big)\,\hi + \big(e^t + e^{-t}\big)\,\hj\\ \vr'(t)\times\vr''(t)&=-2\big(e^t + e^{-t}\big)\,\hi + 2\big(e^t - e^{-t}\big)\,\hj +4\,\hk \end{align*}

\begin{align*} \ka(t) &= \frac{|\vv(t)\times\va(t)|}{\big(\diff{s}{t}\big)^3} =\frac{2\sqrt{4+2e^{2t}+2e^{-2t}}}{[4+2e^{2t}+2e^{-2t}]^{3/2}} =\frac{1}{2+e^{2t}+e^{-2t}} \end{align*}

\begin{align*} \int_0^1 \diff{s}{t}(t) \ \dee{t} &= \sqrt{2} \int_0^1 (e^t+e^{-t}) \ \dee{t} =\sqrt{2}\Big[e^t-e^{-t}\Big]_0^1 =\sqrt{2}\Big[e-\frac{1}{e}\Big] \end{align*}

\begin{align*} \vv(t)&=(1,2t,3t^2) & \vv(2)&=(1,4,12)\\ \va(t)&=(0,2,6t) & \va(2)&=(0,2,12)\\ \diff{\va}{t}(t)&=(0,0,6) & \diff{\va}{t}(2)&=(0,0,6)\\ && \vv(2) \times \va(2)&=(24,-12,2)\\ && |\vv(2) \times \va(2)|&=2\sqrt{181} \end{align*}

\begin{align*} \tau(t)&=\dfrac{(\vv(t) \times \va(t))\cdot\diff{\va}{t}(t)}{|\vv(t) \times \va(t)|^2}\\ \tau(2)&=\frac{(24,-12,2)\cdot(0,0,6)}{(2\sqrt{181})^2}=\frac{3}{{181}} \end{align*}

\begin{align*} \vr(t)&=t\,\hi + \frac{t^2}{2}\,\hj + \frac{t^3}{3}\,\hk\\ \vv(t)=\vr'(t)&= \hi + t\,\hj + t^2\,\hk\\ \va(t)=\vr''(t)&= \hj + 2t\,\hk\\ \vv(t)\times\va(t) &=\det\left[\begin{matrix} \hi & \hj & \hk\\ 1 & t & t^2\\ 0 & 1 & 2t \end{matrix}\right]\\ & = t^2\,\hi-2t\,\hj+\hk\\ \va'(t)&= 2\,\hk \end{align*}

\begin{align*} \hT(t)&=\frac{\vv(t)}{|\vv(t)|} =\frac{\hi + t\,\hj + t^2\,\hk}{\sqrt{1+t^2+t^4}}\\ \ka(t)&=\frac{|\vv(t)\times\va(t)|}{|\vv(t)|^3} =\frac{\sqrt{1+4t^2+t^4}}{[1+t^2+t^4]^{3/2}}\\ \hB(t)&=\frac{\vv(t)\times\va(t)}{|\vv(t)\times\va(t)|} =\frac{t^2\,\hi-2t\,\hj+\hk}{\sqrt{1+4t^2+t^4}}\\ \hN(t)&=\hB(t)\times\hT(t)\\ &= \frac{1}{\sqrt{1+t^2+t^4}\sqrt{1+4t^2+t^4}}\det\left[\begin{matrix} \hi & \hj & \hk\\ t^2 & -2t & 1\\ 1 & t & t^2 \end{matrix}\right]\\ &=\frac{-(t+2t^3)\,\hi+(1-t^4)\,\hj+(2t+t^3)\hk} {\sqrt{1+t^2+t^4}\sqrt{1+4t^2+t^4}}\\ \tau(t)&=\frac{\big(\vv(t)\times\va(t)\big)\cdot\va'(t)}{|\vv(t)\times\va(t)|^2} =\frac{2}{1+4t^2+t^4} \end{align*}

\begin{align*} \vr(t)&=(t^3,t,e^{ct}) & \vr(5)&=(5^3,5,e^{5c})\\ \vv(t)&=(3t^2,1,ce^{ct}) & \vv(5)&=(3\cdot 5^2,1,ce^{5c})\\ \va(t)&=(6t,0,c^2e^{ct}) & \va(5)&=(6\cdot 5,0,c^2e^{5c})\\ \diff{\va}{t}(t)&= (6,0,c^3e^{ct})& \diff{\va}{t}(5)&= (6,0,c^3e^{5c})\\ && \vv(5) \times \va(5)&=(c^2e^{5c},15ce^{5c}(2-5c),-30) \end{align*}

\begin{align*} 0&=\tau(5)=\dfrac{(\vv(5) \times \va(5))\cdot\diff{\va}{t}(5)}{|\vv(5) \times \va(5)|^2}\\ 0&=(\vv(5)\times\va(5))\cdot \diff{\va}{t}(5)\\ &=(c^2e^{5c},15ce^{5c}(2-5c),-30)\cdot (6,0,c^3e^{5c})\\ &=6c^2e^{5c}(1-5c)\\ c&=0 \text{ or } c=\frac15 \end{align*}

\begin{equation*} 2(x-1) +(y-1) +3(z-1)=0\qquad \text{or}\qquad 2x +y +3z= 6 \end{equation*}

\begin{align*} \vv(t)=\vr'(t)&=\big(2t,1,3t^2\big) & \diff{s}{t} &= \sqrt{1+4t^2+9t^4}\\ \va(t)=\vv'(t)&=\big(2,0,6t\big) & \vv(t)\times\va(t)&=\big(6t,-6t^2,-2\big) \end{align*}

\begin{align*} \ka(t) &= \frac{|\vv(t)\times\va(t)|}{\big(\diff{s}{t}\big)^3} =\frac{2\sqrt{1+9t^2+9t^4}}{[1+4t^2+9t^4]^{3/2}} \end{align*}

\begin{align*} \vv(t)&=\vr'(t)=-\sin t\,\hi +\cos t\,\hj + \hk\\ \va(t)&=\vr''(t)= -\cos t\,\hi -\sin t\,\hj \end{align*}

\begin{gather*} \diff{s}{t}=|\vv(t)|=\sqrt{2}\qquad \end{gather*}

\begin{gather*} \ka(t)=\Big(\diff{s}{t}(t)\Big)^{-2}|\va|=\frac{1}{2}\qquad \hat\vN(t) = \frac{\va}{|\va|}=-\cos t\,\hi-\sin t\,\hj \end{gather*}

\begin{align*} \left[\vr(t)\!+\!\frac{1}{\ka(t)}\hat\vN(t)\right]_{t=\frac{\pi}{6}} &=\left[\big(\cos t\,\hi \!+\!\sin t\,\hj \!+\! t\hk \big) +2 \big( -\cos t\,\hi\!-\!\sin t\,\hj\big)\right]_{t=\frac{\pi}{6}}\\ &=\left[-\cos t\,\hi -\sin t\,\hj +t\,\hk \right]_{t=\frac{\pi}{6}}\\ &=-\frac{\sqrt{3}}{2}\,\hi-\frac{1}{2}\,\hj+\frac{\pi}{6}\,\hk \end{align*}

\begin{align*} \vv(t)\times\va(t) &= \det\left[ \begin{matrix} \hi & \hj & \hk\\ -\sin t & \cos t & 1\\ -\cos t &-\sin t & 0\end{matrix} \right] = \sin t\,\hi -\cos t\,\hj + \hk\\ |\vv(t)\times\va(t)|^2 &= 2 \end{align*}

\begin{align*} \hat\vB(t) & = \frac{\vv(t)\times\va(t)}{|\vv(t)\times\va(t)|} =\frac{1}{\sqrt{2}}\sin t\,\hi -\frac{1}{\sqrt{2}} \cos t\,\hj +\frac{1}{\sqrt{2}}\,\hk \end{align*}

\begin{gather*} \hat\vB\big(\frac{\pi}{6}\big) =\frac{1}{2\sqrt{2}}\,\hi -\frac{\sqrt{3}}{2\sqrt{2}}\,\hj +\frac{1}{\sqrt{2}}\,\hk \end{gather*}

\begin{gather*} \vr'(t) = (-\sin(t), \cos(t), 2t) \end{gather*}

\begin{gather*} \vR(t) =\vr(\pi) +t\vT = (-1,0,\pi^2)+t(0,-1,2\pi) \end{gather*}

\begin{gather*} \diff{s}{t} = |\vr'(t)| = \sqrt{1+4t^2} \end{gather*}

\begin{gather*} a_T(t) = \difftwo{s}{t} =\diff{\ }{t}\sqrt{1+4t^2} =\frac{4t}{\sqrt{1+4t^2}} \end{gather*}

\begin{align*} \vr'(t) &= (\cos t - \cos t + t \sin t )\,\hi + (-\sin t + \sin t + t \cos t )\,\hj + 2t\,\hk\\ &= t \sin t \,\hi + t \cos t \,\hj + 2t\,\hk \end{align*}

\begin{align*} \diff{s}{t}=|\vr'(t)| & = \sqrt{t^2\sin^2t + t^2\cos^2t +4t^2} =\sqrt{5}\, t \end{align*}

\begin{equation*} \hat\vT(T) = \frac{\vr'(t)}{|\vr'(t)|} =\frac{1}{\sqrt{5}}\big( \sin t \,\hi + \cos t \,\hj + 2\,\hk\big) \end{equation*}

\begin{equation*} \va_T(t) = \difftwo{s}{t}(t)\ \hat\vT(t) = \sin t \,\hi + \cos t \,\hj + 2\,\hk \end{equation*}

\begin{equation*} \vr''(t) =\diff{\ }{t}\vr'(t) = \big(\sin t + t\cos t\big) \,\hi + \big(\cos t - t\sin t\big) \,\hj + 2\,\hk \end{equation*}

\begin{equation*} \va_N(t) = \va(t) - \va_T(t) = t\cos t \,\hi - t\sin t \,\hj \end{equation*}

\begin{gather*} \ka(t)\left(\diff{s}{t}(t)\right)^2 =\big|t\cos t \,\hi - t\sin t \,\hj \big| =t \end{gather*}

\begin{equation*} \ka(t) = \frac{t}{\left(\diff{s}{t}(t)\right)^2} = \frac{1}{5t} \end{equation*}

\begin{equation*} x^2+y^2 = 8-2x\qquad\text{or}\qquad (x+1)^2+y^2 =9 \end{equation*}

\begin{equation*} (x+1)^2+y^2 =9\qquad\text{and}\qquad z = 8-2x \end{equation*}

\begin{align*} x(\theta) &= -1 + 3\cos\theta\\ y(\theta) &= 3\sin\theta\\ z(\theta) &= 8-2x(\theta) = 10-6\cos\theta\\ \text{ or }\vr(\theta) &= [-1 + 3\cos\theta]\,\hi +3\sin\theta\,\hj +[10-6\cos\theta]\,\hk \end{align*}

\begin{align*} \vv(\theta)&=\vr'(\theta) = -3\sin\theta\,\hi +3\cos\theta\,\hj +6\sin\theta\,\hk & \vv(0)&=3\hj\\ \va(\theta)&=\vv'(\theta) = -3\cos\theta\,\hi -3\sin\theta\,\hj +6\cos\theta\,\hk & \va(0)&= -3\hi+6\hk \end{align*}

\begin{align*} \hat\vT(0) &= \frac{\vv(0)}{|\vv(0)|} =\hj \end{align*}

\begin{align*} \hat\vB(0) &= \frac{\vv(0)\times\va(0)}{|\vv(0)\times\va(0)|} =\frac{2\hi+\hk}{\sqrt{5}}\qquad \kappa(0) = \frac{|\vv(0)\times\va(0)|}{|\vv(0)|^3} =\frac{9\sqrt{5}}{3^3}=\frac{\sqrt{5}}{3} \end{align*}

\begin{gather*} \hat\vN(0) = \hat\vB(0) \times \hat\vT(0) =\frac{1}{\sqrt{5}} (2\hi+\hk)\times\hj =\frac{1}{\sqrt{5}} (2\hk-\hi) \end{gather*}

\begin{align*} \vv(t)&=\vr'(t) = t^2\,\hi +\sqrt{2}\,t\,\hj +\hk & |\vv(t)|&=\sqrt{t^4+2t^2+1}=t^2+1\\ \va(t)&=\vv'(t) = 2t\,\hi +\sqrt{2}\,\hj \end{align*}

\begin{align*} \hat\vT(t) &= \frac{\vv(t)}{|\vv(t)|} = \frac{t^2\,\hi +\sqrt{2}\,t\,\hj +\hk}{t^2+1} \end{align*}

\begin{align*} \vv(t)\times\va(t)&= \det\left[\begin{matrix} \hi & \hj & \hk\\ t^2 & \sqrt{2}\,t & 1\\ 2t & \sqrt{2} & 0 \end{matrix}\right] =-\sqrt{2}\,\hi +2t\,\hj-\sqrt{2}\,t^2\,\hk \end{align*}

\begin{gather*} \kappa(t) = \frac{|\vv(t)\times\va(t)|}{|\vv(t)|^3} =\frac{\sqrt{2+4t^2+2t^4}} {{(t^2+1)}^3} =\frac{\sqrt{2}} {{(t^2+1)}^2} \end{gather*}

\begin{equation*} \vr(\theta) = \cos\theta\,\hi +\sin\theta\,\hj + (1-\cos\theta-\sin\theta)\,\hk \qquad 0\le\theta \lt 2\pi \end{equation*}

\begin{align*} \vv(\theta)=\vr'(\theta)&=\big(-\sin\theta,\cos\theta,\sin\theta-\cos\theta\big)\\ \va(\theta)=\vv'(\theta)&=\big(-\cos\theta,-\sin\theta,\cos\theta+\sin\theta\big) \end{align*}

\begin{align*} \diff{s}{\theta} &= |\vr'(\theta)| =\sqrt{\sin^2\theta+\cos^2\theta+(\sin\theta-\cos\theta)^2}\\ &=\sqrt{2-2\sin\theta\cos\theta}\\ &=\sqrt{2-\sin(2\theta)}\\ \vv(\theta)\times\va(\theta)&=\big(1,1,1\big) \end{align*}

\begin{align*} \ka(\theta) &= \frac{|\vv(\theta)\times\va(\theta)|}{\big(\diff{s}{\theta}\big)^3} =\frac{\sqrt{3}}{[2-\sin(2\theta)]^{3/2}} \end{align*}

\begin{alignat*}{2} \text{maximum curvature }&= \frac{\sqrt{3}}{[2-1]^{3/2}}=\sqrt{3} && \text{at }\frac{\hi}{\sqrt{2}}+\frac{\hj}{\sqrt{2}}+(1-\sqrt{2})\,\hk\\ & && \text{and }-\frac{\hi}{\sqrt{2}}-\frac{\hj}{\sqrt{2}}+(1+\sqrt{2})\,\hk\\ \text{minimum curvature }&= \frac{\sqrt{3}}{[2-(-1)]^{3/2}}=\frac{1}{3}\quad && \text{at }-\frac{\hi}{\sqrt{2}}+\frac{\hj}{\sqrt{2}}+\hk\\ & && \text{and }\frac{\hi}{\sqrt{2}}-\frac{\hj}{\sqrt{2}}+\hk \end{alignat*}

\begin{align*} \vv(t)=\vr'(t)&=2t\,\hi + 2\,\hj + \frac{1}{t}\,\hk & \diff{s}{t} &= \sqrt{4t^2+4+\frac{1}{t^2}} =2t+\frac{1}{t} \end{align*}

\begin{gather*} \hat\vT(t) = \frac{\vr'(t)}{|\vr'(t)|} =\frac{2t\,\hi + 2\,\hj + \frac{1}{t}\,\hk}{2t+\frac{1}{t}} =\frac{2t^2\,\hi + 2t\,\hj + \hk}{2t^2+1} \end{gather*}

\begin{align*} \diff{s}{t}(t)\, \ka(t)\,\hat\vN(t) &= \hat\vT'(t) =\frac{4t\,\hi + 2\,\hj }{2t^2+1} -4t\frac{2t^2\,\hi + 2t\,\hj + \hk}{{(2t^2+1)}^2}\\ &=\frac{4t\,\hi + (-4t^2+2)\,\hj -4t\, \hk}{{(2t^2+1)}^2}\\ &=2\frac{2t\,\hi - (2t^2-1)\,\hj -2t\, \hk}{{(2t^2+1)}^2} \end{align*}

\begin{align*} \sqrt{4t^2 + (2t^2-1)^2+4t^2} &=\sqrt{8t^2 + 4t^4 -4 t^2 + 1} =\sqrt{ 4t^4 +4 t^2 + 1}\\ &=\sqrt{(2t^2+1)^2} =2t^2+1 \end{align*}

\begin{gather*} \hat\vN(t) =\frac{2t\,\hi - (2t^2-1)\,\hj -2t\, \hk}{2t^2+1} \end{gather*}

\begin{gather*} \ka(t) = \frac{|\hat\vT'(t)|}{\diff{s}{t}(t)} =\frac{\frac{2}{2t^2+1}}{2t+\frac{1}{t}} =\frac{2t}{{(2t^2+1)}^2} \end{gather*}

\begin{align*} \vr(t)&= \hi +\frac{t^2}{2}\,\hj + \frac{t^3}{3}\,\hk\\ \vr'(t)&= t\,\hj + t^2\,\hk\\ \diff{s}{t}(t)=|\vr'(t)|&=\sqrt{t^2+t^4}=t\sqrt{1+t^2} \end{align*}

\begin{gather*} \int_0^1 \diff{s}{t}(t)\,\dee{t} =\int_0^1 t\sqrt{1+t^2}\,\dee{t} =\frac{1}{3}{\big[1+t^2\big]}^{3/2}\Big|_0^1 =\frac{1}{3}\big[2^{3/2}-1\big] \end{gather*}

\begin{align*} \vr(t) &= \cos(t)\,\hi + \sin(t)\,\hj + t\,\hk\\ \vr'(t) &= -\sin(t)\,\hi + \cos(t)\,\hj + 1\,\hk\\ \hat\vT(t) &=\frac{-\sin(t)\,\hi + \cos(t)\,\hj + 1\,\hk}{\sqrt{2}}\\ \hat\vT'(t) &=\frac{-\cos(t)\,\hi - \sin(t)\,\hj}{\sqrt{2}}\\ \hat\vT'\big(\frac{\pi}{4}\big) &=\frac{-\hi - \hj}{2}\\ \hat\vN\big(\frac{\pi}{4}\big) &=\frac{\hat\vT'\big(\frac{\pi}{4}\big)} {|\hat\vT'\big(\frac{\pi}{4}\big)|} =\frac{-\hi - \hj}{\sqrt{2}} \end{align*}

\begin{gather*} \hat\vT'(t) =\ka(t) \,\diff{s}{t}(t)\,\hat\vN(t) \end{gather*}

\begin{align*} \ka\big(\frac{\pi}{4}\big) &=\frac{|\hat\vT'\big(\frac{\pi}{4}\big)|} {|\vr'\big(\frac{\pi}{4}\big)|} =\frac{1/\sqrt{2}}{\sqrt{2}} =\frac{1}{2} \end{align*}

\begin{align*} \vr(t) &= \big(t + 2\,,\, 1 - t\,,\, t^2 /2\big)\\ \vv(t) =\vr'(t)&= \big(1\,,\, -1\,,\, t\big)\\ \diff{s}{t}(t)=|\vv(t)|&=\sqrt{2+t^2}\\ \va(t) = \vv'(t)&=\big(0\,,\, 0\,,\, 1\big) \end{align*}

\begin{align*} \ka(t) & = \frac{|\vv(t)\times\va(t)|}{(\diff{s}{t}(t))^3} = \frac{|(-1,-1,0)|}{{[2+t^2]}^{3/2}} = \frac{\sqrt{2}}{{[2+t^2]}^{3/2}} \end{align*}

\begin{align*} \big(0\,,\, 0\,,\, 1\big) =\va(t) &= \difftwo{s}{t} \hat\vT(t) + \kappa(t)\Big(\diff{s}{t}\Big)^2\hat\vN(t)\\ &=\frac{t}{\sqrt{2+t^2}}\,\frac{(1\,,\, -1\,,\, t)}{\sqrt{2+t^2}} + \frac{\sqrt{2}}{{[2+t^2]}^{3/2}}\big(\sqrt{2+t^2}\big)^2\hat\vN(t) \end{align*}

\begin{gather*} \frac{\sqrt{2}}{\sqrt{2+t^2}}\hat\vN(t) =\big(0\,,\, 0\,,\, 1\big) - \frac{(t\,,\, -t\,,\, t^2)}{2+t^2} =\frac{(-t\,,\, t\,,\, 2)}{2+t^2} \end{gather*}

\begin{gather*} \hat\vN(t) = \frac{(-t\,,\, t\,,\, 2)}{\sqrt{2(2+t^2)}} \end{gather*}

\begin{align*} \vr(0) &= (2,1,0)\\ \hat\vT(0)&= \frac{(1,-1,0)}{\sqrt{2}}\\ \hat\vN(0)&= (0,0,1)\\ \hat\vB(0)&= \hat\vT(0)\times\hat\vN(0) = \frac{1}{\sqrt{2}}(1,-1,0)\times (0,0,1) = \frac{1}{\sqrt{2}}(-1 ,-1, 0) \end{align*}

\begin{equation*} \frac{1}{\sqrt{2}}(-1 ,-1, 0)\cdot\big\{(x,y,z) - (2,1,0)\big\}=0 \qquad\text{or}\qquad x+y = 3 \end{equation*}

\begin{equation*} \vr(0)+\frac{1}{\ka(0)}\hat\vN(0) =(2,1,0) +\frac{1}{1/2} (0,0,1) =(2,1,2) \end{equation*}

\begin{align*} \vr(t) &=\frac{t^3}{3}\,\hi + \frac{t^2}{\sqrt{2}}\,\hj + t\,\hk\\ \vr'(t) &=t^2\,\hi + \sqrt{2} t\,\hj + \hk\\ |\vr'(t)|&= \sqrt{t^4+2t^2+1} =t^2+1\\ \vr''(t) &=2t\,\hi + \sqrt{2}\,\hj\\ \vr'(t)\times\vr''(t)&= \det\left[\begin{matrix}\hi&\hj&\hk\\ t^2 & \sqrt{2} t & 1\\ 2t & \sqrt{2} & 0\end{matrix}\right] =-\sqrt{2}\,\hi +2t\,\hj -\sqrt{2} t^2\,\hk \end{align*}

\begin{gather*} \hat\vT(t) = \frac{\vr'(t)}{|\vr'(t)|} =\frac{t^2\,\hi + \sqrt{2} t\,\hj + \hk}{t^2+1} \end{gather*}

\begin{align*} \ka(t) &= \frac{|\vr'(t)\times\vr''(t)|}{|\vr'(t)|^3} =\frac{|-\sqrt{2}\,\hi +2t\,\hj -\sqrt{2} t^2\,\hk|}{{(t^2+1)}^3} =\frac{\sqrt{2+4t^2+2t^4}}{{(t^2+1)}^3}\\ &=\frac{\sqrt{2}}{{(t^2+1)}^2} \end{align*}

\begin{equation*} \ka(0) = \sqrt{2} \end{equation*}

\begin{gather*} \hat\vB(0) = \frac{\vr'(0)\times\vr''(0)}{|\vr'(0)\times\vr''(0)|} = \frac{-\sqrt{2}\,\hi}{\sqrt{2}} =-\hi \end{gather*}

\begin{gather*} \hat\vN(0) = \hat\vB(0)\times\hat\vT(0) =-\hi \times\hk =\hj \end{gather*}

\begin{align*} \vr(t) &= (t^2\,,\, t\,,\, t^3)\\ \vv(t)= \vr'(t) &= (2t\,,\, 1\,,\, 3t^2)\\ \va(t)= \vr''(t) &= (2\,,\, 0\,,\, 6t) \end{align*}

\begin{gather*} (x,y,z) - (1, -1, -1) = t (-2, 1, 3) \end{gather*}

\begin{align*} x&=1-2t\\ y&=-1+t\\ z&=-1+3t \end{align*}

\begin{align*} \vv(1)= \vr'(1) &= (2, 1, 3)\\ \va(1)= \vr''(1) &= (2, 0, 6)\\ \vv(1)\times\va(1) &= (6,-6,-2) \end{align*}

\begin{equation*} \hat\vB(1) = \frac{\vv(1)\times \va(1)}{|\vv(1)\times \va(1)|} =\frac{(3,-3,-1)}{|(3,-3,-1)|} =\frac{1}{\sqrt{19}} (3,-3,-1) \end{equation*}

\begin{equation*} (3,-3,-1)\cdot(x-1\,,\,y-1\,,\,z-1) = 0\qquad\text{or}\qquad 3x-3y-z = -1 \end{equation*}

\begin{align*} \vr'(t) &= t\sin t\,\hi + t\cos t\,\hj +2t\,\hk\\ \diff{s}{t}(t)=|\vr'(t)| &= \sqrt{5} \,t \end{align*}

\begin{gather*} \int_0^\pi \diff{s}{t}(t)\, \dee{t} =\sqrt{5} \int_0^\pi t\, \dee{t} =\frac{\sqrt{5}\,\pi^2}{2} \end{gather*}

\begin{equation*} \hat T(t) = \frac{\vr'(t)}{|\vr'(t)|} = \frac{1}{\sqrt{5}}\big(\sin t\,\hi + \cos t\,\hj +2\,\hk\big) \end{equation*}

\begin{gather*} \ka(t)\,\diff{s}{t}(t)\,\hat\vN(t) = \diff{\hat\vT}{t}(t) =\frac{1}{\sqrt{5}}\big(\cos t\,\hi - \sin t\,\hj \big) \end{gather*}

\begin{gather*} \ka(t)\overbrace{\sqrt{5}\, t}^{\diff{s}{t}(t)} =\frac{1}{\sqrt{5}}\big|\big(\cos t\,\hi - \sin t\,\hj \big)\big| =\frac{1}{\sqrt{5}} \implies \ka(t) = \frac{1}{5t} \end{gather*}

\begin{align*} \vr(t)&=\left(\frac{4\sqrt{2}}{3}t^{3/2},\frac{4\sqrt{2}}{3}t^{3/2}, t(2-t)\right)\\ \vv(t)&=\big(2\sqrt{2}t^{1/2},2\sqrt{2}t^{1/2},2-2t\big)\\ |\vv|&=\sqrt{8t+8t+4-8t+4t^2}=\sqrt{4(1+2t+t^2)}= 2(1+t) \end{align*}

\begin{gather*} \int_0^2|\vv(t)|\,dt=\int_0^2 2(1+t)\,dt = 2\left[t+\frac{t^2}{2}\right]_0^2 =8 \end{gather*}

\begin{equation*} \hat\vT(t) = \frac{\vv(t)}{|\vv(t)|} = \frac{\big(\sqrt{2}t^{1/2},\sqrt{2}t^{1/2},1-t\big)}{1+t} \end{equation*}

\begin{align*} \hat\vT'(t) &= \frac{\big(\sqrt{2}t^{-1/2}/2,\sqrt{2}t^{-1/2}/2,-1\big)}{1+t} -\frac{\big(\sqrt{2}t^{1/2},\sqrt{2}t^{1/2},1-t\big)}{(1+t)^2}\\ \hat\vT'(1) &= \frac{\big(\sqrt{2}/2,\sqrt{2}/2,-1\big)}{2} -\frac{\big(\sqrt{2},\sqrt{2},0\big)}{4}\\ &= \big(0,0, -\frac{1}{2}\big) \end{align*}

\begin{equation*} \hat\vN(1) = \frac{\hat\vT'(1)}{|\hat\vT'(1)|} = (0,0,-1) \end{equation*}

\begin{gather*} \diff{\hat\vT}{t}(1) = \big(0,0, -\frac{1}{2}\big)\qquad \diff{s}{t}(1) = |\vv(1)| = 4 \end{gather*}

\begin{align*} \ka(1) &= \frac{|\hat\vT'(1)|}{|\vv(1)|} =\frac{1}{8} \end{align*}

\begin{align*} \vr(t)&=\big(\cos^3t,\sin^3t,2\sin^2t\big)\\ \vv(t)&=\big(-3\cos^2t\sin t,3\sin^2t\cos t,4\sin t\cos t\big)\\ &=\sin t\cos t (-3\cos t,3\sin t, 4)\\ |\vv(t)|&=\sin t\cos t\sqrt{9\cos^2t+9\sin^2t+16}=5\sin t\cos t \end{align*}

\begin{equation*} \int_0^{\pi/2}|\vv(t)|\,\dee{t}=\int_0^{\pi/2} 5\sin t\cos t\,\dee{t} = \frac{5}{2}\sin^2t\Big|_0^{\pi/2} =\frac{5}{2} \end{equation*}

\begin{align*} \vv(t)&=\sin t\cos t (-3\cos t,3\sin t, 4) & |\vv(t)|&=5\sin t\cos t \end{align*}

\begin{align*} \hat\vT(t) &= \frac{\vv(t)}{|\vv(t)|} =\frac{1}{5}(-3\cos t,3\sin t, 4) & \hat\vT\Big(\frac{\pi}{6}\Big) &=\frac{1}{5}\Big(-\frac{3}{2},\frac{3\sqrt{3}}{2}, 4\Big)\\ \hat\vT'(t) &=\frac{1}{5}(3\sin t,3\cos t, 0) & \hat\vT'\Big(\frac{\pi}{6}\Big) &=\frac{1}{5}\Big(\frac{3\sqrt{3}}{2}, \frac{3}{2}, 0\Big)\\ & & &= \frac{3}{10} \Big(\sqrt{3}, 1, 0\Big)\\ \hat\vN\Big(\frac{\pi}{6}\Big) &= \frac{\hat\vT'\big(\frac{\pi}{6}\big)} {|\hat\vT'\big(\frac{\pi}{6}\big)|} =\frac{1}{2}\big(\sqrt{3},1,0\big) & \hat\vB(\frac{\pi}{6}\big) &= \hat\vT\Big(\frac{\pi}{6}\Big)\times \hat\vN\big(\frac{\pi}{6}\big)\\ & & &=\frac{1}{10}\big(-4,4\sqrt{3},-6)\\ & & &=\frac{1}{5}\big(-2,2\sqrt{3},-3) \end{align*}

\begin{align*} \vr(\theta) &= \cos\theta\,\hi +\sin\theta\,\hj + [\cos^2\theta-\sin^2\theta]\,\hk\\ &=\cos\theta\,\hi +\sin\theta\,\hj + \cos(2\theta)\,\hk \qquad 0\le\theta \lt 2\pi \end{align*}

\begin{align*} \vv(\theta)=\vr'(\theta)&=-\sin\theta\,\hi + \cos\theta\,\hj -2\sin(2\theta)\,\hk\\ \va(\theta)=\vv'(\theta)&=-\cos\theta\,\hi -\sin\theta\,\hj -4\cos(2\theta)\,\hk \end{align*}

\begin{align*} \vv(\pi/4)&=-\frac{1}{\sqrt{2}}\,\hi + \frac{1}{\sqrt{2}}\,\hj -2\,\hk\\ \va(\pi/4)&=-\frac{1}{\sqrt{2}}\,\hi -\frac{1}{\sqrt{2}}\,\hj\\ \diff{s}{\theta}(\pi/4) &= |\vv(\pi/4)|=\sqrt{5}\\ \vv(\pi/4)\times\va(\pi/4)&=-\sqrt{2}\,\hi +\sqrt{2}\,\hj + \hk\\ |\vv(\pi/4)\times\va(\pi/4)|&=\sqrt{5} \end{align*}

\begin{equation*} \ka(\pi/4) = \frac{|\vv(\pi/4)\times\va(\pi/4)|}{|\vv(\pi/4)|^3} =\frac{1}{5} \end{equation*}

\begin{gather*} \hat\vB(\pi/4) =\frac{\vv(\pi/4)\times\va(\pi/4)}{|\vv(\pi/4)\times\va(\pi/4)|} =\frac{-\sqrt{2}\,\hi +\sqrt{2}\,\hj + \hk}{\sqrt{5}} \end{gather*}

\begin{align*} &\big(-\sqrt{2}\,,\,\sqrt{2}\,,\,1\big)\cdot\big(x-1/\sqrt{2}\,,\,y-1/\sqrt{2}\,,\,z-0\big) = 0 \qquad{\rm or}\\ & z =\sqrt{2}\,x-\sqrt{2}\,y \end{align*}

\begin{align*} \hT(\pi/4)&=\frac{\vv(\pi/4)}{|\vv(\pi/4)|} =\frac{-1/\sqrt{2}\,\hi + 1/\sqrt{2}\,\hj -2\,\hk}{\sqrt{5}}\\ \hB(\pi/4)&=\frac{\vv(\pi/4)\times\va(\pi/4)}{|\vv(\pi/4)\times\va(\pi/4)|} =\frac{-\sqrt{2}\,\hi +\sqrt{2}\,\hj + \hk}{\sqrt{5}}\\ \hN(\pi/4)&=\hB(\pi/4)\times\hT(\pi/4) =\frac{-\hi-\hj}{\sqrt{2}} \end{align*}

\begin{align*} \vr_c(\pi/4)&=\vr(\pi/4)+\frac{\hN(\pi/4)}{\ka(\pi/4)} =\big(1/\sqrt{2}\,,1/\sqrt{2}\,,\,0\big) -5\big(1/\sqrt{2}\,,1/\sqrt{2}\,,\,0\big)\\ &=\big(-2\sqrt{2}\,,\,-2\sqrt{2}\,,\,0\big) \end{align*}

\begin{equation*} \va=\difftwo{s}{t}\,\hat\vT +\ka\Big(\diff{s}{t}\Big)^2\hat\vN \end{equation*}

\begin{equation*} \va=9\ka\hat\vN \end{equation*}

\begin{equation*} \vR(y)=y^2\,\hi+ y\,\hj +y^2\,\hk \end{equation*}

\begin{align*} \vR'(y)&=2y\,\hi+ \hj +2y\,\hk\\ \hat\vT(y)&=\frac{\vR'(y)}{|\vR'(y)|} =\frac{2y\,\hi+ \hj +2y\,\hk}{\sqrt{1+8y^2}}\\ \hat\vT'(y)&= \frac{2\,\hi +2\,\hk}{\sqrt{1+8y^2}} -\frac{16y}{2}\ \frac{2y\,\hi+ \hj +2y\,\hk}{[1+8y^2]^{3/2}}\\ \hat\vT'(1)&= \frac{2\,\hi +2\,\hk}{3} -8\frac{2\,\hi+ \hj +2\,\hk}{27} =\frac{2\,\hi-8\,\hj +2\,\hk}{27} \end{align*}

\begin{align*} \ka(1) &=\frac{|\hat\vT'(1)|}{|\vR'(1)|}\\ \hat\vN(1) &=\frac{\hat\vT'(1)}{|\hat\vT'(1)|}\\ \vF&=m\va = 2\times 9\ka(1)\hat\vN(1) =18\frac{\hat\vT'(1)}{|\vR'(1)|} =18\frac{2\,\hi-8\,\hj +2\,\hk}{27\sqrt{1+8\times 1^2}}\\ &=\frac{4}{9}(\hi-4\,\hj+\hk) \end{align*}

\begin{gather*} \vr(t) = 2t\hi + t^2\hj + \sqrt{3} t^2\hk\\ \vr'(t) = 2\hi + 2t\hj + 2\sqrt{3} t\hk \end{gather*}

\begin{align*} \hat\vT(t) &=\frac{\hi + t\hj + \sqrt{3} t\hk }{|\hi + t\hj + \sqrt{3} t\hk |} =\frac{\hi + t\hj + \sqrt{3} t\hk }{\sqrt{1+4t^2}} \end{align*}

\begin{align*} \diff{\hat\vT}{t}(t) &=\frac{ \hj + \sqrt{3} \hk }{\sqrt{1+4t^2}} -4t\frac{\hi + t\hj + \sqrt{3} t\hk }{{(1+4t^2)}^{3/2}} =\frac{-4t\hi + \hj + \sqrt{3} \hk }{{(1+4t^2)}^{3/2}} \end{align*}

\begin{align*} \hat\vN(t) &=\frac{-4t\hi + \hj + \sqrt{3} \hk } {|-4t\hi + \hj + \sqrt{3} \hk|} =\frac{-4t\,\hi + \hj + \sqrt{3} \hk }{2\sqrt{1+4t^2}} \end{align*}

\begin{align*} \hat\vB(t) &= \hat\vT(t)\times \hat\vN(t)\\ &=\frac{1}{2(1+4t^2)}\det\left[\begin{matrix} \hi & \hj & \hk\\ 1 & t &\sqrt{3} t\\ -4t & 1 &\sqrt{3} \end{matrix}\right]\\ &=\frac{-\sqrt{3}(1+4t^2)\hj + (1+4t^2)\hk}{2(1+4t^2)}\\ &=-\frac{\sqrt{3}}{2}\hj +\frac{1}{2}\hk \end{align*}

\begin{equation*} -\sqrt{3} y + z=0 \end{equation*}

\begin{align*} \ka(t) &= \Big|\diff{\hat\vT}{t}(t)\Big|/\Big|\diff{s}{t}\Big| =\frac{|-4t\hi + \hj + \sqrt{3} \hk| }{{(1+4t^2)}^{3/2}} \frac{1}{|2\hi + 2t\hj + 2\sqrt{3} t\hk |}\\ &=\frac{\sqrt{4+16t^2}}{{(1+4t^2)}^{3/2}} \frac{1}{2\sqrt{1+4t^2}} =\frac{1}{{(1+4t^2)}^{3/2}} \end{align*}

\begin{equation*} \hi=\frac{\vu}{2}\qquad \hj=\frac{\vv-\sqrt{3}\,\vw}{4}\qquad \hk=\frac{\sqrt{3}\,\vv+\vw}{4} \end{equation*}

\begin{align*} \vr(t) = t\,\vu + t^2 \,\vv = a(t)\vu + b(t)\vv + c(t)\vw\\ &\qquad\text{ with }a(t)=t,\ b(t)=t^2,\ c(t)=0 \end{align*}

\begin{equation*} \hat\vN = \hat\vB\times\hat\vT \qquad \hat\vN\times\hat\vT = -\hat\vB \qquad \hat\vB\times\hat\vN = -\hat\vT \end{equation*}

\begin{align*} \diff{\hat\vN}{s} & = \diff{\hat\vB}{s}\times\hat\vT +\hat\vB\times \diff{\hat\vT}{s} = -\tau\,\hat\vN\times\hat\vT +\hat\vB\times\big(\ka\hat\vN\big) = \tau\,\hat\vB - \ka\hat\vT \end{align*}

\begin{align*} \vr(\theta) &= \big(\sin(2\theta),1-\cos(2\theta),2\cos\theta\big),\\ \vv =\vr'(\theta)&= \big(2\cos(2\theta),2\sin(2\theta),-2\sin\theta\big),\\ \va = \vr''(\theta)&=\big(-4\sin(2\theta),4\cos(2\theta),-2\cos\theta\big). \end{align*}

\begin{equation*} \vr = (1,1,\sqrt2), \quad \vv = \big(0,2,-\sqrt2\big), \quad v = |\vv| = \sqrt6, \quad \va = \big(-4,0,-\sqrt2\big). \end{equation*}

\begin{equation*} \vv\times\va = \left|\begin{matrix}\hi & \hj & \hk \\ 0 & 2 & -\sqrt2 \\ -4 & 0 & -\sqrt2 \end{matrix}\right| = \big(-2\sqrt2,4\sqrt2,8\big), \qquad |\vv\times\va| = \sqrt{104} = 2\sqrt{26}. \end{equation*}

\begin{align*} \hat\vN &= \hat\vB\times\hat\vT =\frac{1}{\sqrt{78}} \left|\begin{matrix}\hi & \hj & \hk \\ -1 & 2 & 2\sqrt2 \\ 0 & 2 & -\sqrt2 \end{matrix}\right| = -\frac{1}{\sqrt{78}}\big(6\sqrt2,\sqrt2,2\big)\\ &= -\frac{1}{\sqrt{39}}\big(6,1,\sqrt2\big). \end{align*}

\begin{equation*} \kappa = \frac{|\vv\times\va|}{v^3} = \frac{2\sqrt{26}}{{(\sqrt{6})}^3} = \frac{2\sqrt2\,\sqrt{13}}{6\sqrt2\sqrt3} = \frac{\sqrt{13}}{3\sqrt3} = \frac{\sqrt{39}}{9}. \end{equation*}

\begin{align*} \va\cdot \hT &= \left(\diff{v}{t}\hat\vT + v^2\kappa\hat\vN\right)\cdot \hT\\ &=\diff{v}{t}\hT\cdot\hT+(v^2\kappa)\hN\cdot\hT\\ \end{align*}

Since \(\hT\) is a unit vector, \(\hT \cdot \hT=\|\hT\|^2=1\text{;}\) since \(\hT\) and \(\hN\) are perpendicular, \(\hT \cdot \hN =0\text{.}\)

\begin{align*} &=\diff{v}{t}\\ \end{align*}

This gives us a nice way to compute \(\diff{v}{t}\text{,}\) the rate of change of speed.

\begin{align*} \diff{v}{t} &= \va\cdot\hat\vT = (-2,3,-2\sqrt2)\cdot\frac{1}{\sqrt6}\big(0,2,-\sqrt2\big)\\ &= \frac{1}{\sqrt6}[0 + 6 + 4] = \frac{10}{\sqrt6} = \frac{5}{3}\sqrt6. \end{align*}

\begin{equation*} v^2 = \frac{1}{\kappa}\va\cdot\hat\vN = \frac{9}{\sqrt{39}}\frac{-1}{\sqrt{39}}(-2,3,-2\sqrt2) \cdot\big(6,1,\sqrt2\big) = \frac{9\times 13}{39} = 3. \end{equation*}

\begin{align*} \vr(t) &= \big(\cos t\,,\, \sin t\,,\, c \sin t\big)\\ \vv(t) =\vr'(t) &= \big(-\sin t\,,\, \cos t\,,\, c \cos t\big)\\ \va(t) =\vr''(t)&= \big(-\cos t\,,\, -\sin t\,,\, -c \sin t\big) \end{align*}

\begin{gather*} v(t) = |\vv(t)| = \sqrt{1+c^2\cos^2 t} \end{gather*}

\begin{gather*} \difftwo{s}{t} = \diff{\ }{t} \sqrt{1+c^2\cos^2 t} = \frac{-c^2\sin t\cos t}{\sqrt{1+c^2\cos^2 t}} \end{gather*}

\begin{align*} \vr(\theta) &= \Big(4 \cos\theta\,,\, 2\sin\theta\,,\, \frac{1}{4}\cos(2\theta)\Big)\\ \vv(\theta)=\vr'(\theta) &= \Big(-4 \sin\theta\,,\, 2\cos\theta\,,\, -\frac{1}{2}\sin(2\theta)\Big)\\ \va(\theta) = \vr''(\theta) &= \Big(-4 \cos\theta\,,\, -2\sin\theta\,,\, -\cos(2\theta)\Big)\\ \vv(\pi) &= \big(0\,,\, -2\,,\, 0\big)\\ \va(\pi) &= \big(4\,,\, 0\,,\, -1\big)\\ \vv(\pi) \times \va(\pi) &= \big(2\,,\, 0\,,\, 8\big) \end{align*}

\begin{align*} \ka(\pi)&=\frac{|\vv(\pi) \times \va(\pi)|}{|\vv(\pi)|^3} =\frac{|(2,0,8)|}{|(0,-2,0)|^3} =\frac{\sqrt{17}}{4} \end{align*}

\begin{equation*} \frac{1}{\ka(\pi)} = \frac{4}{\sqrt{17}} \end{equation*}

\begin{align*} \vR'(t)&=2t\,\vr'(t^2)\\ \vR''(t) &= 2\,\vr'(t^2) +4t^2\,\vr''(t^2) \end{align*}

\begin{align*} \vR(\sqrt{\pi}) &= \Big(-4\,,\, 0\,,\, \frac{1}{4}\Big)\\ \vR'(\sqrt{\pi}) &= 2\sqrt{\pi}\,\vv(\pi) =\big(0\,,\, -4\sqrt{\pi}\,,\, 0\big)\\ \text{speed} = \big|\vR'(\sqrt{\pi})\big| &=4\sqrt{\pi}\\ \text{acceleration}=\vR''(\sqrt{\pi}) &= 2\,\vv(\pi)+4\pi\,\va(\pi) = \big( 16\pi\,,\, -4\,,\, -4\pi\big) \end{align*}

\begin{gather*} \ka\Big(\diff{s}{t}\Big)^2 = \frac{\sqrt{17}}{4}\big(4\sqrt{\pi}\big)^2 =4\sqrt{17}\,\pi \end{gather*}

\begin{align*} \int_a^b |\vr'(t)|\dee{t}&=\int_a^b \sqrt{x'(t)^2+y'(t)^2+z'(t)^2}\dee{t} \end{align*}

\begin{align*} \vv(\theta) =\diff{\vr}{\theta}(\theta)&= x'(\theta)\hi+ y'(\theta)\hj\\ &=\big[r'(\theta)\cos\theta-r(\theta)\sin\theta\big]\hi+ \big[r'(\theta)\sin\theta+r(\theta)\cos\theta\big]\hj\\ \implies \diff{s}{\theta}(\theta) &=\sqrt{\big[r'(\theta)\cos\theta-r(\theta)\sin\theta\big]^2 + \big[r'(\theta)\sin\theta+r(\theta)\cos\theta\big]^2}\cr &=\sqrt{r'(\theta)^2+r(\theta)^2} \end{align*}

\begin{align*} \int_\cC f(x,y)\,ds &= \int_{\theta_1}^{\theta_2}\!\!\!f\big(x(\theta), y(\theta)\big) \diff{s}{\theta}\,\dee{\theta} \cr &=\int_{\theta_1}^{\theta_2}\!\!\!f\big(r(\theta)\cos\theta, r(\theta)\sin\theta\big) \sqrt{r(\theta)^2 +\left(\diff{r}{\theta}(\theta)\right)^2}\, \dee{\theta} \end{align*}

\begin{align*} \int_C \,ds &=\int_0^{2\pi} \sqrt{[1+\cos\theta]^2+[-\sin\theta]^2}\,\dee{\theta} =\int_0^{2\pi} \sqrt{2(1+\cos\theta)}\,\dee{\theta}\cr &=\int_0^{2\pi} \sqrt{4\cos^2\frac{\theta}{2}}\,\dee{\theta} =2\int_0^{2\pi} \left|\cos\frac{\theta}{2}\right|\,\dee{\theta} =4\int_0^{\pi} \cos\frac{\theta}{2}\ \dee{\theta}\\ &=8 \sin\frac{\theta}{2}\bigg|_0^{\pi} =8 \end{align*}

\begin{align*} \int_C \left(\frac{xy}{z}\right)\dee{s}&=\int_1^2 \left(\frac{\frac23t^3 \cdot \sqrt{3}t^2}{ 3t} \right)\sqrt{(2t^2)^2+(2\sqrt3 t)^2+(3)^2}\,\dee{t}\\ &=\int_1^2 \left(\frac{2}{ 3\sqrt 3} \,t^4\right)(2t^2+3)\,\dee{t} \\ &= \frac{4}{21\sqrt 3}(2^7-1) + \frac{2}{5\sqrt 3}(2^5-1) \end{align*}

\begin{align*} \int_C x^2\,\dee{s}&=\int_0^{2\pi} \cos^2 t \sqrt{(-\sin t)^2+(\cos t)^2}\,\dee{t}=\int_0^{2\pi} \cos^2 t \,\dee{t}\\ &=\int_0^{2\pi} \frac{1+\cos(2t)}{2}\ \dee{t} =\left[\frac{t}{2} +\frac{\sin(2t)}{4} \right]_0^{2\pi}=\pi \mathrm{kg} \end{align*}

\begin{align*} \int_C(xy+z)\dee{s}&=\int_0^1 \left((1+t)(2+2t)+(3+2t)\right)\sqrt{1^1+2^2+2^2}\dee{t}\\ &= \int_0^1 3\big(5 + 6t + 2t^2\big)\ \dee{t}=26 \end{align*}

\begin{align*} \int_\cC f(x,y,z)\,\dee{s} &=\int_0^1 x(t)\cos z(t) \diff{s}{t}\,\dee{t} =\int_0^1 t(\cos 0) \sqrt{1+4t^2}\,\dee{t}\\ &=\frac{1}{8}\frac{ {(1+4t^2)}^{3/2}}{3/2}\bigg|_0^1 =\frac{5^{3/2}-1}{12} \end{align*}

\begin{align*} \int_\cC f(x,y,z)\,\dee{s} &=\int_1^2 \frac{x(t)+y(t)}{y(t)+z(t)} \, \diff{s}{t}\,\dee{t} =\int_1^2 \frac{t+{2\over 3}t^{3/2}}{{2\over3}t^{3/2}+t} \sqrt{2+t}\,\dee{t}\\ &=\frac{(2+t)^{3/2}}{3/2}\bigg|_1^2 =\frac{8-3^{3/2}}{3/2} \end{align*}

\begin{align*} \int_C \sin x \,\dee{s}&=\int_{1}^{\sqrt 2}\sin\left(\arcsec t \right)\sqrt{\left(\frac{1}{t\sqrt{t^2-1}} \right)^2+\left(\frac1t \right)^2}\,\dee{t}\\ &=\int_1^{\sqrt{2}} \frac{\sqrt{t^2-1}}{t}\sqrt{\frac{1}{t^2(t^2-1)}+\frac{1}{t^2}} \,\dee{t}\\ &=\int_1^{\sqrt 2}\frac1t\,\dee{t}=\frac{1}{2}\ln 2 \end{align*}

\begin{align*} \int_C \vF\cdot\hn\,\dee{s} &= \int_{L_1} \vF\cdot(-\hj)\,\dee{s} \!+\!\int_{L_2} \vF\cdot \hi\,\dee{s} \!+\!\int_{L_3} \vF\cdot(\hj)\,\dee{s} \!+\!\int_{L_4} \vF\cdot(-\hi)\,\dee{s}\\ &= \int_0^3 -\overbrace{(-1)}^{y}e^x\,\dee{x} +\int_{-1}^1 \overbrace{(3)}^{x}y^2\,\dee{y} +\int^0_3 \overbrace{(1)}^{y}e^x\,\overbrace{(-\dee{x})}^{\dee{s}}\\ &\hskip1in +\int^{-1}_1 \overbrace{(0)}^{x}y^2\,\overbrace{(-\dee{y})}^{\dee{s}}\\ &=\big[e^3-1\big]+\big[1^3-(-1)^3\big] + \big[e^3-1\big]+0\\ &=2e^3 \end{align*}

\begin{equation*} \vr(t) = (3,1) + t\big\{(0,1)-(3,1)\big\} = \big(3-3t,1\big)\qquad 0\le t\le 1 \end{equation*}

\begin{equation*} \vr'(t) = (-3,0)\qquad \diff{s}{t}(t) = |\vr'(t)| = 3 \end{equation*}

\begin{align*} \int_{L_3} \vF\cdot \hj\,\dee{s} &=\int_0^1 \vF\big(\vr(t)\big)\cdot \hj\,\diff{s}{t}(t)\,\dee{t} =\int_0^1\overbrace{e^{3-3t}}^{y(t)e^{x(t)}}\, \overbrace{3}^{\diff{s}{t}(t)}\,\dee{t}\\ &=-e^{3-3t}\Big|_0^1 =e^3-1 \end{align*}

\begin{equation*} \vr(t) = (0,1) + t\big\{(3,1)-(0,1)\big\} = \big(3t,1\big)\qquad 0\le t\le 1 \end{equation*}

\begin{equation*} \vr'(t) = (3,0)\qquad \diff{s}{t}(t) = |\vr'(t)| = 3 \end{equation*}

\begin{equation*} \int_{L_3} \vF\cdot \hj\,\dee{s} =\int_0^1 \vF\big(\vr(t)\big)\cdot \hj\,\diff{s}{t}(t)\,\dee{t} =\int_0^1\overbrace{e^{3t}}^{y(t)e^{x(t)}}\, \overbrace{3}^{\diff{s}{t}(t)}\,\dee{t} =e^{3t}\Big|_0^1 =e^3-1 \end{equation*}

\begin{align*} \vr'(t)&=\big(\cos t-t\sin t\big)\hi+\big(\sin t+t\cos t\big)\,\hj+2t\,\hk\\ \diff{s}{t} &=|\vr'(t)| =\sqrt{\big(\cos t-t\sin t\big)^2+\big(\sin t+t\cos t\big)^2+(2t)^2}\\ &=\sqrt{1+5t^2}\\ \vr'(\pi)&=-\hi-\pi\,\hj+2\pi\,\hk\\ \hat\vT(\pi)&=\frac{\vr'(t)}{|\vr'(t)|} =\frac{1}{\sqrt{1+5\pi^2}}\big(-\hi-\pi\,\hj+2\pi\,\hk\big) \end{align*}

\begin{align*} \int_\cC \sqrt{x^2+y^2}\ \dee{s} &=\int_0^\pi \sqrt{x^2(t)+y^2(t)}\ \diff{s}{t}\ \dee{t} =\int_0^\pi t\ \sqrt{1+5t^2}\ \dee{t}\\ &=\left[\frac{1}{15}(1+5t^2)^{3/2}\right]_0^\pi =\frac{1}{15}\big[(1+5\pi^2)^{3/2}-1\big] \end{align*}

\begin{align*} x(t)&= t+ \frac12 t^2 & x'(t)&=1+t\\ y(t)&= t- \frac12 t^2 & y'(t)&=1-t\\ z(t)&=\frac43 t^{3/2} & z'(t)&=2\sqrt{t} \end{align*}

\begin{align*} {\color{blue}{\sqrt{x'(t)^2+y'(t)^2+z'(t)^2}}} &=\sqrt{1+2t+t^2+1-2t+t^2+4t}\\ &=\sqrt{2(t^2+2t+1)}={\color{blue}{\sqrt{2}(t+1)}}\\ {\color{red}{\rho(x(t),y(t),z(t))}}&= \frac{x(t)+y(t)}{2}={\frac{(t+t^2/2 )+( t - t^2/2)}{2}}={\color{red}{t}} \end{align*}

\begin{align*} \int_C {\color{red}{\rho}} \,\dee{s}&=\int_0^4 {\color{red}{t}}\,{\color{blue}{\sqrt{2}(t+1)}}\,\dee{t} =\frac{2^3\cdot 11\sqrt{2}}{3}\\ \int_C x {\color{red}{\rho}} \,\dee{s}&=\int_0^4 \left(t+\frac12t^2\right) {\color{red}{t}}\,{\color{blue}{\sqrt{2}(t+1)}}\,\dee{t} =\sqrt{2}\int_0^{2^2} \left(\frac{t^4}{2}+\frac{3}{2}t^3 +t^2\right)\ \dee{t}\\ &=\sqrt{2}\left(\frac{2^9}{5} +3(2^5)+\frac{2^6}{3}\right) =\frac{2^5\cdot103\sqrt 2}{15}\\ \int_C y {\color{red}{\rho}} \,\dee{s}&=\int_0^4 \left(t-\frac12t^2\right){\color{red}{t}}\, {\color{blue}{\sqrt{2}(t+1)}}\,\dee{t}=\sqrt2\int_0^{2^2}\left(-\frac{t^4}{2}+\frac{t^3}{2}+t^2 \right)\dee{t}\\ &=\sqrt{2}\left(-\frac{2^9}{5}+{2^5}+\frac{2^6}{3} \right) =-\frac{2^5\cdot 23\sqrt 2}{15}\\ \int_C z {\color{red}{\rho}}\, \dee{s}&=\int_0^4 \left(\frac{4}{3}t^{3/2}\right) {\color{red}{t}}\, {\color{blue}{\sqrt{2}(t+1)}}\,\dee{t}=\frac{4\sqrt2}{3}\int_0^{2^2} \left( t^{7/2}+t^{5/2}\right)\dee{t}\\ &=\frac{4\sqrt2}{3}\left( \frac{2^{10}}{9}+\frac{2^8}{7}\right)=\frac{2^{10}\cdot37\sqrt{2}}{7\cdot 3^3}\\ \overline{x}&=\frac{\int x \rho\,\dee{s}}{\int \rho\,\dee{s}}=\frac{\frac{2^5\cdot103\sqrt 2}{15}}{\frac{2^3\cdot 11\sqrt{2}}{3}}=\frac{412}{55}\approx 7.5\\ \overline{y}&=\frac{\int y \rho\,\dee{s}}{\int \rho\,\dee{s}}=\frac{-\frac{2^5\cdot 23\sqrt 2}{15}}{\frac{2^3\cdot 11\sqrt{2}}{3}}=-\frac{92}{55}\approx-1.7\\ \overline{z}&=\frac{\int z \rho\,\dee{s}}{\int \rho\,\dee{s}}=\frac{\frac{2^{10}\cdot37\sqrt{2}}{7\cdot 3^3}}{\frac{2^3\cdot 11\sqrt{2}}{3}}=\frac{4736}{693}\approx 6.8 \end{align*}

\begin{align*} F=m\va&= W\hN -mg\hj\\ \end{align*}

We dot both sides with \(\hN\text{.}\)

\begin{align*} W\hN \cdot \hN - mg\hj\cdot\hN&=m\va\cdot\hN\\ \end{align*}

Using the equation \(\va(t) = \ddiff{2}{s}{t}\hT + \ka\left(\diff{s}{t}\right)^2\hN\text{,}\)

\begin{align*} W-mg\hj\cdot\hN&=m\ka\left(\diff{s}{t}\right)^2\\ W&=mg\hn\cdot\hN+m\ka\left(\diff{s}{t}\right)^2\\ &=mg\cos\theta+m\ka\left(\diff{s}{t}\right)^2 \end{align*}

\begin{align*} \frac{25}{2}&\le\frac{E}{mg}=\frac{\frac12m|\vv|^2}{m\cdot9.8}\\ |\vv|&\ge5\sqrt{9.8} \end{align*}

\begin{align*} \vr(\theta)&=(3 \cos \theta, 5\sin\theta, 4+4\cos\theta)\\ \vv(\theta)&=(-3\sin\theta,5\cos\theta,-4\sin\theta)\\ |\vv(\theta)|=\diff{s}{\theta}&=\sqrt{9\sin^2\theta+25\cos^2\theta+16\sin^2\theta}=5\\ \va(\theta)&=(-3\cos\theta,-5\sin\theta,-4\cos\theta)\\ \vv \times \va &=5(-4,0,3)\\ \ka(\theta)&=\frac{|\vv \times \va|}{\left(\diff{s}{\theta}\right)^3}=\frac{25}{5^3}=\frac15\\ \end{align*}

Since \(\ddiff{2}{s}{\theta}=0\text{,}\) we use the following theorem to find \(\hN\text{:}\)

\begin{align*} \va(\theta)&=\ddiff{2}{s}{\theta}\hT+\kappa\left(\diff{s}{\theta}\right)^2\hN\\ (-3\cos\theta,-5\sin\theta,-4\cos\theta)&=0+\frac{25}{5}\hN\\ \hN(\theta)&=\left(-\frac35\cos\theta,-\sin\theta,-\frac45\cos\theta\right)\\ \end{align*}

Using the given quantity \(|\vv(t)|=5\) at the specified point,

\begin{align*} W\Big|_{\theta=\pi/4}&=m\ka|\vv|^2+mg\hk\cdot \hat\vN\\ &=(1)\frac15 5^2+1(9.8)\left(- \frac45 \cos (\pi/4)\right)=5-\frac{39.2}{5\sqrt 2}\\ W\hN\Big|_{\theta=\pi/4}&=\left(5-\frac{39.2}{5\sqrt 2}\right)\left(-\frac35\cos(\pi/4),-\sin(\pi/4),-\frac45\cos(\pi/4)\right)\\ &=\left(5-\frac{39.2}{5\sqrt 2}\right)\left(-\frac3{5\sqrt 2},-\frac{1}{\sqrt 2},-\frac{4}{5\sqrt2}\right)\\ &=\left( -\frac{3}{\sqrt2}+2.352 , -\frac{5}{\sqrt 2}+{3.92} , -{2\sqrt 2}+3.136\right) \end{align*}

\begin{align*} \vr(\theta)&=(\sin \theta , \sin \theta - \theta)\\ \vv(\theta)&=(\cos \theta, \cos \theta -1)\\ |\vv(\theta)|=\diff{s}{\theta}&=\sqrt{2\cos^2\theta-2\cos \theta+1}\\ \va(\theta)&=(-\sin \theta, -\sin \theta)\\ |\vv \times \va|&=|\sin \theta|\\ \ka(\theta)&=\frac{|\vv\times\va|}{\left(\diff{s}{\theta}\right)^3}=\frac{|\sin \theta|}{(2\cos^2\theta-2\cos \theta+1)^{3/2}}\\ \ddiff{2}{s}{\theta}&=\frac{\sin \theta (1-2\cos \theta)}{\sqrt{2\cos^2\theta-2\cos \theta+1}} \end{align*}

\begin{align*} \va(13\pi/3)&=\left(-\sqrt{3}/2,-\sqrt{3}/2\right)\\ \ddiff{2}{s}{\theta}(13\pi/3)&=0\\ \ka(13\pi/3)&=\sqrt{6}\\ \diff{s}{\theta}(13\pi/3)&=1/\sqrt{2}\\ \va(\theta)\cdot\hj&=\left(\ddiff{2}{s}{\theta}\hT+\kappa\left(\diff{s}{\theta}\right)^2\hN\right)\cdot\hj\\ -\frac{\sqrt 3}{2}&=0+\sqrt{6}(1/2)\hN\cdot\hj\\ \hN\cdot\hj&=-\frac{1}{\sqrt2} \end{align*}

\begin{align*} W&=m\ka|\vv|^2+mg\hj\cdot\hN\\ \pm 100 &=\left(\frac{1}{9.8} \right)\sqrt{6}|\vv|^2+\frac{9.8}{9.8}\left(-\frac{1}{\sqrt 2} \right)\\ |\vv|&=\sqrt{\frac{9.8}{\sqrt6}\left(100+\frac{1}{\sqrt 2} \right)} \approx20 \mathrm{m}/\mathrm{s} \approx 72 \mathrm{kph} \end{align*}

\begin{equation*} |\vv| \gt \sqrt{\frac{g}{\kappa}|\hj\cdot\hN|} \end{equation*}

\begin{align*} \vr(t)&=(\ln t,1-t)\\ \vr'(t)&=\vv(t)=(t^{-1}, -1)\qquad\qquad \diff{s}{t}=|\vv(t)|=\sqrt{1+t^{-2}}\\ \vr''(t)&=\va(t)=(-t^{-2},0)\\ \ka(t)&=\frac{|\vv \times \va|}{\left(\diff{s}{t}\right)^{3}}=\frac{t^{-2}}{\sqrt{1+t^{-2}}^3}=\frac{|t|}{(1+t^{2})^{3/2}}=\frac{t}{(1+t^{2})^{3/2}}\\ \end{align*}

Note \(t\) is positive in the interval in question.

\begin{align*} \hT(t)&=\frac{\vv(t)}{|\vv(t)|}=\frac{1}{\sqrt{1+t^{-2}}}(t^{-1},-1)=\left( \frac{1}{\sqrt{1+t^{2}}},\frac{-t}{\sqrt{1+t^{2}}}\right)\\ \hT'(t)&= \left( \frac{-t}{({1+t^2})^{3/2}},\frac{-1}{({1+t^2})^{3/2}}\right)\qquad\qquad |\hT'(t)|=\frac{1}{t^2+1}\\ \hN(t)&=\frac{\hT'(t)}{|\hT'(t)|}=\left( \frac{-t}{\sqrt{1+t^2}},\frac{-1}{\sqrt{1+t^2}}\right)\\ |\hN \cdot \hj|&=\frac{1}{\sqrt{1+t^2}} \end{align*}

\begin{align*} |\vv|&=\sqrt{\frac{g}{\kappa}|\hN \cdot \hj|}=\sqrt{\frac{g\cdot (1+t^{2})^{3/2}}{t(1+t^2)^{1/2}}}=\sqrt{\frac{g(1+t^2)}{t}} \end{align*}

\begin{align*} \sqrt{\frac{g(1+t^2)}{|t|}}\Bigg|_{t=1}&=\sqrt{2g}=\sqrt{2^6\cdot3^4\cdot7^2}=2^3\cdot 3^2\cdot 7 = 504\text{ kph} \end{align*}

\begin{align*} F=m\va&=U\hT + W\hN - mg\hj\\ \end{align*}

To focus on the force in the direction of \(\hT\text{,}\) we dot both sides of the equation with \(\hT(s)=\left(\diff{x}{s},\diff{y}{s} \right)\text{.}\) (Recall \(\vr(s)\) was parametrized with respect to arclength, so \(\hT(s)=\diff{\vr}{s}\) everywhere.) Since the speed of the bead is constant, the tangential component of its acceleration, \(\va \cdot \hT\text{,}\) is 0 (see Theorem 1.3.3.c).

\begin{align*} 0&=(U\hT + W\hN - mg\hj)\cdot \hT\\ &=(U\hT\cdot \hT) +( W\hN\cdot \hT)-mg\hj\cdot\hT\\ &=U+0-mg\diff{y}{s}\\ U&=mg\diff{y}{s} \end{align*}

\begin{align*} m\va&=W\hN+M\hT-mg\hj\\ \end{align*}

To isolate \(M\text{,}\) we dot both sides of the equation with \(\hT\text{.}\) Remember \(\hT\) is a unit vector, and it is perpendicular to \(\hN\text{.}\)

\begin{align*} m\va \cdot \hT&=W\hN\cdot\hT+M\hT\cdot\hT-mg\hj\cdot\hT\\ &=0+M-mg\hj\cdot\hT\\ \\ \end{align*}

Since the speed of the snowmachine is constant, the tangential component of its acceleration, \(\va \cdot \hT\text{,}\) is 0 (see Theorem 1.3.3.c).

\begin{align*} 0&=M-mg\hj\cdot\hT\\ M&=mg\hj\cdot\hT \end{align*}

\begin{align*} \vr(x)&=(x,1+\cos x) & \vr'(x)&=(1,-\sin x)\\ |\vr'(x)|&=\sqrt{1+\sin^2 x} &\hT(x)&=\frac{1}{\sqrt{1+\sin^2 x}}(1,-\sin x)\\ \hT(3\pi/4)&=\left( \sqrt{\frac{2}{3}},-\frac{1}{\sqrt{3}}\right) \end{align*}

\begin{equation*} M=(200 \mathrm{ kg})( 9.8 \mathrm{m}/\mathrm{s}^2) \left(-\frac1{\sqrt3} \mathrm{m}\right)=-\frac{1960}{\sqrt 3} \mathrm{N} \approx - 1131.6 \mathrm{N} \end{equation*}

\begin{align*} \vr(\theta)&=(4\cos\theta,3(1+\sin\theta))\\ \vv(\theta)=\vr'(\theta)&=(-4\sin\theta,3\cos\theta) \\ |\vv(\theta)|=\diff{s}{\theta}&=\sqrt{16\sin^2\theta+9\cos^2\theta}=\sqrt{9+7\sin^2\theta}\\ \va(\theta)&=(-4\cos\theta,-3\sin\theta) \\ |\vv(\theta) \times \va(\theta)|&=12\\ \ka(\theta)&=\frac{|\vv(\theta)\times\va(\theta)|}{\left(\diff{s}{\theta} \right)^3}=\frac{12}{(9+7\sin^2\theta)^{3/2}}\\ \hT(\theta)&=\frac{(-4\sin\theta,3\cos\theta)}{\sqrt{9+7\sin^2\theta}}\\ \hT'(\theta)&=\frac{(36\cos\theta,48\sin\theta)}{-(9\cos^2\theta+16\sin^2\theta)^{3/2}}\\ |\hT'(\theta)|&=\frac{12}{9\cos^2\theta+16\sin^2\theta} \\ \hN(\theta)&= \frac{(3\cos\theta,4\sin\theta)}{-\sqrt{9\cos^2\theta+16\sin^2\theta}} \end{align*}

\begin{align*} \text{Equation } \knowl{./knowl/xref/eqn_consEnergy.html}{\text{1.7.1}}:\quad E&=\frac12m|\vv|^2+mgy\\ \text{If } |\vv|=0: \qquad E&=mgy_S \implies y_S=\frac{E}{mg}\\ \end{align*}

This answers part a.

\begin{align*} \text{Equation }\knowl{./knowl/xref/eqn_normalForce.html}{\text{1.7.2}}:\quad W &=2\ka(E-mgy)+mg\hj\cdot \hat\vN\\ \mbox{If } W=0 :\qquad 0&=2\ka(E-mgy_A)+mg\hj\cdot \hat\vN\\ &=\frac{24\left(E-mgy_A\right)}{(9+7\sin^2\theta)^{3/2}} -mg\left(4\frac{\sin\theta}{\sqrt{9+7\sin^2\theta}} \right)\\ \end{align*}

Using \(y=3+3\sin\theta\text{:}\)

\begin{align*} &=\frac{24\left(E-mgy_A\right)}{\left(9+7\left( \frac{y_A-3}{3}\right)^2 \right)^{3/2}} -4mg\left(\frac{ \frac{y_A-3}{3}}{\sqrt{9+7\left( \frac{y_A-3}{3}\right)^2}} \right) \end{align*}

\begin{equation*} \frac{24\left(E-mgy_A\right)}{\left(9+7\left( \frac{y_A-3}{3}\right)^2 \right)^{3/2}} =4mg\left(\frac{ \frac{y_A-3}{3}}{\sqrt{9+7\left( \frac{y_A-3}{3}\right)^2}} \right) \end{equation*}

\begin{align*} \frac{24\left(E-mgy_A\right)}{\left(9+7\left( \frac{y_A-3}{3}\right)^2 \right)^{3/2}} &=4mg\left(\frac{ \frac{y_A-3}{3}}{\sqrt{9+7\left( \frac{y_A-3}{3}\right)^2}} \right)\\ \Leftrightarrow \qquad \frac{6\left(\frac{E}{mg}-y_A\right)}{\left(9+7\left( \frac{y_A-3}{3}\right)^2 \right)^{3/2}} &=\frac{ \frac{y_A-3}{3}}{\sqrt{9+7\left( \frac{y_A-3}{3}\right)^2}}\\ \Leftrightarrow \qquad \frac{6\left(\frac{11^2}{2\cdot9.8}-y_A\right)}{\left(9+7\left( \frac{y_A-3}{3}\right)^2 \right)^{3/2}} &=\frac{ \frac{y_A-3}{3}}{\sqrt{9+7\left( \frac{y_A-3}{3}\right)^2}}\\ \end{align*}

To simplify to a more standard form, we multiply both sides by \(\left(9+7\left(\frac{y_A-3}{3}\right)^2\right)^{3/2}\text{:}\)

\begin{align*} 6\left(\frac{11^2}{2\cdot9.8}-y_A\right) &=\left( \frac{y_A-3}{3}\right)\left(9+7\left( \frac{y_A-3}{3}\right)^2 \right)\\ \end{align*}

Now, we simplify to

\begin{align*} 0&=\frac{7}{9}y_A^3 -7 y_A^2 +48 y_A- \frac{7797}{49} \end{align*}

\begin{align*} E'(t)&=\diff{}{t}\left[\frac{1}{2} m |\vv(t)|^2 +mg \vr(t)\cdot\hk\right] =m\vv(t)\cdot\vv'(t)+mg\vv(t)\cdot\hk\\ &=\vv(t)\cdot\big[\vN\big(\vr(t)\big) - mg\hk\big]+mg\vv(t)\cdot\hk\\ &=0 \end{align*}

\begin{align*} E(t)=E(0) &\implies \half m |\vv(t)|^2 +mg b\theta(t) = mg(2\pi b)\\ &\implies|\vv(t)|^2=2gb\big(2\pi-\theta(t)\big) \end{align*}

\begin{align*} &\vv=\diff{\vr}{t}=\diff{\vr}{\theta}\diff{\theta}{t} =\big(-a\sin\theta,a\cos\theta,b\big)\diff{\theta}{t}\\ &\implies |\vv|^2=\big[a^2+b^2]\left(\diff{\theta}{t}\right)^2\\ &\implies \diff{\theta}{t}=-\left[\frac{|\vv|^2}{a^2+b^2}\right]^{1/2} =-\left[\frac{2gb(2\pi-\theta)}{a^2+b^2}\right]^{1/2} \end{align*}

\begin{align*} \int dt&=\int_{2\pi}^0 \frac{dt}{d\theta}d\theta =-\left[\frac{a^2+b^2}{2gb}\right]^{1/2}\int_{2\pi}^0 \frac{1}{(2\pi-\theta)^{1/2}}d\theta\\ &=\left[\frac{a^2+b^2}{2gb}\right]^{1/2}\int^{2\pi}_0 \frac{1}{(2\pi-\theta)^{1/2}}d\theta\\ &=\left[\frac{a^2+b^2}{2gb}\right]^{1/2} \left[-2(2\pi-\theta)^{1/2}\right]_0^{2\pi} =2\left[\frac{a^2+b^2}{gb}\pi\right]^{1/2} \end{align*}

\begin{alignat*}{2} (x_1,y_1) &= (3,0) &&\implies \quad r_1=3,\ \tan\theta_1=0\\ &&&\implies \quad \theta_1=0 \text{ as $(x_1,y_1)$ is on the positive $x$-axis}\\ (x_2,y_2) &= (1,1) &&\implies \quad r_2=\sqrt{2},\ \tan\theta_2=1\\ &&&\implies \quad \theta_2=\frac{\pi}{4} \text{ as $(x_2,y_2)$ is in the first octant}\\ (x_3,y_3) &= (0,1) &&\implies \quad r_3=1,\ \cos\theta_3=0\\ &&&\implies \quad \theta_3=\frac{\pi}{2} \text{ as $(x_3,y_3)$ is on the positive $y$-axis}\\ (x_4,y_4) &= (-1,1) &&\implies \quad r_4=\sqrt{2},\ \tan\theta_4=-1\\ &&&\implies \quad \theta_4=\frac{3\pi}{4} \text{ as $(x_4,y_4)$ is in the third octant}\\ (x_5,y_5) &= (-2,0) &&\implies \quad r_5=2,\ \tan\theta_5=0\\ &&&\implies \quad \theta_5=\pi \text{ as $(x_5,y_5)$ is on the negative $x$-axis} \end{alignat*}

\begin{equation*} \sqrt{r^2\cos^2\theta + r^2\sin^2\theta} =\sqrt{r^2} =|r| \end{equation*}

\begin{align*} |\he_r(\theta)| &= \sqrt{\cos^2\theta+\sin^2\theta} = 1\\ |\he_\theta(\theta)| &= \sqrt{(-\sin\theta)^2+\cos^2\theta} = 1 \end{align*}

\begin{equation*} \he_r(\theta) \cdot \he_\theta(\theta) = (\cos\theta)(-\sin\theta) +(\sin\theta)(\cos\theta)=0 \end{equation*}

\begin{align*} \he_r(\theta) \times \he_\theta(\theta) &=\det\left[\begin{matrix} \hi & \hj & \hk \\ \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \end{matrix}\right] =\hk \end{align*}

\begin{align*} r=2\cos(\theta) &\iff r^2 =2r\cos(\theta) \iff x^2+y^2=2x\\ &\iff (x-1)^2+y^2=1 \end{align*}

\begin{equation*} \diff{}{\theta}\big(e^{\theta/10}+e^{-\theta/10}\big) =\frac{1}{10}\big(e^{\theta/10}-e^{-\theta/10}\big) \gt 0\qquad\text{for all }\theta \gt 0 \end{equation*}

\begin{equation*} \diff{}{\theta} \theta =1 \qquad\text{for all }\theta \end{equation*}

\begin{align*} x(\theta)&= f(\theta)\cos\theta\\ y(\theta)&= f(\theta)\sin\theta\\ \vr(\theta)&= f(\theta)\big[\cos\theta\ \hi + \sin\theta\ \hj\big]\\ \vv(\theta)=\vr'(\theta)&= f'(\theta)\big[\cos\theta\ \hi + \sin\theta\ \hj\big] +f(\theta)\big[-\sin\theta\ \hi + \cos\theta\ \hj\big]\\ \va(\theta)=\vr''(\theta)&= \big\{f''(\theta)\!-\!f(\theta)\big\}\big[\cos\theta\,\hi \!+\! \sin\theta\,\hj\big] \!+\!2f'(\theta)\big[-\sin\theta\ \hi\!+\! \cos\theta\,\hj\big] \end{align*}

\begin{align*} \vv(\theta)&= f'(\theta)\,\he_r(\theta) +f(\theta)\,\he_\theta(\theta)\\ \va(\theta)&= \big\{f''(\theta)-f(\theta)\big\}\,\he_r(\theta) +2f'(\theta)\,\he_\theta(\theta) \end{align*}

\begin{align*} |\vv(\theta)|^2&=\vv(\theta)\!\cdot\!\vv(\theta) \!=\!\big[f'(\theta)\,\he_r(\theta)\!+\!f(\theta)\,\he_\theta(\theta)\big]\!\cdot\! \big[f'(\theta)\,\he_r(\theta) \!+\!f(\theta)\,\he_\theta(\theta)\big]\\ &=\!f'(\theta)^2\,\he_r(\theta)\cdot\he_r(\theta) \!+\!f(\theta)^2\,\he_\theta(\theta)\cdot\he_\theta(\theta) \!+\!2f'(\theta)\,f(\theta)\,\he_r(\theta)\cdot\he_\theta(\theta)\\ &=f'(\theta)^2+f(\theta)^2\\ |\vv(\theta)|& = \sqrt{f'(\theta)^2+f(\theta)^2}\\ \vv(\theta)\!\times\!\va(\theta) &=\big[f'(\theta)\,\he_r(\theta)\!+\!f(\theta)\,\he_\theta(\theta)\big]\!\times\! \big[\big\{f''(\theta)\!-\!f(\theta)\big\}\,\he_r(\theta) \!+\!2f'(\theta)\,\he_\theta(\theta)\big]\\ &= 2f'(\theta)^2\,\he_r(\theta)\times\he_\theta(\theta) +f(\theta)[f''(\theta)-f(\theta)]\,\he_\theta(\theta)\times\he_r(\theta)\\ &= \big\{2f'(\theta)^2-f(\theta)[f''(\theta)-f(\theta)]\big\} \,\hk \end{align*}

\begin{align*} \ka(\theta)&=\frac{|\vv(\theta)\times\va(\theta)|}{|\vv(\theta)|^3} =\frac{\big|f(\theta)^2+2f'(\theta)^2-f(\theta)f''(\theta)\big|} {{[f(\theta)^2+f'(\theta)^2]}^{3/2}} \end{align*}

\begin{gather*} f(\theta) = a(1-\cos\theta)\qquad f'(\theta) = a\sin\theta\qquad f''(\theta) =a\cos\theta \end{gather*}

\begin{align*} \ka(\theta) &=\frac{\big|f(\theta)^2+2f'(\theta)^2-f(\theta)f''(\theta)\big|} {[f(\theta)^2+f'(\theta)^2]^{3/2}}\\ &=\frac{\big|a^2-2a^2\cos\theta+a^2\cos^2\theta+2a^2\sin^2\theta- a^2\cos\theta+a^2\cos^2\theta\big|} {[a^2-2a^2\cos\theta+a^2\cos^2\theta+a^2\sin^2\theta]^{3/2}}\\ &=\frac{3a^2-3a^2\cos\theta}{[2a^2-2a^2\cos\theta]^{3/2}} =\frac{3}{2^{3/2}a\sqrt{1-\cos\theta}} =\frac{3}{2\sqrt{2ar(\theta)}} \end{align*}

\begin{equation*} \vOm(t) = \vr(t)\times\vv(t) \end{equation*}

\begin{align*} \diff{\vOm}{t}(t) &=\diff{\ }{t}\big(\vr(t)\times\vv(t)\big) =\vv(t)\times\vv(t) + \vr(t)\times\va(t)\\ &=m\vr(t)\times f\big(r(t))\vr(t)\big) =\vZero \end{align*}

\begin{equation*} \vr(t)\cdot\vOm =0 \end{equation*}

\begin{align*} &\text{constant} = \big|\vr\times\vv\big| = \Big|r\hat\vr\times\Big(\diff{r}{t}\ \hat\vr + r\ \diff{\theta}{t}\ \hat\vth\Big)\Big| =r^2\diff{\theta}{t}\\ &\text{since}\quad |\hat\vr\times\hat\vr|=0,\ |\hat\vr\times\hat\vth|=1 \end{align*}

\begin{align*} \vr''(t) &= f\big(r(t)\big)\vr(t) \end{align*}

\begin{align*} g''(t)\hat\vT & = f\big(r(t)\big)\,\big[\vr_0+ g(t)\hat\vT\big] = f\big(r(t)\big)\,\vr_0 + f\big(r(t)\big) g(t)\hat\vT \end{align*}

\begin{align*} g''(t)\hj & = f\big(r(t)\big)\,\hi + f\big(r(t)\big) g(t)\hj \end{align*}

\begin{equation*} \diff{\ }{t}\big(\vr(t)\times\vv(t)\big) =\vv(t)\times\vv(t) + \vr(t)\times\va(t) =\vZero +\vZero =\vZero \end{equation*}

\begin{equation*} \vr(t)\times\vv(t) =r(t)^2\dot\theta(t)\,\hat\vr\big(\theta(t)\big)\times \hat\vth\big(\theta(t)\big) \end{equation*}

\begin{equation*} |\vr(t)\times\vv(t)| = r(t)^2|\dot\theta(t)| \end{equation*}

\begin{align*} \va(t) &= \Big(\difftwo{r}{t}(t)-r(t)\Big(\diff{\theta}{t}(t)\Big)^2\Big) \hat\vr\big(\theta(t)\big)\\ &\hskip1in +\Big(r(t)\ \difftwo{\theta}{t}(t) + 2 \diff{r}{t}(t)\diff{\theta}{t}(t)\Big) \hat\vth\big(\theta(t)\big) \end{align*}

\begin{equation*} \va(t) = \Big(\difftwo{r}{t}(t)-r(t)\Big(\diff{\theta}{t}(t)\Big)^2\Big) \hat\vr\big(\theta(t)\big) \end{equation*}

\begin{equation*} |\va(t)| = \Big|\difftwo{r}{t}(t)-r(t)\Big(\diff{\theta}{t}(t)\Big)^2\Big| \end{equation*}

\begin{gather*} \dot r(t) = - \frac{1}{[\theta(t)+\al]^2}\dot\theta(t) =- r(t)^2 \dot\theta(t) =-h \end{gather*}

\begin{gather*} |\va(t)| = r(t)\dot\theta(t)^2 =\frac{r(t)^4\dot\theta(t)^2}{r(t)^3} =\frac{h^2}{r(t)^3} \end{gather*}

\begin{equation*} \vv(x,y)\cdot \hi \begin{cases} \gt 0 & \text{ when } \fbox{$x \gt 0$}\\ =0 &\text{ when } \fbox{$x=0$}\\ \lt 0&\text{ when } \fbox{$x \lt 0$} \end{cases} \end{equation*}

\begin{equation*} \vv(x,y)\cdot \hj \begin{cases} \gt 0 & \text{ when } \fbox{$-2 \lt x \lt 2$}\\ =0 &\text{ when } \fbox{$x \in \{-2,2\}$}\\ \lt 0&\text{ when } \fbox{$x \lt -2$ or $x \gt 2$} \end{cases} \end{equation*}

\begin{equation*} \vv(x,y)\cdot \hi \begin{cases} \gt 0 & \mbox{ when } \fbox{$y \gt -x$}\\ =0 &\mbox{ when } \fbox{$y=-x$}\\ \lt 0&\mbox{ when } \fbox{$y \lt -x$} \end{cases} \quad \text{and} \quad \vv(x,y)\cdot \hj \begin{cases} \gt 0 & \mbox{ when } \fbox{$y \lt x$}\\ =0 &\mbox{ when } \fbox{$y=x$}\\ \lt 0&\mbox{ when } \fbox{$y \gt x$} \end{cases} \end{equation*}

\begin{equation*} (1,1) + 0.01(1,1) = (1.01\,,\,1.01) \end{equation*}

\begin{equation*} (0,0) + 0.01(0,0) = (0\,,\,0) \end{equation*}

\begin{equation*} (0,0)+10(0,-1) = (0\,,\,-10) \end{equation*}

\begin{equation*} \vv(x,y,z)=-\frac{\alpha}{{x^2+y^2+z^2}}(x,y,z) \end{equation*}

\begin{equation*} \vv(x,y)\cdot\hi=x^2 \begin{cases} \gt 0 & x \neq 0\\ =0 & x=0 \end{cases} \end{equation*}

\begin{equation*} \vv(x,y)\cdot\hj=y \begin{cases} \gt 0 & y \gt 0\\ =0 & y=0\\ \lt 0 & y \lt 0 \end{cases}. \end{equation*}

\begin{align*} \vf(x,y)&=\vf_1(x,y)+\vf_2(x,y)+\vf_3(x,y)\\ &=\frac{-5G(x,y)}{(x^2+y^2)^{3/2}}+\frac{3G(2-x,3-y)}{((x-2)^2+(y-3)^2)^{3/2}}+\frac{7G(4-x,-y)}{((x-4)^2+y^2)^{3/2}} \end{align*}

\begin{alignat*}{2} x'(t) &= -e^{-t}\cos t - e^{-t}\sin t &&= -x(t)-y(t)\\ y'(t) &= -e^{-t}\sin t + e^{-t}\cos t &&= -y(t)+x(t) \end{alignat*}

\begin{align*} \diff{x}{t} &= v_1(x,y) = -x-y\\ \diff{y}{t} &= v_2(x,y) = \phantom{-}x-y \end{align*}

\begin{gather*} \frac{\dee{x}}{y} = \frac{\dee{y}}{x} \iff x\,\dee{x} = y\,\dee{y} \iff \frac{x^2}{2} =\frac{y^2}{2} +C \end{gather*}

\begin{align*} \hi\cdot\vF(x,y) = y \left.\begin{cases} \gt 0 &\text{if } y \gt 0 \\ =0 &\text{if } y=0 \\ \lt 0 &\text{if } y \lt 0 \end{cases}\right\}\\ \hj\cdot\vF(x,y) = x \left.\begin{cases} \gt 0 &\text{if } x \gt 0 \\ =0 &\text{if } x=0 \\ \lt 0 &\text{if } x \lt 0 \end{cases}\right\} \end{align*}

\begin{equation*} \frac{\dee{x}}{2y}=\frac{\dee{y}}{x/y^2}=\frac{\dee{z}}{e^y} \qquad\text{ if } x,y\ne 0 \end{equation*}

\begin{gather*} \frac{\dee{x}}{2y}=\frac{y^2\,\dee{y}}{x} \implies x\,\dee{x}=2y^3\,\dee{y} \implies \frac{1}{2}x^2=\frac{1}{2}y^4+C \end{gather*}

\begin{equation*} \frac{\dee{y}}{x/y^2}=\frac{\dee{z}}{e^y} \implies e^ydy=dz \implies z=e^y+D \end{equation*}

\begin{equation*} x=y^2 \qquad z=e^y \end{equation*}

\begin{align*} &\frac{\dee{x}}{x}=\frac{\dee{y}}{3y}\qquad\text{ if } x,y\ne 0\\ &\implies 3\ln |x|=\ln |y|+C\\ &\implies |x|^3=e^C|y|\\ &\implies y=\pm e^{-C}x^3\\ &\implies y=C'x^3 \end{align*}

\begin{align*} \vF&=x\hi + z\hj + y\hk\\ \vnabla \times \vF&=\Big(\pdiff{F_3}{y} -\pdiff{F_2}{z} \Big)\hi +\Big(\pdiff{F_1}{z} -\pdiff{F_3}{x} \Big)\hj +\Big(\pdiff{F_2}{x} -\pdiff{F_1}{y} \Big)\hk\\ &=(1-1)\hi+(0-0)\hj+(0-0)\hk = \mathbf0 \end{align*}

\begin{align*} \vF&=y^2z\hi + x^2z\hj + x^2y\hk\\ \vnabla \times \vF&=\Big(\pdiff{F_3}{y} -\pdiff{F_2}{z} \Big)\hi +\Big(\pdiff{F_1}{z} -\pdiff{F_3}{x} \Big)\hj +\Big(\pdiff{F_2}{x} -\pdiff{F_1}{y} \Big)\hk\\ &=(x^2-x^2)\hi+(y^2-2xy)\hj+(2xz-2yz)\hk \neq \mathbf0 \end{align*}

\begin{align*} \vF&=(ye^{xy}+1)\hi + (xe^{xy}+z)\hj + \left( \frac1z+y\right)\hk\\ \vnabla \times \vF&=\Big(\pdiff{F_3}{y} -\pdiff{F_2}{z} \Big)\hi +\Big(\pdiff{F_1}{z} -\pdiff{F_3}{x} \Big)\hj +\Big(\pdiff{F_2}{x} -\pdiff{F_1}{y} \Big)\hk\\ &=(1-1)\hi+(0-0)\hj+(e^{xy}(xy+1)-e^{xy}(xy+1))\hk = \mathbf0 \end{align*}

\begin{align*} \vF&=y\cos(xy)\hi + x\sin(xy)\hj\\ \pdiff{F_2}{x}&=xy\cos(xy)+\sin(xy)\\ \pdiff{F_1}{y}&=-xy\sin(xy)+\cos(xy)\\ \pdiff{F_2}{x}&\neq \pdiff{F_1}{y} \end{align*}

\begin{alignat*}{2} \pdiff{\varphi}{x}(x,y) &= \frac{-\frac{y}{x^2}}{1+\big(\frac{y}{x}\big)^2} &&= -\frac{y}{x^2+y^2}\\ \pdiff{\varphi}{y}(x,y) &= \frac{\frac{1}{x}}{1+\big(\frac{y}{x}\big)^2} &&= \phantom{-} \frac{x}{x^2+y^2} \end{alignat*}

\begin{align*} \pdiff{\varphi}{x}(x,y,z) &= x\\ \pdiff{\varphi}{y}(x,y,z) &= -2y\\ \pdiff{\varphi}{z}(x,y,z) &= 3z \end{align*}

\begin{equation*} \varphi(x,y,z) = \frac{x^2}{2} + f(y,z) \end{equation*}

\begin{equation*} \pdiff{f}{y}(y,z) = -2y \end{equation*}

\begin{equation*} f(y,z) = -y^2+ g(z) \end{equation*}

\begin{equation*} g'(z) = 3z \end{equation*}

\begin{equation*} g(z) = \frac{3}{2} z^2 +C \end{equation*}

\begin{equation*} \pdiff{F_1}{y}=\pdiff{F_2}{x} \end{equation*}

\begin{equation*} \pdiff{F_1}{y} =\frac{\partial }{\partial y}\Big(\frac{x}{x^2+y^2}\Big) =-\frac{2xy}{{(x^2+y^2)}^2} \end{equation*}

\begin{equation*} \pdiff{F_2}{x} =\frac{\partial }{\partial x}\Big(\frac{-y}{x^2+y^2}\Big) =\frac{2xy}{{(x^2+y^2)}^2} \end{equation*}

\begin{gather*} \pdiff{F_1}{y}=\pdiff{F_2}{x}\qquad \pdiff{F_1}{z}=\pdiff{F_3}{x}\qquad \pdiff{F_2}{z}=\pdiff{F_3}{y} \end{gather*}

\begin{align*} \pdiff{}{y}\big(e^{(z^2)}\big) &=\pdiff{}{x}\big(2Byz^3\big) & &\iff & 0 & =0\\ \pdiff{}{z}\big(e^{(z^2)}\big) &=\pdiff{}{x}\big(Axze^{(z^2)}+3By^2z^2\big) & &\iff & 2ze^{(z^2)} & =Aze^{(z^2)}\\ \pdiff{}{z}\big(2Byz^3\big) &=\pdiff{}{y}\big(Axze^{(z^2)}+3By^2z^2\big) & &\iff & 6Bye^{(z^2)}& =6Bye^{(z^2)} \end{align*}

\begin{equation*} \vF=e^{(z^2)}\,\hi+2Byz^3\,\hj +\big(2xze^{(z^2)}+3By^2z^2\big)\,\hk \end{equation*}

\begin{align*} \pdiff{\varphi}{x}(x,y,z) &= e^{(z^2)}\\ \pdiff{\varphi}{y}(x,y,z) &= 2Byz^3\\ \pdiff{\varphi}{z}(x,y,z) &= 2xze^{(z^2)}+3By^2z^2 \end{align*}

\begin{equation*} \varphi(x,y,z) = xe^{(z^2)} + f(y,z) \end{equation*}

\begin{equation*} \pdiff{f}{y}(y,z) = 2Byz^3 \end{equation*}

\begin{equation*} f(y,z) = By^2 z^3 + g(z) \end{equation*}

\begin{equation*} 2xze^{(z^2)} + 3By^2z^2 + g'(z) = 2xze^{(z^2)}+3By^2z^2\qquad\text{or}\quad g'(z) = 0 \end{equation*}

\begin{equation*} g(z) = C \end{equation*}

\begin{gather*} \vv=m\frac{x\hi+y\hj}{x^2+y^2} \end{gather*}

\begin{align*} \pdiff{\varphi}{x}(x,y) &= m\frac{x}{x^2+y^2}\\ \pdiff{\varphi}{y}(x,y) &= m\frac{y}{x^2+y^2} \end{align*}

\begin{equation*} \varphi(x,y) = \half m\ln(x^2+y^2) + f(y) \end{equation*}

\begin{equation*} m\frac{y}{x^2+y^2} + f'(y) = m\frac{y}{x^2+y^2}\qquad\text{or}\quad f'(y) = 0 \end{equation*}

\begin{equation*} f(y) = C \end{equation*}

\begin{equation*} \varphi=\half m\ln(x^2+y^2) \end{equation*}

\begin{equation*} \Set{(x,y,z)}{\varphi(x,y,z)\ge -20} \end{equation*}

\begin{equation*} \Set{(x,y,z)}{x^2+y^2+z^2 \le 20} \end{equation*}

\begin{align*} E&=\frac{1}{2}\cdot\frac{1}{2}|1|^2-\varphi\big(0,0,0\big)=\frac{1}{4}\\ \end{align*}

When the particle is at \((1,1,1)\text{:}\)

\begin{align*} \frac{1}{4}&=\frac{1}{2}\cdot\frac{1}{2}|\vv|^2-\varphi(1,1,1)=\frac{|\vv|^2}{4}-\left(1+\frac{9}{4}\right)\\ |\vv|&=\sqrt{14} \end{align*}

\begin{align*} \mathbf\vnabla \times \vF &=\Big(\pdiff{F_3}{y} -\pdiff{F_2}{z} \Big)\hi +\Big(\pdiff{F_1}{z} -\pdiff{F_3}{x} \Big)\hj +\Big(\pdiff{F_2}{x} -\pdiff{F_1}{y} \Big)\hk\\ &=\Big(g'(y)h'(z)-g'(y)h'(z) \Big)\hi +\Big(0 -0 \Big)\hj +\Big(0 -0 \Big)\hk=\mathbf{0} \end{align*}

\begin{align*} \vnabla\times\vF &=\Big(\pdiff{F_3}{y} -\pdiff{F_2}{z} \Big)\hi +\Big(\pdiff{F_1}{z} -\pdiff{F_3}{x} \Big)\hj +\Big(\pdiff{F_2}{x} -\pdiff{F_1}{y} \Big)\hk\\ \end{align*}

When \(\vF=\left \lt xy, xz,y^2+z \right \gt \text{,}\)

\begin{align*} \vnabla\times\vF&=(2y-x)\hi+(0-0)\hj+(z-x)\hk \end{align*}

\begin{align*} \int_\cC x^2y^2\,\dee{x}+x^3y\,\dee{y} &=\int_0^1 x^2\times 0^2\,\dee{x} +\int_0^1 1^3\times y\,\dee{y} +\int_1^0 x^2\times 1^2\,\dee{x}\\ &\hskip2in+\int_1^0 0^3\times y\,\dee{y}\\ &=\frac{1}{2}-\frac{1}{3} =\frac{1}{6} \end{align*}

\begin{align*} \vF&=x\hi + z\hj + y\hk\\ \vnabla \times \vF&=\Big(\frac{\partial F_3}{\partial y} -\frac{\partial F_2}{\partial z} \Big)\hi +\Big(\frac{\partial F_1}{\partial z} -\frac{\partial F_3}{\partial x} \Big)\hj +\Big(\frac{\partial F_2}{\partial x} -\frac{\partial F_1}{\partial y} \Big)\hk\\ &=(1-1)\hi+(0-0)\hj+(0-0)\hk = \mathbf0 \end{align*}

\begin{align*} \vF&=y^2z\hi + x^2z\hj + x^2y\hk\\ \vnabla \times \vF&=\Big(\frac{\partial F_3}{\partial y} -\frac{\partial F_2}{\partial z} \Big)\hi +\Big(\frac{\partial F_1}{\partial z} -\frac{\partial F_3}{\partial x} \Big)\hj +\Big(\frac{\partial F_2}{\partial x} -\frac{\partial F_1}{\partial y} \Big)\hk\\ &=(x^2-x^2)\hi+(y^2-2xy)\hj+(2xz-2yz)\hk \neq \mathbf0 \end{align*}

\begin{align*} \vF&=(ye^{xy}+1)\hi + (xe^{xy}+z)\hj + \left( z+y\right)\hk\\ \vnabla \times \vF&=\Big(\frac{\partial F_3}{\partial y} -\frac{\partial F_2}{\partial z} \Big)\hi +\Big(\frac{\partial F_1}{\partial z} -\frac{\partial F_3}{\partial x} \Big)\hj +\Big(\frac{\partial F_2}{\partial x} -\frac{\partial F_1}{\partial y} \Big)\hk\\ &=(1-1)\hi+(0-0)\hj+\{e^{xy}(xy+1)-e^{xy}(xy+1)\}\hk = \mathbf0 \end{align*}

\begin{align*} \vF&=y\cos(xy)\hi + x\sin(xy)\hj\\ \frac{\partial F_2}{\partial x}&=xy\cos(xy)+\sin(xy)\\ \frac{\partial F_1}{\partial y}&=-xy\sin(xy)+\cos(xy)\\ \frac{\partial F_2}{\partial x}&\neq \frac{\partial F_1}{\partial y} \end{align*}

\begin{align*} \vZero=\vnabla\times\vF &=\det\left[\begin{matrix}\hi&\hj&\hk \\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z} \\ e^x\sin y & ae^x\cos y+bz & cx\end{matrix}\right]\\ &=(0-b)\hi-(c-0)\hj+(ae^x\cos y-e^x\cos y)\hk \end{align*}

\begin{align*} \oint_{\cC}\vF\cdot \dee{\vr} &= \int_{L_1}\vF\cdot \dee{\vr} +\int_{L_2}\vF\cdot \dee{\vr} +\int_{L_3}\vF\cdot \dee{\vr} +\int_{L_4}\vF\cdot \dee{\vr}\\ & =0+0+0+\int_{L_4}\vF\cdot \dee{\vr} \gt 0 \end{align*}

\begin{align*} \oint_{\cC}\vF\cdot \dee{\vr} &= \int_{L_1}\vF\cdot \dee{\vr} +\int_{L_2}\vF\cdot \dee{\vr} +\int_{L_3}\vF\cdot \dee{\vr} +\int_{L_4}\vF\cdot \dee{\vr}\\ & =0+\int_{L_2}\vF\cdot \dee{\vr}+0 +\int_{L_4}\vF\cdot \dee{\vr} \gt 0 \end{align*}

\begin{equation*} \vF(x,y) = a(y)\,\hj \end{equation*}

\begin{align*} \frac{\partial }{\partial y}\left(\frac{x-2y}{x^2+y^2}\right) &=\frac{-2}{x^2+y^2}-\frac{2y(x-2y)}{{(x^2+y^2)}^2} =\frac{-2x^2+2y^2-2xy}{{(x^2+y^2)}^2}\\ \frac{\partial }{\partial x}\left(\frac{2x+y}{x^2+y^2}\right) &=\frac{2}{x^2+y^2}-\frac{2x(2x+y)}{{(x^2+y^2)}^2} =\frac{-2x^2+2y^2-2xy}{{(x^2+y^2)}^2} \end{align*}

\begin{align*} \vnabla\times\vF &=\det\left[\begin{matrix} \hi &\hj &\hk\\ \pdiff{}{x} & \pdiff{}{y} & \pdiff{}{z}\\ \frac{x-2y}{x^2+y^2} & \frac{2x + y}{x^2+y^2} & z \end{matrix} \right]\\ &=\left(\frac{-2x^2+2y^2-2xy}{{(x^2+y^2)}^2} -\frac{-2x^2+2y^2-2xy}{{(x^2+y^2)}^2}\right)\hk\\ &=\vZero \end{align*}

\begin{equation*} \vr(t) = 2\cos t\,\hi +2\sin t\,\hj +3\,\hk\qquad \vr'(t) = -2\sin t\,\hi +2\cos t\,\hj \end{equation*}

\begin{align*} \int_C \vF \cdot \dee{\vr} &=\int_0^{2\pi}\!\!\bigg\{ \overbrace{\frac{2\cos t-4\sin t}{4}}^{\frac{x-2y}{x^2+y^2}}\,\hi +\overbrace{\frac{4\cos t + 2\sin t}{4}}^{\frac{2x+y}{x^2+y^2}}\,\hj + \overbrace{3}^{z}\,\hk\bigg\} \cdot\\ &\hskip3in \Big\{\overbrace{-2\sin t\,\hi +2\cos t\,\hj}^{\vr'(t)}\Big\}\dee{t}\\ &=\int_0^{2\pi} \frac{-4\sin t\cos t+8\sin^2 t+8\cos^2t+4\sin t\cos t}{4}\,\dee{t}\\ &= 2\int_0^{2\pi}\dee{t} =4\pi \end{align*}

\begin{equation*} \phi=\frac{x^2}{2}+yx-yz+\frac{z^2}{2} \end{equation*}

\begin{align*} \int_\cC\vF\cdot \dee{\vr} =\phi\big(\vr(t_1)\big)-\phi\big(\vr(t_0)\big) &=\phi(0, -2, 3)-\phi(1,0,-1)\\ &=\left[0+0+6+\frac{9}{2}\right]-\left[\frac{1}{2}+0-0+\frac{1}{2}\right]\\ &=9\frac{1}{2} \end{align*}

\begin{align*} \int_C \vV\cdot \dee{\vr} &=\int_0^1 \vV(x,0)\cdot\hi\ \dee{x}+\int_0^\pi \vV(1,y)\cdot\hj\ \dee{y} +\int_1^0 \vV(x,\pi)\cdot\hi\ \dee{x}\\ &=\int_0^1 (e^x+x^2)\ \dee{x} +\int_0^\pi (y+3)\ \dee{y}+\int_1^0 (-e^x+x^2)\ \dee{x}\cr &=2\int_0^1 e^x\ \dee{x}+\int_0^\pi (y+3)\ \dee{y}\\ &=2(e-1)+\frac{\pi^2}{2}+3\pi \end{align*}

\begin{align*} \int_\cC\vF\cdot \dee{\vr} &=\int_0^1 \vF\big(x(t),y(t)\big)\cdot \diff{\vr}{t}(t)\ \dee{t} =\int_0^1\big[t^3\,\hi-t^2\,\hj\big] \cdot\big[\hi+2t\,\hj\big]\,\dee{t}\\ &=\int_0^1\big[-t^3\big]\,\dee{t}\\ &=-\frac{1}{4} \end{align*}

\begin{equation*} \int_\cC\vF\cdot \dee{\vr}=\int_0^1 x\,\dee{x} +\int_0^1 y\,\dee{y}+\int_0^1(-2)\,\dee{z} =\frac{1}{2}+\frac{1}{2}-2 =-1 \end{equation*}

\begin{align*} \vr(t)&=2t\,\hi+t\,\hj+\frac{4}{t^2}\,\hk,\qquad 1\le t\le 2\\ \vr'(t)&=2\,\hi+\hj-\frac{8}{t^3}\,\hk\\ \vF(\vr(t))&= 4t^2\,\hi+4t^3\,\hk\\ \vF(\vr(t))\cdot \vr'(t)&=8t^2-32 \end{align*}

\begin{align*} \int_\cC \vF\cdot \dee{\vr} &=\int_1^2 \vF(\vr(t))\cdot \vr'(t)\ \dee{t} =\int_1^2 \big(8t^2-32\big)\ \dee{t} = \left[\frac{8}{3}t^3-32t\right]_1^2\\ &=-\frac{40}{3} \end{align*}

\begin{align*} &\vnabla\times\vF = \det\left[\begin{matrix} \hi & \hj & \hk\\ \pdiff{}{x} & \pdiff{}{y} & \pdiff{}{z}\\ 6x^2yz^2 & 2x^3z^2 + 2y - xz & 4x^3yz \end{matrix}\right]\\ &\hskip0.25in=(4x^3z-4x^3z+x)\,\hi-(12x^2yz-12x^2yz)\,\hj+(6x^2z^2-z-6x^2z^2)\,\hk\\ &\hskip0.25in=x\,\hi-z\,\hk\\ &\vnabla\times\vG =\det\left[\begin{matrix} \hi & \hj & \hk\\ \pdiff{}{x} & \pdiff{}{y} & \pdiff{}{z}\\ yz & 0 & xy \end{matrix}\right]\\ &\hskip0.25in=x\hi-z\,\hk \end{align*}

\begin{equation*} \vnabla\times\big(\vF+\la\vG\big)=(x+\la x)\,\hi-(z+\la z)\,\hk \end{equation*}

\begin{align*} \vF-\vG &=(6x^2yz^2-yz)\,\hi+(2x^3z^2 + 2y - xz)\,\hj+(4x^3yz-xy)\,\hk\\ &=\vnabla\big(2x^3yz^2-xyz+y^2\big) \end{align*}

\begin{align*} \pdiff{\phi}{x}(x,y,z) &= 6x^2yz^2-yz\\ \pdiff{\phi}{y}(x,y,z) &= 2x^3z^2 + 2y - xz\\ \pdiff{\phi}{z}(x,y,z) &= 4x^3yz-xy \end{align*}

\begin{equation*} \phi(x,y,z) = 2x^3yz^2-xyz + f(y,z) \end{equation*}

\begin{equation*} 2x^3z^2-xz + \pdiff{f}{y}(y,z) = 2x^3z^2 + 2y - xz \qquad\hbox{or}\qquad \pdiff{f}{y}(y,z) = 2y \end{equation*}

\begin{equation*} f(y,z) = y^2+ g(z) \end{equation*}

\begin{equation*} 4x^3yz-xy + g'(z) = 4x^3yz-xy\qquad\hbox{or}\qquad g'(z) = 0 \end{equation*}

\begin{equation*} g(z) = K \end{equation*}

\begin{align*} &\int_C\vF\cdot \dee{\vr} =\int_C(\vF-\vG)\cdot \dee{\vr}+\int_C\vG\cdot \dee{\vr}\\ &\hskip0.25in=\Big[2x^3yz^2-xyz+y^2\Big]_{(0,1,0)}^{(1,e,1)} +\int_0^1\overbrace{[xe^{x^2}\hi + xe^{x^2}\hk]}^{\vG(\vr(x))}\cdot \overbrace{[\hi+2xe^{x^2}\,\hj+\hk]}^{\diff{\vr}{x}}\ \dee{x}\\ &\hskip0.25in=e+e^2-1+\int_0^1 2xe^{x^2}\ \dee{x} =e+e^2-1+\Big[e^{x^2}\Big]_0^1 =e^2+2e-2 \end{align*}

\begin{equation*} \vr(t) = (0,0,1) + t\big\{(2,1,0)-(0,0,1)\big\} = \big(2t,t,1-t\big)\qquad 0\le t\le 1 \end{equation*}

\begin{equation*} \vr'(t) = (2,1,-1)\qquad \end{equation*}

\begin{align*} \int\vF\cdot\dee{\vr} &=\int_0^1 \vF\big(\vr(t)\big)\cdot \vr'(t)\,\dee{t}\\ &=\int_0^1\big(2t-t^2\,,\, t-(1-t)^2 \,,\,(1-t)-4t^2\big)\cdot(2,1,-1)\,\dee{t}\\ &=\int_0^1\big(4t-2t^2 \ +\ t-1+2t-t^2 \ -\ 1+t+4t^2\big)\,\dee{t}\\ &=\int_0^1\big(t^2+8t-2\big)\,\dee{t}\\ &=\frac{1}{3}+4-2 =\frac{7}{3} \end{align*}

\begin{equation*} \vr(\theta)=\cos\theta\,\hi+\sin\theta\,\hj-\ln(\cos\theta)\,\hk\qquad 0\le\theta\le\frac{\pi}{4} \end{equation*}

\begin{align*} \vr\,'(\theta)&=-\sin\theta\,\hi+\cos\theta\,\hj+\tan\theta\,\hk\\ \vF\big(\vr(\theta)\big)&=\cos\theta\,\hi+\sin\theta\,\hj+\cos^3\theta\,\hk\\ \vF\big(\vr(\theta)\big)\cdot\vr\,'(\theta)&=\sin\theta\cos^2\theta \end{align*}

\begin{align*} \text{Work} &=\int_P\vF\cdot \dee{\vr} =\int_0^{\pi/4}\vF\big(\vr(\theta)\big)\cdot\vr\,'(\theta)\ \dee{\theta} =\int_0^{\pi/4}\sin\theta\cos^2\theta\ \dee{\theta}\\ &=-\frac{1}{3}\cos^3\theta\Big|_0^{\pi/4}\\ &=\frac{1}{3}\big[1-\frac{1}{2^{3/2}}\big]\approx 0.2155 \end{align*}

\begin{align*} \frac{\partial \varphi}{\partial x}(x,y,z) &= yz\cos x\\ \frac{\partial \varphi}{\partial y}(x,y,z) &= z\sin x+2yz\\ \frac{\partial \varphi}{\partial z}(x,y,z) &= y\sin x+y^2-\sin z \end{align*}

\begin{equation*} \varphi(x,y,z) = yz\sin x + f(y,z) \end{equation*}

\begin{equation*} z\sin x + \pdiff{f}{y}(y,z) = z\sin x+2yz\qquad\text{or}\quad \pdiff{f}{y}(y,z) = 2yz \end{equation*}

\begin{equation*} f(y,z) = y^2z + g(z) \end{equation*}

\begin{equation*} y\sin x+y^2 + g'(z) = y\sin x+y^2-\sin z\qquad\text{or}\quad g'(z) = -\sin z \end{equation*}

\begin{equation*} g(z) = \cos z + C \end{equation*}

\begin{gather*} \int_C\vF\cdot\dee{\vr} = \varphi(\vr)\Big|_{\vr(0)}^{\vr(\pi/2)} = \varphi(\pi/2\,,\,\pi/2\,,\,\pi/2) -\varphi (0,0,0) = \frac{\pi^3}{8}+\frac{\pi^2}{4}-1 \end{gather*}

\begin{equation*} \vr(\theta) = \big(x(\theta)\,,\,y(\theta)\big) =(1-\cos\theta)\,\hi +\sin\theta\,\hj\qquad 0\le\theta\le\pi \end{equation*}

\begin{gather*} \int_C xy\,\dee{y} =\int_0^\pi x(\theta)\,y(\theta)\, y'(\theta)\ \dee{\theta} =\int_0^\pi (1-\cos\theta)\,\sin\theta\,\cos\theta \ \dee{\theta} \end{gather*}

\begin{align*} \int_C xy\,\dee{y} &=\int_1^{-1} (1-u)\, u\, (-\dee{u}) =\int_{-1}^1 (u-u^2)\,\dee{u}\\ &=-2 \int_0^1 u^2\,\dee{u} =-2 \frac{1^3}{3}=-\frac{2}{3} \end{align*}

\begin{align*} \int_C xy\dee{y} &=\int_{C_1} xy\dee{y}+\int_{c_2}xy\dee{y}\\ &=\int_0^1\left(1-\sqrt{1-y^2}\right)y\,\dee{y}+\int_1^0\left(1+\sqrt{1-y^2}\right)y\,\dee{y}\\ &=\int_0^1 y\,\dee{y}-\int_0^1y\sqrt{1-y^2}\,\dee{y}+\int_1^0y\,\dee{y}+\int_1^0 y\sqrt{1-y^2}\,\dee{y}\\ &=-2\int_0^1 y\sqrt{1-y^2}\,\dee{y}\\ \end{align*}

Using the substitution \(u=1-y^2\text{,}\) \(\dee{u} =-2y\,\dee{y}\text{:}\)

\begin{align*} &=\int_1^0u^{1/2}\dee{u}=-\frac23 \end{align*}

\begin{align*} \frac{\partial \varphi}{\partial x}(x,y) &= ye^x + \sin y\\ \frac{\partial \varphi}{\partial y}(x,y) &= e^x + \sin y + x \cos y \end{align*}

\begin{equation*} \varphi(x,y) = ye^x + x\sin y + f(y) \end{equation*}

\begin{equation*} e^x + x\cos y + f'(y) = e^x + \sin y + x \cos y\qquad\text{or}\quad f'(y) = \sin y \end{equation*}

\begin{equation*} f(y) = -\cos y + C \end{equation*}

\begin{align*} &\int_C (ye^x + \sin y)\,\dee{x} + (e^x + \sin y + x \cos y)\,\dee{y} =\int_C \vF\cdot\dee{\vr} =\varphi(x,y)\Big|^{(0,\pi/2)}_{(1,0)}\\ &\hskip1in=\Big[ye^x + x\sin y -\cos y\Big]^{(0,\pi/2)}_{(1,0)}\\ &\hskip1in=1+\frac{\pi}{2} \end{align*}

\begin{align*} \int_{C} xy \,\dee{x} + yz \,\dee{y} + zx \,\dee{z} &=\sum_{\ell=1}^3\int_{C_\ell} xy \,\dee{x} + yz \,\dee{y} + zx \,\dee{z} =3\times\frac{1}{6}=\frac{1}{2} \end{align*}

\begin{align*} \frac{\partial\varphi}{\partial x} &= y + ze^x\\ \frac{\partial\varphi}{\partial y} &= x + e^y \sin z\\ \frac{\partial\varphi}{\partial z} &= z + e^x + e^y \cos z \end{align*}

\begin{equation*} \varphi(x,y,z) = xy +ze^x +e^y\sin z +\frac{z^2}{2} \end{equation*}

\begin{equation*} \int_C \vF\cdot \dee{\vr} =\varphi\big(\vr(\pi)\big) -\varphi\big(\vr(0)\big) =\varphi\big(\pi,e^\pi,0\big) -\varphi\big(0,1,0\big) =\pi e^\pi \end{equation*}

\begin{align*} 0&=\vnabla\times\vF =\vnabla\times\big(\alpha e^y\,\hi+(xe^y+\beta\cos z)\,\hj -\gamma y\sin z\,\hk\big)\\ &=(-\gamma\sin z+\beta\sin z)\hi+(e^y-\alpha e^y)\hk \end{align*}

\begin{equation*} \int_C \vF \cdot \dee{\vr} = \varphi(P_1)-\varphi(P_0) \end{equation*}

\begin{align*} \frac{\partial \varphi}{\partial x}(x,y,z) &= e^y & \implies \varphi(x,y,z) &=xe^y+\psi_1(y,z)\\ \frac{\partial \varphi}{\partial y}(x,y,z) &= xe^y+\beta\cos z & \implies \varphi(x,y,z) &=xe^y+\beta y \cos z + \psi_2(x,z)\\ \frac{\partial \varphi}{\partial z}(x,y,z) &= -\beta y\sin z & \implies \varphi(x,y,z) &=\beta y \cos z + \psi_3(x,y) \end{align*}

\begin{equation*} \varphi(x,y,z)\stackrel{?}{=} xe^y+\beta y \cos z \end{equation*}

\begin{equation*} \int_C \vF \cdot \dee{\vr} = \varphi(1,e,\pi)-\varphi(0,1,0) = \big(e^e-\beta e\big)-\beta =e^e-\beta(e+1) \end{equation*}

\begin{align*} \vnabla\times\vF &=\det\left[\begin{matrix} \hi &\hj &\hk\\ \pdiff{}{x} & \pdiff{}{y} & \pdiff{}{z}\\ \cos x & 2+\sin y & e^z \end{matrix} \right] =\vZero \end{align*}

\begin{equation*} \vF=\vnabla\big(\sin x +2y-\cos y +e^z\big) \end{equation*}

\begin{align*} \pdiff{f}{x}(x,y,z) &= \cos x &\implies f(x,y,z)&=\sin x + \psi_1(y,z)\\ \pdiff{f}{y}(x,y,z) &= 2 + \sin y &\implies f(x,y,z)&=2y-\cos y + \psi_2(x,z)\\ \pdiff{f}{z}(x,y,z) &= e^z&\implies f(x,y,z)&=e^z+\psi_3(x,y) \end{align*}

\begin{equation*} f(x,y,z)\stackrel{?}{=} \sin x +2y-\cos y +e^z \end{equation*}

\begin{align*} \int_C\vF\cdot\dee{\vr} &= f\big(\vr(3\pi)\big) -f\big(\vr(0)\big)\\ &= f(3\pi,-1,0) - f(0,1,0)\\ &=\big(0-2-\cos (-1)+1\big)-\big(0+2-\cos 1+1\big)\\ \end{align*}

Since cosine is an even function, \(\cos(-1)=\cos1\text{.}\)

\begin{align*} &=-4 \end{align*}

\begin{align*} \vnabla\times\vF &=\det\left[\begin{matrix} \hi &\hj &\hk\\ \pdiff{}{x} & \pdiff{}{y} & \pdiff{}{z}\\ z+e^y & xe^y-e^z\sin y & 1+x+e^z\cos y \end{matrix} \right] =\vZero \end{align*}

\begin{align*} \frac{\partial \varphi}{\partial x}(x,y,z) &= z + e^y\\ &\hskip1in \implies \varphi(x,y,z)=zx+xe^y+\psi_1(y,z)\\ \frac{\partial \varphi}{\partial y}(x,y,z) &= xe^y - e^z \sin y\\ &\hskip1in \implies \varphi(x,y,z)=xe^y+e^z\cos y+\psi_2(x,z)\\ \frac{\partial \varphi}{\partial z}(x,y,z) &= 1 + x + e^z \cos y\\ &\hskip1in \implies \varphi(x,y,z)=z+zx+e^z\cos y+\psi_3(x,y) \end{align*}

\begin{equation*} \varphi(x,y,z)\stackrel{?}{=} zx+xe^y+e^z\cos y+z \end{equation*}

\begin{align*} \int_C \vF \cdot \dee{\vr} &=\varphi\big(\vr(\pi)\big)-\varphi\big(\vr(0)\big)=\Big[xz+xe^y+e^z\cos y +z\Big]_{\vr(0)}^{\vr(\pi)}\\ &=\Big[xz+xe^y+e^z\cos y +z\Big]_{(0,0,1)}^{(\pi^2,0,1)}\\ &=\left(\pi^2+\pi^2+e+1 \right)-\left(0+0+e+1 \right) =2\pi^2 \end{align*}

\begin{align*} \vZero = \vnabla\times\vF &=\det\left[\begin{matrix} \hi &\hj &\hk\\ \pdiff{}{x} & \pdiff{}{y} & \pdiff{}{z}\\ (x - a)ye^x & xe^x + z^3 & byz^2 \end{matrix} \right]\\ &= \big(bz^2-3z^2)\,\hi -\big(0-0\big)\,\hj +\big(e^x+xe^x-(x-a)e^x\big)\,\hk \end{align*}

\begin{align*} \pdiff{f}{x}(x,y,z) &= (x + 1)ye^x\\ \pdiff{f}{y}(x,y,z) &= xe^x + z^3\\ \pdiff{f}{z}(x,y,z) &= 3yz^2 \end{align*}

\begin{equation*} f(x,y,z) = xye^x + yz^3 +g(x,z) \end{equation*}

\begin{equation*} 3yz^2+\frac{\partial g}{\partial z}(x,z) = 3yz^2\qquad\text{or}\qquad \frac{\partial g}{\partial z}(x,z)=0 \end{equation*}

\begin{equation*} g(x,z) = h(x) \end{equation*}

\begin{equation*} xye^x + ye^x + h'(x) =(x + 1)ye^x \quad\text{or}\quad h'(x) = 0 \end{equation*}

\begin{align*} \int_C\vF\cdot\dee{\vr} &=\int_C\vnabla f\cdot\dee{\vr} =f\big(\vr(\pi)) - f(\big(\vr(0)\big) =f(\pi,1,-1)-f(0,1,1)\\ &=\big[\pi e^\pi -1 \big]-\big[ 1\big] =\pi e^\pi-2 \end{align*}

\begin{equation*} \int_C\vF\cdot\dee{\vr} =\int_C (x + 1)ye^x \,\dee{x} + (xe^x + z^3 ) \,\dee{y} + 3yz^2 \,\dee{z} \end{equation*}

\begin{align*} I &= \int_C\vF\cdot\dee{\vr} + \int_C yz^2\dee{z}\\ &= \pi e^\pi-2 +\int_0^\pi \overbrace{(\cos 2t)}^{y} \overbrace{\cos^2 t}^{z^2} \overbrace{(-\sin t)\,\dee{t}}^{\dee{z}}\\ &= \pi e^\pi-2 +\int_0^\pi (2\cos^2t-1) \cos^2 t (-\sin t)\,\dee{t}\\ &= \pi e^\pi-2 +\int_1^{-1} (2u^2-1)u^2\,\dee{u} \qquad\text{with } u=\cos t,\ \dee{u}=-\sin t\,\dee{t}\\ &= \pi e^\pi-2 +\Big[\frac{2u^5}{5} -\frac{u^3}{3}\Big]_{1}^{-1}\\ &= \pi e^\pi-2 +\Big[-\frac{4}{5} +\frac{2}{3}\Big]\\ &= \pi e^\pi -\frac{32}{15} \end{align*}

\begin{align*} \vZero &= \vnabla\times\vF=\det\left|\begin{matrix}\hi&\hj&\hk\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z} \\ y^2 e^{3z} +Axy^3 & 2xye^{3z}+3x^2y^2 & Bxy^2e^{3z}\end{matrix}\right|\\ &=\big(2Bxy e^{3z}-6xy e^{3z}\big)\,\hi -\big(By^2 e^{3z}-3y^2e^{3z}\big)\,\hj\\ &\hskip1in +\big(2ye^{3z}+6xy^2- 2ye^{3z}-3Axy^2\big)\,\hk \end{align*}

\begin{align*} \frac{\partial \varphi}{\partial x}(x,y,z) &= y^2 e^{3z} +2xy^3\\ &\hskip1in \implies \varphi(x,y,z)=xy^2e^{3z}+x^2y^3+\psi_1(y,z)\\ \frac{\partial \varphi}{\partial y}(x,y,z) &= 2xye^{3z}+3x^2y^2\\ &\hskip1in \implies \varphi(x,y,z)=xy^2e^{3z}+x^2y^3+\psi_2(x,z)\\ \frac{\partial \varphi}{\partial z}(x,y,z) &= 3xy^2e^{3z}\\ &\hskip1in \implies \varphi(x,y,z)=xy^2e^{3z}+\psi_3(x,y) \end{align*}

\begin{equation*} \varphi(x,y,z)=xy^2e^{3z}+x^2y^3 \end{equation*}

\begin{equation*} \vG = (y^2 e^{3z} + xy^3)\,\hi + (2xye^{3z} + 3x^2 y^2 )\,\hj + 3xy^2 e^{3z}\,\hk = \vF - xy^3\,\hi \end{equation*}

\begin{align*} &\int_C (y^2 e^{3z} + xy^3)\,\dee{x} + (2xye^{3z} + 3x^2 y^2 )\,\dee{y} + 3xy^2 e^{3z}\,\dee{z}\\ &\hskip0.25in = \int_C \vF\cdot\dee{\vr} - \int_C xy^3\,\dee{\vr}\\ &\hskip0.25in = \varphi\big(\vr(1)\big) - \varphi\big(\vr(0)\big) -\int_0^1 \overbrace{e^{2t}\big(e^{-t}\big)^3\,\hi}^{xy^3\,\hi}\cdot \Big(\overbrace{2e^{2t}\,\hi-e^{-t}\,\hj+\frac{1}{1+t}\,\hk}^ {\vr'(t)}\Big)\ \dee{t}\\ &\hskip0.25in = \varphi\big(e^2,\frac{1}{e},\ln 2\big) - \varphi\big(1,1,0\big) - \int_0^1 2e^t\,\dee{t}\\ &\hskip0.25in = \big\{e^2\big(\frac{1}{e}\big)^2 e^{3\ln 2} +e^4\big(\frac{1}{e}\big)^3\big\} -\big(1+1\big) -2(e-1)\\ &\hskip0.25in = 2^3 + e -2 -2e+2\\ &\hskip0.25in = 8-e \end{align*}

\begin{equation*} \frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x}\qquad \frac{\partial F_1}{\partial z}=\frac{\partial F_3}{\partial x}\qquad \frac{\partial F_2}{\partial z}=\frac{\partial F_3}{\partial y} \end{equation*}

\begin{align*} \frac{\partial }{\partial y}\left(2x\sin(\pi y)-e^z\right) &=\frac{\partial }{\partial x}\left(ax^2\cos(\pi y)-3e^z\right)\\ &\hskip1in \iff \quad 2\pi x\cos(\pi y) =2a x\cos(\pi y)\\ \frac{\partial }{\partial z}\left(2x\sin(\pi y)-e^z\right) &=-\frac{\partial }{\partial x}\left(x+by\right)e^z\\ &\hskip1in \iff \quad-e^z =-e^z\\ \frac{\partial }{\partial z}\left(ax^2\cos(\pi y)-3e^z\right) &=-\frac{\partial }{\partial y}\left(x+by\right)e^z\\ &\hskip1in \iff \quad -3e^z =-be^z \end{align*}

\begin{align*} \vF&=\left(2x\sin(\pi y)-e^z\right)\hi +\left(\pi x^2\cos(\pi y)-3e^z\right)\hj -\left(x+3y\right)e^z\hk\\ &=\vnabla\big( x^2\sin(\pi y)-xe^z-3ye^z+C\big) \end{align*}

\begin{align*} \frac{\partial \varphi}{\partial x}(x,y,z) &= 2x\sin(\pi y)-e^z\\ &\implies \varphi(x,y,z)=x^2\sin(\pi y)-xe^z+\psi_1(y,z)\\ \frac{\partial \varphi}{\partial y}(x,y,z) &= \pi x^2\cos(\pi y)-3e^z\\ &\implies \varphi(x,y,z)=x^2\sin(\pi y)-3ye^z+\psi_2(x,z)\\ \frac{\partial \varphi}{\partial z}(x,y,z) &= -(x+3y)e^z \\ &\implies \varphi(x,y,z)=-xe^z-3ye^z+\psi_3(x,y) \end{align*}

\begin{equation*} \varphi(x,y,z) = x^2\sin(\pi y)-xe^z-3ye^z \end{equation*}

\begin{align*} \int _C\vF\cdot \dee{\vr} &=\varphi(1,1,\ln 2)-\varphi(0,0,0)\\ &=\left(\sin\pi-e^{\ln 2}-3e^{\ln 2}\right)-\left(\sin(0)-0-0\right)\\ &=-8 \end{align*}

\begin{equation*} \int _C\vG\cdot \dee{\vr}=\int _C\vF\cdot \dee{\vr} +\int _C3ye^z\,\hk\cdot \dee{\vr} =-8+\int _C3ye^z\,\hk\cdot \dee{\vr} \end{equation*}

\begin{equation*} \int _C\vG\cdot \dee{\vr} =-8+\int_0^13t\,\dee{t} =-8+\frac{3}{2}=-\frac{13}{2} \end{equation*}

\begin{align*} \pdiff{f}{x}(x,y,z) &= -2y \cos x \sin x\\ \pdiff{f}{y}(x,y,z) &= \cos^2 x + (1 + yz) e^{yz}\\ \pdiff{f}{z}(x,y,z) &= y^2 e^{yz} \end{align*}

\begin{equation*} f(x,y,z) = ye^{yz} +g(x,y) \end{equation*}

\begin{equation*} e^{yz} + yze^{yz} +\frac{\partial g}{\partial y}(x,y) = \cos^2 x + (1 + yz) e^{yz} \quad\text{or}\quad \frac{\partial g}{\partial y}(x,y)=\cos^2 x \end{equation*}

\begin{equation*} g(x,y) = y\cos^2 x + h(x) \end{equation*}

\begin{equation*} -2y\sin x\cos x + h'(x) =-2y \cos x \sin x \quad\text{or}\quad h'(x) = 0 \end{equation*}

\begin{align*} \int_C \vF \cdot \dee{\vr} &= \int_C \vnabla f \cdot \dee{\vr} =f\big(\pi, e^\pi , 0\big) -f\big(0, 1, -\pi^2\big)\\ &=\Big[ye^{yz} + y\cos^2 x\Big]^{(\pi, e^\pi , 0)}_{(0, 1, -\pi^2 )}\\ &=\big(2e^\pi\big)-\big(e^{-\pi^2}+1\big) \end{align*}

\begin{align*} \vnabla\times\vF &=\det\left[\begin{matrix} \hi &\hj &\hk\\ \pdiff{}{x} & \pdiff{}{y} & \pdiff{}{z}\\ 2x & 2y & 2z \end{matrix} \right]=(0-0)\hi+(0-0)\hj+(0-0)\hk =\vZero \end{align*}

\begin{align*} \int_C \vF \cdot \dee{\vr} &=\Big[x^2+y^2+z^2\Big]^{(a_1,a_2,a_3)}_{(0,0,0)} =a_1^2 + a_2^2 + a_3^2 =\va\cdot\va \end{align*}

\begin{align*} \vnabla\times\vF &=\det\left[\begin{matrix} \hi &\hj &\hk\\ \pdiff{}{x} & \pdiff{}{y} & \pdiff{}{z}\\ e^{yz} & xze^{yz} + ze^y & xye^{yz} + e^y \end{matrix} \right]\\ &= \big[(xe^{yz}+xyze^{yz}+e^y)-(xe^{yz}+xyze^{yz}+e^y)\big]\,\hi -\big[ye^{yz}-ye^{yz}\big]\,\hj\\ &\hskip1in + \big[ze^{yz}-ze^{yz}\big]\,\hk\\ &= \vZero \end{align*}

\begin{align*} \frac{\partial \varphi}{\partial x}(x,y,z) &= e^{yz} &\implies \varphi(x,y,z)&=xe^{yz}+\psi_1(y,z)\\ \frac{\partial \varphi}{\partial y}(x,y,z) &= xze^{yz} + ze^y&\implies \varphi(x,y,z)&=xe^{yz}+ze^y+\psi_2(x,z)\\ \frac{\partial \varphi}{\partial z}(x,y,z) &= xye^{yz} + e^y&\implies \varphi(x,y,z)&=xe^{yz}+ze^y+\psi_3(x,y) \end{align*}

\begin{gather*} \int_C\vF\cdot\dee{\vr} =\varphi\big(\vr(\pi/2)\big) - \varphi\big(\vr(0)\big) =\varphi\big(0,1,\pi/2\big) - \varphi\big(1,0,0\big) =\frac{\pi e}{2} - 1 \end{gather*}

\begin{align*} \pdiff{f}{x}(x,y) &= 2xy \cos(x^2)\\ \pdiff{f}{y}(x,y) &= \sin(x^2) - \sin(y) \end{align*}

\begin{equation*} f(x,y) = y\sin(x^2) + g(y) \end{equation*}

\begin{equation*} \sin(x^2)+g'(y) = \sin(x^2) - \sin(y) \qquad\text{or}\qquad g'(y) = -\sin(y) \end{equation*}

\begin{equation*} g(y) = \cos(y) + C \end{equation*}

\begin{equation*} \vr(t) = \sin(t)\,\hi + t\,\hj\qquad \frac{\pi}{2}\le t\le\pi \end{equation*}

\begin{align*} \int_C\vF\cdot\dee{\vr} &=f\big(\vr(\pi)\big) -f\big(\vr(\frac{\pi}{2})\big) =f\big(0,\pi\big) -f\big(1,\frac{\pi}{2}\big)\\ &=\big(-1\big)-\Big(\frac{\pi}{2}\sin(1)\Big)\\ &=-1 -\frac{\pi}{2}\sin(1) \end{align*}

\begin{align*} \vZero &= \nabla\times\vF = \left|\begin{matrix} \hi & \hj & \hk\\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ (mxyz + z^2 - ny^2) & (x^2 z - 4xy) & (x^2y + pxz + qz^3) \end{matrix}\right|\\ &= \hi\,\big(x^2-x^2\big) - \hj\,(2xy+pz-mxy-2z) + \hk\,(2xz-4y-mxz+2ny). \end{align*}

\begin{equation*} \vF = (2xyz + z^2 - 2y^2)\,\hi + (x^2 z - 4xy)\,\hj + (x^2y + 2xz + qz^3)\,\hk. \end{equation*}

\begin{align*} \frac{\partial \varphi}{\partial x}(x,y,z) &= 2xyz + z^2 - 2y^2\\ &\implies \varphi(x,y,z)=x^2yz+xz^2-2xy^2+\psi_1(y,z)\\ \frac{\partial \varphi}{\partial y}(x,y,z) &= x^2 z - 4xy\\ &\implies \varphi(x,y,z)=x^2yz-2xy^2+\psi_2(x,z)\\ \frac{\partial \varphi}{\partial z}(x,y,z) \varphi(x,y,z)&= x^2y + 2xz + qz^3\\ &\implies \varphi(x,y,z)=x^2yz+xz^2+\frac{q}{4}z^4+\psi_3(x,y) \end{align*}

\begin{equation*} \varphi(x,y,z) = x^2 y z + x z^2 - 2 x y^2 + \frac{1}{4}q z^4 + C \end{equation*}

\begin{equation*} W = \int_\cC \vF\cdot \dee{\vr} = \int_\cC \nabla\varphi\cdot \dee{\vr} = \varphi(0,0,2) - \varphi(0,0,0) = 4q \end{equation*}

\begin{equation*} \cC:\qquad \vr = (0,0,t),\quad 0\le t\le 2 \end{equation*}

\begin{equation*} \int_\cC \vF\cdot \dee{\vr} = \int_{t=0}^2 \vF\big(\vr(t)\big)\cdot\vr'(t)\,\dee{t} = \int_{t=0}^2 qt^3\,\dee{t} = \Big[\frac{1}{4}qt^4\Big]_{t=0}^2 = 4q \end{equation*}

\begin{equation*} \vr''(t) = \hj - \sin t\,\hk \end{equation*}

\begin{equation*} \vr'(t) = t\,\hj + \cos t\,\hk + \mathbf{C} \end{equation*}

\begin{gather*} \hi+\hk = \vr'(0) = \hk + \mathbf{C} \implies \mathbf{C} = \hi \end{gather*}

\begin{equation*} \vr'(t) = \hi + t\,\hj + \cos t\,\hk \end{equation*}

\begin{equation*} \vr(t) = t\,\hi + \frac{t^2}{2}\,\hj + \sin t\,\hk +\hj = t\,\hi +\Big(1+\frac{t^2}{2}\Big)\,\hj + \sin t\,\hk \end{equation*}

\begin{equation*} \vr(\pi/2) = \frac{\pi}{2}\,\hi +\Big(1+\frac{\pi^2}{8}\Big)\,\hj + \hk \end{equation*}

\begin{align*} \text{Work} &= \int_0^{\pi/2} \vF(t)\cdot\vr'(t)\ \dee{t}\\ &= \int_0^{\pi/2} \big(\hj - \sin t\,\hk\big)\cdot \big(\hi + t\,\hj + \cos t\,\hk\big)\ \dee{t}\\ &= \int_0^{\pi/2} \big(t-\sin t\cos t\big)\,\dee{t}\\ &= \Big[\frac{t^2}{2} +\frac{1}{2}\cos^2 t\Big]_0^{\pi/2}\\ & = \frac{\pi^2}{8}-\frac{1}{2} \end{align*}

\begin{align*} \vr(x)&=x\,\hi+x^2\,\hj+x^3\,\hk\qquad 0\le x\le 1\\ \vr'(x)&=\hi+2x\,\hj+3x^2\,\hk\\ \diff{s}{x}(x)&=\sqrt{1+4x^2+9x^4} \end{align*}

\begin{align*} \rho(x)\ \diff{s}{x}(x) &=\big(8x+36x^3\big)\sqrt{1+4x^2+9x^4}\\ \int_C\rho\ \dee{s} &=\int_0^1 \big(8x+36x^3\big)\sqrt{1+4x^2+9x^4}\,\dee{x} \end{align*}

\begin{align*} \int_C\rho\ \dee{s} &=\int_1^{14} \sqrt{u}\ \dee{u} =\frac{2}{3}u^{3/2}\bigg|_1^{14}\\ &=\frac{2}{3}\big[14^{3/2}-1\big]\approx 34.26 \end{align*}

\begin{equation*} \int_C \vF\cdot \dee{\vr}=f(1,1,1)-f(0,0,0) =\sin 1+\frac{3}{2} \approx 2.3415 \end{equation*}

\begin{align*} \pdiff{f}{x}(x,y,z) &= \sin y\\ \pdiff{f}{y}(x,y,z) &= x\cos y+z\\ \pdiff{f}{z}(x,y,z) &= y+z \end{align*}

\begin{equation*} f(x,y,z) = x\sin y +g(y,z) \end{equation*}

\begin{equation*} x\cos y+\frac{\partial g}{\partial y}(y,z) = x\cos y+z\qquad\hbox{or}\qquad \frac{\partial g}{\partial y}(x,z)=z \end{equation*}

\begin{equation*} g(y,z) = yz + h(z) \end{equation*}

\begin{equation*} y + h'(z) =y+z \quad\hbox{or}\quad h'(z) = z \end{equation*}

\begin{equation*} \vr(t) = \cos t\,\hi +\sin t\,\hj +\frac{t}{2\pi}\,\hk \end{equation*}

\begin{align*} \vr'(t) &= -\sin t\,\hi +\cos t\,\hj +\frac{1}{2\pi}\,\hk\\ \vF\big(x(t),y(t),z(t)\big)& = -\sin t\,\hi + \cos t\,\hj + \frac{t^2}{4\pi^2} \hk\\ \vF\big(x(t),y(t),z(t)\big)\cdot\vr'(t)& = 1 + \frac{t^2}{8\pi^3} \end{align*}

\begin{align*} \int_C \vF\cdot\dee{\vr} &= \int_0^{2\pi} \vF\big(x(t),y(t),z(t)\big)\cdot\vr'(t)\ \dee{t} = \int_0^{2\pi}\left(1 + \frac{t^2}{8\pi^3} \right)\ \dee{t}\\ &=2\pi+\frac{1}{3} \end{align*}

\begin{equation*} \vr(x) = x\,\hi +(9-x^2)\,\hj \end{equation*}

\begin{align*} \int_C (x^2 + y)\,\dee{x} + x\,\dee{y} &=\int_{-3}^3 \big(x^2\ +\ \overbrace{9-x^2}^{y}\ +x\overbrace{(-2x)}^{\diff{y}{x}}\big)\,\dee{x}\\ &=\int_{-3}^3 \big(9-2x^2\big)\,\dee{x}\\ &=2\int_0^3 \big(9-2x^2\big)\,\dee{x} =2\left(27-2\frac{3^3}{3}\right)\\ &=18 \end{align*}

\begin{gather*} \int_{L_1} \vF \cdot\hn\, \dee{s} =\int_0^1 \vF(s,0) \cdot(-\hj)\, \dee{s} =\int_0^1 (-0)\, \dee{s} =0 \end{gather*}

\begin{gather*} \int_{L_2} \vF \cdot\hn\, \dee{s} =\int_0^1 \vF(1,s) \cdot \hi\, \dee{s} =\int_0^1 2\, \dee{s} =2 \end{gather*}

\begin{gather*} \int_{L_3} \vF \cdot\hn\, \dee{s} =\int_0^1 \vF(1-s,1) \cdot\hj\, \dee{s} =\int_0^1 e^{1-s}\, \dee{s} =\Big[-e^{1-s}\Big]_0^1 =e-1 \end{gather*}

\begin{gather*} \int_{L_2} \vF \cdot\hn\, \dee{s} =\int_0^1 \vF(0,1-s) \cdot(- \hi)\, \dee{s} =\int_0^1 (-0)\, \dee{s} =0 \end{gather*}

\begin{align*} \int_C \vF \cdot\hn\, \dee{s} &= \int_{L_1} \vF \cdot\hn\, \dee{s} +\int_{L_2} \vF \cdot\hn\, \dee{s} +\int_{L_3} \vF \cdot\hn\, \dee{s} +\int_{L_4} \vF \cdot\hn\, \dee{s}\\ &= 0+2+(e-1) +0 =e+1 \end{align*}

\begin{gather*} \va(t) = \vv'(t) = \vF(t) = \hj - \sin t\, \hk \end{gather*}

\begin{equation*} \vv(t) = t\,\hj +\cos t\,\hk +\vc \end{equation*}

\begin{gather*} \vr'(t) = \vv(t) = \hi + t\,\hj +\cos t\,\hk \end{gather*}

\begin{equation*} \vr(t) = t\,\hi+\frac{t^2}{2}\,\hj +\sin t\,\hk +\vc \end{equation*}

\begin{gather*} \vr(t) = t\,\hi+\left(1+\frac{t^2}{2}\right)\,\hj +\sin t\,\hk \end{gather*}

\begin{gather*} \vr_1=\vr(\pi/2) = \frac{\pi}{2}\,\hi+\left(1+\frac{\pi^2}{8}\right)\,\hj + \hk \end{gather*}

\begin{align*} \int_0^{\pi/2} \vF(t)\cdot \dee{\vr} &=\int_0^{\pi/2} \vF(t)\cdot \diff{\vr}{t}(t)\,\dee{t}\\ &=\int_0^{\pi/2} [\hj-\sin t\,\hk]\cdot [\hi+t\,\hj+\cos t\,\hk]\,\dee{t}\\ &= \int_0^{\pi/2} [t -\sin t\cos t]\,\dee{t} =\Big[\frac{t^2}{2}+\frac{1}{2}\cos^2 t\Big]_0^{\pi/2} =\frac{\pi^2}{8}-\frac{1}{2} \end{align*}

\begin{equation*} \vr(t) = \big(x(t),y(t)\big) =(t,t), \end{equation*}

\begin{align*} \int_L \vF\cdot\dee{\vr} &=\int_2^1 \vF\big(x(t),y(t)\big)\cdot(x'(t),y'(t))\ \dee{t} =\int_2^1 \big(3t\,,\,t-1\big)\cdot\big(1,1\big)\ \dee{t}\\ &=\int_2^1 (4t-1)\ \dee{t} =-5 \end{align*}

\begin{align*} \int_{C_Y} \vF\cdot\dee{\vr} &= \int_{L_1}\hskip-5pt \big\{(3y\,\dee{x} + (x-1)\dee{y}\big\} + \int_{L_2}\hskip-5pt \big\{(3y\,\dee{x} + (x-1)\dee{y}\big\}\\ & \hskip1in + \int_{L_3}\hskip-5pt \big\{(3y\,\dee{x} + (x-1)\dee{y}\big\}\\ &= \int_2^Y \dee{y} + \int_2^1 3Y\,\dee{x} + \int_Y^1 0\,\dee{y}\\ &= (Y-2) + 3Y(1-2) =-2Y-2 \end{align*}

\begin{alignat*}{2} &&(4a+2b+c)-(a+b+c)&=(2)-(1)\\ \implies&& 3a+b&=1\\ \implies&& b&=1-3a\\ \end{alignat*}

Using \(b=1-3a\text{,}\)

\begin{alignat*}{2} && a+b+c&=1\\ \implies&& a+(1-3a)+c&= 1\\ \implies&& c&=2a \end{alignat*}

\begin{align*} \vr(x)&=\big(x,\ ax^2+(1-3a)x+2a\big)\\ \vF(\vr(x))&=\big( 3ax^2+3(1-3a)x+6a,\ x-1 \big)\\ \vr'(x)&=\big( 1,\ 2ax+1-3a\big)\\ \vF(\vr(x))\cdot \vr'(x)&=\big(3ax^2+3(1-3a)x+6a\big)\\ &\hskip1in + \big( 2ax^2+(1-3a)x-2ax+(3a-1)\big)\\ &=5ax^2+(4-14a)x+(9a-1)\\ \end{align*}

So, if \(C\) is a portion of this parabola from \((2,2)\) to \((1,1)\text{,}\) then

\begin{align*} \int_C \vF \cdot \dee{\vr}&=\int_2^1\big(5ax^2+(4-14a)x+(9a-1)\big)\ \dee{x}\\ &=\left[\frac{5a}{3}x^3+(2-7a)x^2+(9a-1)x\right]_2^1\\ &=\frac{a}3-5 \end{align*}

\begin{align*} \int_{C_Y} \vF\cdot\dee{\vr} &= \int_{L_1}\hskip-5pt (2y+2)\,\dee{x} + \int_{L_2}\hskip-5pt (2y+2)\,\dee{x} + \int_{L_3}\hskip-5pt (2y+2)\,\dee{x}\\ &= 0 + \int_0^2 (2Y+2)\,\dee{x} + 0\\ &= 2(2Y+2) \end{align*}

\begin{align*} \vF(\vr(t))\cdot\vr'(t)&=-\sin t (2A\sin 2 +2)=-A(2\sin^2t)-2\sin t\\ &=-A(1-\cos2t)-2\sin t\\ \int \vF \cdot \dee\vr&=\int_{\pi}^0\big( A(\cos2t-1)-2\sin t \big)\dee{t}\\ &=\left[A\left(\frac12\sin(2t)-t\right) +2\cos t\right]_{\pi}^0\\ &=A\pi+4 \end{align*}

\begin{equation*} \vF=\vnabla\varphi\qquad \varphi(x,y) = x+ \int_0^y \tilde y g(\tilde y)\ \dee{\tilde y} \end{equation*}

\begin{align*} \int_C\vF\cdot\dee{\vr} &=\varphi(Q)-\varphi(P) =x_1 + \int_0^{0} \tilde y g(\tilde y)\ \dee{\tilde y} - x_0 - \int_0^{0} \tilde y g(\tilde y)\ \dee{\tilde y}\\ &=x_1-x_0 \end{align*}

\begin{equation*} \left|\int_C\vF\cdot\dee{\vr}\right| =|x(Q)-x(P)| = \text{distance between } P \text{ and } Q \end{equation*}

\begin{align*} &\vnabla\times\vF =\det\left[\begin{matrix} \hi &\hj &\hk\\ \pdiff{}{x} & \pdiff{}{y} & \pdiff{}{z}\\ (1 + ax^2 )ye^{3x^2} - bxz \cos(x^2 z) & xe^{3x^2} & x^2 \cos(x^2 z) \end{matrix} \right]\\ &\hskip0.25in= 0\,\hi +[-bx\cos(x^2 z)+bx^3z\sin(x^2 z) -2x\cos(x^2 z) +2x^3z\sin(x^2 z)]\,\hj\\ &\hskip1in + [e^{3x^2} +6x^2e^{3x^2}-(1 + ax^2 )e^{3x^2}]\,\hk\\ &\hskip0.25in= [-(b+2)x\cos(x^2 z)+(b+2)x^3z\sin(x^2 z)]\,\hj + (6-a)x^2e^{3x^2}\,\hk \end{align*}

\begin{align*} \pdiff{f}{x}(x,y,z) &= (1 + 6x^2 )ye^{3x^2} + 2xz \cos(x^2 z)\\ \pdiff{f}{y}(x,y,z) &= xe^{3x^2}\\ \pdiff{f}{z}(x,y,z) &= x^2 \cos(x^2 z) \end{align*}

\begin{equation*} f(x,y,z) = xye^{3x^2} +g(x,z) \end{equation*}

\begin{equation*} \frac{\partial g}{\partial z}(x,z) = x^2 \cos(x^2 z) \end{equation*}

\begin{equation*} g(x,z) = \sin(x^2 z) + h(x) \end{equation*}

\begin{align*} (1+6x^2)ye^{3x^2} +2xz\cos(x^2 z) + h'(x) &=(1 + 6x^2 )ye^{3x^2} + 2xz \cos(x^2 z)\\ &\text{or}\quad h'(x) = 0 \end{align*}

\begin{align*} &\int_C \Big(ye^{3x^2} + 2xz\cos(x^2 z)\Big) \dee{x} + xe^{3x^2} \dee{y} + x^2 \cos(x^2 z) \dee{z}\\ &\hskip1in = \int_C \vnabla f\cdot\dee{\vr} -6\int_Cx^2ye^{3x^2}\,\dee{x}\\ &\hskip1in= f(1,1,1) -f(0,0,0) -6\int_0^1 t^3e^{3t^2}\,\dee{t}\\ &\hskip1in= e^3 +\sin 1 -\frac{1}{3}\int_0^3 ue^u\,\dee{u}\qquad \text{with }u=3t^2,\ \dee{u} = 6t\,\dee{t}\\ &\hskip1in= e^3 +\sin 1 -\frac{1}{3}\Big[ue^u-e^u\Big]_0^3\qquad\text{integration by parts}\\ &\hskip1in= \frac{1}{3} e^3 +\sin 1 -\frac{1}{3} \end{align*}

\begin{align*} \vr(x)&=x\,\hi+x^2\,\hj+x^3\,\hk\qquad 0\le x\le 1\\ \vr'(x)&=\hi+2x\,\hj+3x^2\,\hk\\ \vF\big(\vr(x)\big)\cdot\vr'(x) &=\big((x^4-x^2)\,\hi +(x+x^3)\,\hj+x^2\,\hk\big)\cdot \big(\hi+2x\,\hj+3x^2\,\hk\big)\\ &=x^4-x^2+2x^2+2x^4+3x^4 =x^2+6x^4\\ \int_C\vF\cdot \dee{\vr} &=\int_0^1 \big[x^2+6x^4\big]\,\dee{x} =\Big[\frac{x^3}{3}+\frac{6x^5}{5}\Big]_0^1 =\frac{23}{15} = 1.5\dot 3 \end{align*}

\begin{align*} \diff{s}{x}&=\left|\diff{\vr}{x}\right|=\sqrt{1+4x^2+9x^4}\\ \rho(x,x^2,x^3)\ \diff{s}{x} &=\big(8x+36x^3\big)\sqrt{1+4x^2+9x^4}\\ \int_C\rho\ \dee{s} &=\int_0^1 \big(8x+36x^3\big)\sqrt{1+4x^2+9x^4}\,\dee{x}\\ \end{align*}

Using the substitution \(u=1+4x^2+9x^4\text{,}\) \(\dee{u} = (8x+36x^3)\,\dee{x}\text{:}\)

\begin{align*} &=\frac{2}{3}\big[1+4x^2+9x^4\big]^{3/2}\Big|_0^1\\ &=\frac{2}{3}\big[14^{3/2}-1\big]\approx 34.26 \end{align*}

\begin{equation*} \int_C \vF\cdot \dee{\vr}=f(1,1,1)-f(0,0,0) =\sin 1+\frac{3}{2} \approx 2.3415 \end{equation*}

\begin{align*} \pdiff{f}{x}(x,y,z) &= \sin y &\implies f(x,y,z)&=x\sin y + \psi_1(y,z)\\ \pdiff{f}{y}(x,y,z) &= x\cos y+z &\implies f(x,y,z)&=x\sin y+yz+\psi_2(x,z)\\ \pdiff{f}{z}(x,y,z) &= y+z&\implies f(x,y,z)&=yz+\frac12z^2+\psi_3(x,y) \end{align*}

\begin{equation*} \frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x}\qquad \frac{\partial F_1}{\partial z}=\frac{\partial F_3}{\partial x}\qquad \frac{\partial F_2}{\partial z}=\frac{\partial F_3}{\partial y} \end{equation*}

\begin{alignat*}{3} \frac{\partial }{\partial y}(Ax^3y^2z) &=\frac{\partial }{\partial x}\big(z^3+Bx^4yz\big) & &\ \iff\ & 2Ax^3yz & =4Bx^3yz\\ \frac{\partial }{\partial z}(Ax^3y^2z) &=\frac{\partial }{\partial x}\big(3yz^2-x^4y^2\big) & &\ \iff\ & Ax^3y^2 & =-4x^3y^2\\ \frac{\partial }{\partial z}\big(z^3+Bx^4yz\big) &=\frac{\partial }{\partial y}\big(3yz^2-x^4y^2\big) & &\ \iff\ & 3z^2+Bx^4y& =3z^2-2x^4y \end{alignat*}

\begin{equation*} \vF=-4x^3y^2z\,\hi+\big(z^3-2x^4yz\big)\,\hj +\big(3yz^2-x^4y^2\big)\,\hk \end{equation*}

\begin{align*} \frac{\partial \varphi}{\partial x}(x,y,z) &= -4x^3y^2z &\implies \varphi(x,y,z)&= -x^4y^2z+\psi_1(y,z)\\ \frac{\partial \varphi}{\partial y}(x,y,z) &= z^3-2x^4yz &\implies \varphi(x,y,z)&=yz^3-x^4y^2z+\psi_2(x,z)\\ \frac{\partial \varphi}{\partial z}(x,y,z) &= 3yz^2-x^4y^2&\implies \varphi(x,y,z)&=yz^3-x^4y^2z+\psi_3(x,y) \end{align*}

\begin{align*} \vG &= (z-4x^3y^2z)\hi+(z^3-x^4yz)\hj+(3yz^2-x^4y^2)\hk\\ &=\vF +z\,\hi+x^4yz\,\hj \end{align*}

\begin{align*} J &=\int_\cC(z\hi+x^4yz\,\hj+\vF)\cdot \dee{\vr} =-2+\int_\cC(z\hi+x^4yz\,\hj)\cdot \dee{\vr} \end{align*}

\begin{align*} \int_\cC(z\hi+x^4yz\,\hj)\cdot \dee{\vr} &=\int_0^1(x^2\hi-x^7\,\hj)\cdot \big(\hi-\hj+2x\,\hk\big)\ \dee{x}\\ &=\int_0^1(x^2+x^7)\ \dee{x}\\ &=\frac{1}{3}+\frac{1}{8}\\ &=\frac{11}{24}\\ &\implies J=-2+\frac{11}{24}=-\frac{37}{24}\approx-1.5417 \end{align*}

\begin{align*} \int_\cT(z\hi+\vF)\cdot \dee{\vr} &=\int_{\cT_3}z\hi\cdot \dee{\vr}=\int_{0}^1\overbrace{(1-t)\hi}^{z\hi}\cdot\overbrace{(\hi-\hk)\ \dee{t}}^{\dee{\vr}}\\ &=\int_0^1 (1-t) \ \dee{t} =\frac{1}{2} \end{align*}

\begin{equation*} m\va =\vF\ \implies\ 2\vv'(t) = \big(4t\,,\,6t^2\,,\,-4t\big) \ \implies\ \vv'(t) = \big(2t\,,\,3t^2\,,\,-2t\big) \end{equation*}

\begin{align*} \vv(t) &= \vv(0) +\int_0^t \vv'(u)\, \dee{u} = (0,0,0) + \int_0^t \big(2u\,,\,3u^2\,,\,-2u\big) \, \dee{u}\\ &= \big(t^2\,,\,t^3\,,\,-t^2\big) \end{align*}

\begin{align*} \vr(t) &= \vr(0) +\int_0^t \vr'(u)\, \dee{u} = (1,2,3) + \int_0^t \big(u^2\,,\,u^3\,,\,-u^2\big) \, \dee{u}\\ &= (1,2,3) + \big(t^3/3\,,\,t^4/4\,,\,-t^3/3\big) =\left(\frac{t^3}{3}+1\,,\,\frac{t^4}{4}+2\,,\,-\frac{t^3}{3}+3\right) \end{align*}

\begin{equation*} |\vr'(t)|= \big|t^2(1,t,-1)\big| = t^2 \sqrt{2+t^2} \end{equation*}

\begin{align*} \vr'(t)\times\vr''(t)&= \det\left[\begin{matrix}\hi&\hj&\hk \\ t^2 & t^3 & -t^2 \\ 2t & 3t^2 & -2t\end{matrix}\right]\\ &=\big(-2t^4+3t^4\big)\,\hi -\big(-2t^3+2t^3\big)\,\hj + \big(3t^4-2t^4\big)\,\hk\\ & = t^4\,\hi +t^4\,\hk\\ \implies \big|\vr'(t)\times\vr''(t)\big| &= \sqrt{2}\, t^4 \end{align*}

\begin{align*} \ka(t) &= \frac{|\vr'(t)\times\vr''(t)|}{|\vr'(t)|^3} =\frac{\sqrt{2}\, t^4}{{\big(t^2 \sqrt{2+t^2}\big)}^3}\\ &=\frac{\sqrt{2}}{t^2{(2+t^2)}^{3/2}} \end{align*}

\begin{align*} \int_{t=0}^{t=T} \vF(t)\cdot \dee{\vr} &=\int_0^T \vF(t)\cdot \diff{\vr}{t}(t)\,\dee{t}\\ &=\int_0^T \big(4t\,,\,6t^2\,,\,-4t\big)\cdot \big(t^2\,,\,t^3\,,\,-t^2\big)\,\dee{t}\\ &= \int_0^T \big(8t^3+6t^5\big)\,\dee{t} =2T^4 + T^6 \end{align*}

\begin{align*} \vr(t)&=\Big(\frac{4\sqrt{2}}{3}t^{3/2},\frac{4\sqrt{2}}{3}t^{3/2}, t(2-t)\Big)\\ \vv(t)=\vr'(t)&=\big(2\sqrt{2}t^{1/2},2\sqrt{2}t^{1/2},2-2t\big)\\ |\vv|&=\sqrt{8t+8t+4-8t+4t^2}=\sqrt{4(1+2t+t^2)}\\ &= 2|1+t|=2(1+t) \end{align*}

\begin{equation*} \int_0^2|\vv(t)|\,\dee{t}=\int_0^2 2(1+t)\,\dee{t} = 2\Big[t+\frac{t^2}{2}\Big]_0^2 =8 \end{equation*}

\begin{align*} \vv(t)=\vr'(t)&=\big(2\sqrt{2}t^{1/2},2\sqrt{2}t^{1/2},2-2t\big) & \vv(1)&=2\sqrt{2}\big(1,1,0\big)\\ \va(t)=\vv'(t)&=\big(\sqrt{2}t^{-1/2},\sqrt{2}t^{-1/2},-2\big) & \va(1)&=\sqrt{2}\big(1,1,-\sqrt{2}\big)\\ \vv(1)\times\va(1)&=4\big(-\sqrt{2},\sqrt{2},0\big) & |\vv(1)|&=4 \end{align*}

\begin{align*} \ka(1) &= \frac{|\vv(1)\times\va(1)|}{|\vv(1)|^3} =\frac{8}{4^3} =\frac{1}{8} \end{align*}

\begin{equation*} \varphi\big(\vr(2)\big)-\varphi\big(\vr(0)\big) =\varphi\big(\frac{16}{3}\,,\,\frac{16}{3}\,,\,0\big) -\varphi(0,0,0)=0 \end{equation*}

\begin{align*} \int_0^2\!\! \vF\cdot \dee{\vr} &=\int_0^2\!\! \vF(t)\cdot \diff{\vr}{t}(t)\,\dee{t} =-\int_0^2 |\vv(t)|^2\vv(t)\cdot \vv(t)\,\dee{t}\\ &=-\int_0^2 |\vv(t)|^4\,\dee{t}\\ &=-2^4\int_0^2 (1+t)^4\,\dee{t} =-\frac{16}{5}(1+t)^5\Big|_0^2\\ &=-\frac{16}{5}(3^5-1)\approx-774.4 \end{align*}

\begin{equation*} \va(t) = \difftwo{s}{t}\hat\vT + \ka\Big(\diff{s}{t}\Big)^2\hat\vN \end{equation*}

\begin{equation*} \va(1) = 0\hat\vT + \frac{1}{8}\big(3\big)^2(-\hk) = -\frac{9}{8}\hk \end{equation*}

\begin{equation*} \vr(u)=\left(\frac{4\sqrt{2}}{3}u^{3/2},\frac{4\sqrt{2}}{3}u^{3/2}, u(2-u)\right) \end{equation*}

\begin{equation*} \vR(t) = \vr\big(u(t)\big) \end{equation*}

\begin{align*} &3=\Big|\diff{\vR}{t}(t)\Big|=\Big|\diff{\vr}{u}(u(t))\diff{u}{t}\Big| =2\big(1+u(t)\big)\diff{u}{t}\\ &\implies \diff{u}{t}=\frac{3}{2(1+u(t))}\\ & \implies \difftwo{u}{t}=-\frac{3}{2(1+u(t))^2}\diff{u}{t} =-\frac{9}{4(1+u(t))^3} \end{align*}

\begin{align*} \difftwo{\vR}{t}(t)&=\diff{}{t}\Big(\diff{\vR}{t}(t)\Big) =\diff{ }{t}\Big(\diff{\vr}{u}(u(t))\diff{u}{t}(t)\Big)\\ &=\difftwo{\vr}{u}\Big(\diff{u}{t}\Big)^2 +\diff{\vr}{u}\difftwo{u}{t} \end{align*}

\begin{align*} \diff{\vr}{u}&=\big(2\sqrt{2}u^{1/2},2\sqrt{2}u^{1/2},2-2u\big)\\ \difftwo{\vr}{u}&=\big(\sqrt{2}u^{-1/2},\sqrt{2}u^{-1/2},-2\big) \end{align*}

\begin{align*} \diff{\vr}{u}&=\big(2\sqrt{2},2\sqrt{2},0\big)\\ \difftwo{\vr}{u}&=\big(\sqrt{2},\sqrt{2},-2\big) \end{align*}

\begin{align*} \difftwo{\vR}{t} &=\difftwo{\vr}{u}\Big(\diff{u}{t}\Big)^2 +\diff{\vr}{u}\difftwo{u}{t}\\ & =\big(\sqrt{2},\sqrt{2},-2\big)\Big(\frac{3}{4}\Big)^2 +\big(2\sqrt{2},2\sqrt{2},0\big)\Big(-\frac{9}{32}\Big)\\ &=\Big(0,0,-\frac{9}{8}\Big) \end{align*}

\begin{align*} x(u,v) &= u+v\\ y(u,v) &= u^2+v^2\\ z(u,v) &= u-v \end{align*}

\begin{align*} y(u,v) &= u^2+v^2 = \frac{1}{4}\big(x(u,v)+z(u,v)\big)^2 +\frac{1}{4}\big(x(u,v)-z(u,v)\big)^2\\ &= \frac{1}{2} x(u,v)^2 + \frac{1}{2} z(u,v)^2 \end{align*}

\begin{align*} x(\theta, \phi)^2 + y(\theta, \phi)^2 \le 1 &\iff 4\sin^2\phi \le 1 \iff \sin\phi \le \frac{1}{2}\\ &\iff 0\le\phi\le \frac{\pi}{6} \end{align*}

\begin{align*} x(\theta, z)^2 + y(\theta, z)^2 \le 1 &\iff 4-z^2 \le 1 \iff z \ge \sqrt{3} \end{align*}

\begin{align*} x(\theta, \phi)^2 + y(\theta, \phi)^2 \le z(\theta,\phi)^2 &\iff 4\sin^2\phi \le 4\cos^2\phi\\ &\iff \tan\phi \le 1\\ &\iff 0\le\phi\le \frac{\pi}{4} \end{align*}

\begin{align*} x(\theta, z)^2 + y(\theta, z)^2 \le z(\theta, z)^2 &\iff 4-z^2 \le z^2 \iff z \ge \sqrt{2} \end{align*}

\begin{align*} x(\theta, z)^2 + y(\theta, z)^2 \le 1 &\iff z^2 \le 1 \iff 0\le z\le 1 \end{align*}

\begin{align*} x(x, y)^2 + y(x, y)^2 \le 1 &\iff x^2+y^2 \le 1 \end{align*}

\begin{align*} x(\theta, z)^2 + y(\theta, z)^2+ z(\theta, z)^2 \le 4 &\iff 2z^2 \le 4 \iff 0\le z\le \sqrt{2} \end{align*}

\begin{align*} x(x, y)^2 + y(x, y)^2 + z(x, y)^2 \le 4 &\iff 2x^2+2y^2 \le 4 \end{align*}

\begin{align*} d&=\sqrt{(x-2)^2+(y-2)^2+(z-4)^2}\\ &=\sqrt{\left(\frac1{\sqrt 2}\cos\theta\right)^2+\left(\frac1{\sqrt 2}\cos\theta\right)^2+\left(\sin\theta\right)^2}\\ &=\sqrt{\frac12\cos^2\theta+\frac12\cos^2\theta+\sin^2\theta}=\sqrt{\cos^2\theta+\sin^2\theta}\\ &=1 \end{align*}

\begin{align*} R&=\sqrt{(x-4)^2+(y-4)^2+(z-z)^2}\\ &=\sqrt{(\frac1{\sqrt 2}\cos\theta-2)^2+(\frac1{\sqrt 2}\cos\theta-2)^2+0)}\\ &=\cos\theta-2\sqrt2 \end{align*}

\begin{align*} \vnabla F(x,y,z) &=\left(\pdiff{F}{x}(x,y,z)\,,\, \pdiff{F}{y}(x,y,z)\,,\, \pdiff{F}{z}(x,y,z)\right)\\ &=\left( 2x\,,\,2y\,,\,2(z-1)\right)\\ \vnabla G(x,y,z) &=\left(\pdiff{G}{x}(x,y,z)\,,\, \pdiff{G}{y}(x,y,z)\,,\, \pdiff{G}{z}(x,y,z)\right)\\ &=\left( 2x\,,\,2y\,,\,2(z+1)\right) \end{align*}

\begin{equation*} \vnabla F(0,0,0)=\left( 0,0,-2\right)\qquad \vnabla G(0,0,0)=\left( 0,0,2\right) \end{equation*}

\begin{equation*} \hk\cdot\left(x-0,y-0,z-0\right)=0\qquad\text{or}\qquad z=0 \end{equation*}

\begin{gather*} G\big(x(t),y(t),z(t)\big) =0 \end{gather*}

\begin{align*} 0&= \diff{}{t}G\big(x(t),y(t),z(t)\big)\\ &=\pdiff{G}{x}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ x'(t) +\pdiff{G}{y}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ y'(t)\\ &\hskip1in+\pdiff{G}{z}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ z'(t) \end{align*}

\begin{align*} &\pdiff{G}{x}\big( x_0\,,\,y_0\,,\,z_0\big)\ x'(t_0) +\pdiff{G}{y}\big( x_0\,,\,y_0\,,\,z_0\big)\ y'(t_0)\\ &\hskip2in+\pdiff{G}{z}\big( x_0\,,\,y_0\,,\,z_0\big)\ z'(t_0) = 0 \tag{E3} \end{align*}

\begin{align*} &\pdiff{G}{x}\big( x_0\,,\,y_0\,,\,z_0\big)\ {\color{blue}{\{x-x_0\}}} +\pdiff{G}{y}\big( x_0\,,\,y_0\,,\,z_0\big)\ {\color{blue}{\{y-y_0\}}}\\ &\hskip2in+\pdiff{G}{z}\big( x_0\,,\,y_0\,,\,z_0\big)\ {\color{blue}{\{z-z_0\}}}\\ &=\pdiff{G}{x}\big( x_0\,,\,y_0\,,\,z_0\big)\ {\color{blue}{\big\{ t\,x'(t_0)\big\}}} +\pdiff{G}{y}\big( x_0\,,\,y_0\,,\,z_0\big)\ {\color{blue}{\big\{ t\,y'(t_0)\big\}}}\\ &\hskip2in+\pdiff{G}{z}\big( x_0\,,\,y_0\,,\,z_0\big)\ {\color{blue}{\big\{ t\,z'(t_0)\big\}}}\\ &={\color{blue}{t}}\Big[\pdiff{G}{x}\big( x_0\,,\,y_0\,,\,z_0\big)\ {\color{blue}{x'(t_0)}} +\pdiff{G}{y}\big( x_0\,,\,y_0\,,\,z_0\big)\ {\color{blue}{y'(t_0)}}\\ &\hskip2in+\pdiff{G}{z}\big( x_0\,,\,y_0\,,\,z_0\big)\ {\color{blue}{z'(t_0)}} \Big]\\ &=0 \end{align*}

\begin{gather*} \vn =-f_x(x_0,y_0)\,\hi - f_y(x_0,y_0)\,\hj + \hk \end{gather*}

\begin{align*} &(x-x_0\,,\,y-y_0\,,\,z-z_0) = t\big(-f_x(x_0,y_0)\,,\,- f_y(x_0,y_0)\,,\, 1\big) \qquad\text{or}\\ &x=x_0-tf_x(x_0,y_0)\quad y=y_0-tf_y(x_0,y_0)\quad z=f(x_0,y_0)+t \end{align*}

\begin{gather*} \vt = \vnabla F(x_0,y_0,z_0)\times \vnabla G(x_0,y_0,z_0) \end{gather*}

\begin{align*} &\vn\cdot\left( x-x_0\,,\,y-y_0\,,\,z-z_0\right) =0 \\ &\qquad\text{where } \vn=\vt=\vnabla F(x_0,y_0,z_0)\times \vnabla G(x_0,y_0,z_0) \end{align*}

\begin{align*} &\vt = \big[-f_x(x_0,y_0)\,\hi - f_y(x_0,y_0)\,\hj+\hk\big]\times \big[-g_x(x_0,y_0)\,\hi - g_y(x_0,y_0)\,\hj+\hk\big]\\ &=\det\left[\begin{matrix} \hi & \hj & \hk\\ -f_x(x_0,y_0) & -f_y(x_0,y_0) & 1\\ -g_x(x_0,y_0) & -g_y(x_0,y_0) & 1 \end{matrix}\right]\\ &=\Big( g_y(x_0,y_0)-f_y(x_0,y_0)\,,\, f_x(x_0,y_0)-g_x(x_0,y_0)\,,\,\\ &\hskip2in f_x(x_0,y_0)g_y(x_0,y_0)-f_y(x_0,y_0)g_x(x_0,y_0)\Big) \end{align*}

\begin{equation*} \left( x-x_0\,,\,y-y_0\,,\,z-z_0\right) = t\,\vd \end{equation*}

\begin{align*} x&=x_0+t\big[g_y(x_0,y_0)-f_y(x_0,y_0)\big]\\ y&=y_0+t\big[f_x(x_0,y_0)-g_x(x_0,y_0)\big]\\ z&=z_0+ t\big[f_x(x_0,y_0)g_y(x_0,y_0)-f_y(x_0,y_0)g_x(x_0,y_0)\big] \end{align*}

\begin{alignat*}{2} f_x(x,y)&=\frac{2xy}{x^4+2y^2}-\frac{x^2y(4x^3)}{{(x^4+2y^2)}^2}\qquad & f_x(-1,1)&=-\frac{2}{3} +\frac{4}{3^2}=-\frac{2}{9}\\ f_y(x,y)&=\frac{x^2}{x^4+2y^2}-\frac{x^2y(4y)}{{(x^4+2y^2)}^2}\qquad & f_y(-1,1)&=\frac{1}{3} -\frac{4}{3^2}=-\frac{1}{9} \end{alignat*}

\begin{align*} \frac{2}{9}(x+1) +\frac{1}{9}(y-1) +\Big(z-\frac{1}{3}\Big) &=0\\ \frac{2}{9}x +\frac{1}{9}y + z &=-\frac{2}{9}+\frac{1}{9}+\frac{1}{3} =\frac{2}{9} \end{align*}

\begin{align*} \vnabla G(2,1,1) &=-\frac{1}{2}\ \frac{27}{(x^2+y^2+z^2+3)^{3/2}}\big(2x\,,\,2y\,,\,2z\big) \bigg|_{(x,y,z)=(2,1,1)}\\ &=-( 2\,,\,1\,,\,1) \end{align*}