The curve crosses itself at all points \((0,(\pi n)^2)\) where \(n\) is an integer. It passes such a point twice, \(2\pi n\) time units apart.
1.1.4.
Answer.
(a) \((a+a\theta,a)\)
(b)\((a+a\theta+a\sin\theta,a+a\cos\theta)\)
1.1.5.
Answer.
\(z=-\frac12\sqrt{1-\frac{y^2}{2}}-\frac{y}{4}\)
1.1.6.
Answer.
The particle is moving upwards from \(t=1\) to \(t=2\text{,}\) and from \(t=3\) onwards. The particle is moving downwards from \(t=0\) to \(t=1\text{,}\) and from \(t=2\) to \(t=3\text{.}\)
The particle is moving faster when \(t=1\) than when \(t=3\text{.}\)
1.1.7.
Answer.
The red vector is \(\vr(t+h)-\vr(t)\text{.}\) The arclength of the segment indicated by the blue line is the (scalar) \(s(t+h)-s(t)\text{.}\)
Remark: as \(h\) approaches 0, the curve (if it’s differentiable at \(t\)) starts to resemble a straight line, with the length of the vector \(\vr(t+h)-\vr(t)\) approaching the scalar \(s(t+h)-s(t)\text{.}\) This step is crucial to understanding Lemma 1.1.4.
1.1.8.
Answer.
Velocity is a vector-valued quantity, so it has both a magnitude and a direction. Speed is a scalar — the magnitude of the velocity. It does not include a direction.
(b) \(\frac{s}{\sqrt{2}}\left(\cos\Big(\ln\Big(\frac{s}{\sqrt{2}}\Big)\Big)\,,\,
\sin\Big(\ln\Big(\frac{s}{\sqrt{2}}\Big)\Big)\right)\) with \(s \gt 0\)
1.2.6.
Answer.
\(\left(\cos z, z\sin z, z\right)\) for \(0 \le z \lt \pi/2\text{.}\) The curve is (the first quarter-turn of) a spiral, with width in the \(x\)-direction 2, and increasing width in the \(y\)-direction. The parameter \(z\) is the height, as well as a radian measure for the spiral.
(a) One possible parametrization is \(\vr(\theta) = \cos\theta\,\hi +\sin\theta\,\hj + (1-\cos\theta-\sin\theta)\,\hk\) with \(0\le\theta\le 2\pi\text{.}\)
(c) \(\text{maximum curvature }=\sqrt{3}\) at \(\frac{\hi}{\sqrt{2}}+\frac{\hj}{\sqrt{2}}+(1-\sqrt{2})\,\hk\) and \(-\frac{\hi}{\sqrt{2}}-\frac{\hj}{\sqrt{2}}+(1 + \sqrt{2})\,\hk\) and \(\text{minimum curvature }= \frac{1}{3}\) at \(-\frac{\hi}{\sqrt{2}}+\frac{\hj}{\sqrt{2}}+\hk\) and \(\frac{\hi}{\sqrt{2}}-\frac{\hj}{\sqrt{2}}+\hk\)
The curve \(\big(a(t),b(t)\big)=(t,t^2)\) is the curve \(y=x^2\text{.}\) It is “curviest” at the origin, which is consistent with part (f). It becomes flatter and flatter as \(|t|\) increases, but never achieves “perfect flatness”, which is consistent with (g).
(c) The skateboarder makes it up to the ceiling, but falls off rather than making it all the way around. Ouch.
1.7.4.13.
Answer.
(a), (b) See the solution.
(c) \(2\left[\frac{a^2+b^2}{gb}\pi\right]^{1/2}\)
1.8Optional — Polar Coordinates
Exercises
1.8.1.
Answer.
The upper sketch below contains the points, \((x_1,y_1)\text{,}\)\((x_3,y_3)\text{,}\)\((x_5,y_5)\text{,}\) that are on the axes. The lower sketch below contains the points, \((x_2,y_2)\text{,}\)\((x_4,y_4)\text{,}\) that are not on the axes.
(a) \(\big(r=2\,,\,\theta= n\pi,\ n\text{ odd integer }\big)\) or \(\big(r=-2\,,\,\theta= n\pi,\ n\text{ even integer }\big)\)
(b) \(\big(r=\sqrt{2}\,,\,\theta= \frac{\pi}{4} + 2n\pi\big)\) or \(\big(r=-\sqrt{2}\,,\,\theta= \frac{5\pi}{4} + 2n\pi\big)\text{,}\) with \(n\) integer.
(c) \(\big(r=\sqrt{2}\,,\,\theta= \frac{5\pi}{4} + 2n\pi\big)\) or \(\big(r=-\sqrt{2}\,,\,\theta= \frac{\pi}{4} + 2n\pi\big)\text{,}\) with \(n\) integer.
1.8.3.
Answer.
(a) Both \(\he_r(\theta)\) and \(\he_\theta(\theta)\) have length 1. The angle between them is \(\frac{\pi}{2}\text{.}\) The cross product is \(\he_r(\theta) \times \he_\theta(\theta)=\hk\text{.}\)
(b) Here is a sketch of \((x_i,y_i)\text{,}\)\(\he_r(\theta_i)\text{,}\)\(\he_\theta(\theta_i)\) for \(i =1,3,5\) (the points on the axes)
and here is a sketch (to a different scale) of \((x_i,y_i)\text{,}\)\(\he_r(\theta_i)\text{,}\)\(\he_\theta(\theta_i)\) for \(i =2,4\) (the points off the axes).
a. \(\vv(p)=\left( \left(1-\frac{p}{2}\right)\frac{1}{2\sqrt{3}} , -\frac{p}{4} \right)\)
b. \(\vV(x,y,z)=\left(
-\frac{x}{6}, -\frac{y}{6},\frac{z}{2}\right)\) or equivalent
2.2Optional — Field Lines 2.2.2Exercises
2.2.2.1.
Answer.
2.2.2.2.
Answer.
\(\vv(x,y) = (-x-y\,,\,x-y)\)
2.2.2.3.(✳).
Answer.
(a) \(\frac{x^2}{2} =\frac{y^2}{2} +C\)
(b)
2.2.2.4.(✳).
Answer.
\(x=y^2\text{,}\)\(z=e^y\)
2.2.2.5.(✳).
Answer.
The field lines are \(y=C'x^3\) with \(C'\) a nonzero constant, as well as \(x=0\) and \(y=0\text{.}\)
2.3Conservative Vector Fields
Exercises
2.3.1.
Answer.
In general, false.
2.3.2.
Answer.
a. C
b. B
c. C
d. B
2.3.3.
Answer.
Let \(\varphi\) be a potential for \(\vF\text{.}\) Define \(\phi=\varphi+ax+by+cz\text{.}\) Then \(\vnabla \phi = \vnabla\varphi+(a,b,c)=\vF+(a,b,c)\text{.}\)
2.3.4.
Answer.
If \(\vF+\vG\) is conservative for any particular \(\vF\) and \(\vG\text{,}\) then by definition, there exists a potential \(\varphi\) with \(\vF+\vG = \vnabla \varphi\text{.}\)
Since \(\vF\) is conservative, there also exists a potential \(\psi\) with \(\vF = \vnabla \psi\text{.}\)
But now \(\vG = (\vF+\vG)-\vF=\vnabla \varphi - \vnabla \psi = \vnabla(\varphi-\psi)\text{.}\) That means the function \((\varphi-\psi)\) is a potential for \(G\text{.}\) However, this is impossible: since \(\vG\) is non-conservative, no function with this property exists.
So it is not possible that \(\vF+\vG\) is conservative. It must be non-conservative.
Counterexample: if \(\vF = -\vG\text{,}\) then \(\vF+\vG = \mathbf 0 = \vnabla c\) for any constant \(c\text{.}\)
Since both fields are conservative, they both have potentials, say \(\vF=\vnabla \varphi\) and \(\vG = \vnabla \psi\text{.}\) Then \(\vF+\vG = \vnabla\varphi+\vnabla\psi=\vnabla(\varphi+\psi)\text{.}\) That is, \((\varphi+\psi)\) is a potential for \(\vF+\vG\text{,}\) so \(\vF+\vG\) is conservative.
2.3.5.(✳).
Answer.
Yes, \(\vF\) is conservative on \(D\text{.}\) A potential is \(\varphi(x,y) = \arctan\frac{y}{x}\text{.}\)
2.3.6.
Answer.
\(\varphi=\frac{1}{2}x^2+xy-\frac{1}{2}y^2\)
2.3.7.
Answer.
\(\varphi=\ln |x| - \frac{x}{y}\)
2.3.8.
Answer.
None exists: \(\pdiff{F_2}{z}=\frac13x^3\text{,}\) while \(\pdiff{F_3}{y}=\frac{1}{3}x^3+1\text{,}\) so \(\vF\) fails the screening test, Theorem 2.3.9.
2.3.9.
Answer.
\(\varphi(x,y,z) = \frac12\ln(x^2+y^2+z^2)\)
2.3.10.
Answer.
(a) \(\vF\) is conservative with potential \(\phi(x,y,z)=\half x^2-y^2+\frac{3}{2}z^2+C\) for any constant \(C\text{.}\)
(b) \(\vF\) is not conservative.
2.3.11.
Answer.
(a) \(A=2\text{,}\)\(B\) is arbitrary.
(b) \(\varphi(x,y,z)=xe^{(z^2)}+By^2 z^3+C\) for any constant \(C\text{.}\)
2.3.12.
Answer.
\(\vv=m\frac{x\hi+y\hj}{x^2+y^2}\)
\(\varphi=\half m\ln(x^2+y^2)+C\) for any constant \(C\)
2.3.13.
Answer.
It can never escape the sphere centred at the origin with radius \(\sqrt{20}\text{.}\)
2.3.14.
Answer.
\(\sqrt{14}\)
2.3.15.
Answer.
\(\varphi=f^2(x)+g(y)h(z)\) is a potential for \(\vF\text{,}\) so \(\vF\) is conservative.
2.3.16.
Answer.
The line through the origin in the direction of the vector \((2,1,2)\text{.}\)
2.4Line Integrals 2.4.2Exercises
2.4.2.1.
Answer.
\(\frac{1}{6}\)
2.4.2.2.
Answer.
a. A
b. B
c. A
d. B
2.4.2.3.
Answer.
0
2.4.2.4.
Answer.
5
2.4.2.5.(✳).
Answer.
\(a=1\text{,}\)\(b=c=0\)
2.4.2.6.
Answer.
(a) Not conservative
(b) Not conservative
(c) Not conservative
(d) Conservative
2.4.2.7.(✳).
Answer.
(a) The (largest possible) domain is \(D=\Set{(x,y,z)}{x^2+y^2\ne 0}\text{.}\)
(b) \(\vnabla\times\vF=\vZero\) on \(D\)
(c) \(\int_C \vF \cdot \dee{\vr}=4\pi\)
(d) \(\vF\) is not conservative.
2.4.2.8.
Answer.
\(9\frac{1}{2}\) for all paths from \((1,0,-1)\) to \((0,-2,3)\)
2.4.2.9.
Answer.
\(2(e-1)+\frac{\pi^2}{2}+3\pi\)
2.4.2.10.
Answer.
(a) \(-\frac{1}{4}\)
(b) \(-1\)
2.4.2.11.(✳).
Answer.
\(-\frac{40}{3}\)
2.4.2.12.
Answer.
(a) \(\la=-1\)
(b) \(\phi(x,y,z)=2x^3yz^2-xyz+y^2+K\text{,}\) for any constant \(K\)
The line integral is independent of path because it is of the form \(\int_C \vF\cdot\dee{\vr}\) with \(\vF\) being a conservative field. The value of the integral is \(1+\frac{\pi}{2}\text{.}\)
2.4.2.18.(✳).
Answer.
\(\frac{1}{2}\)
2.4.2.19.(✳).
Answer.
\(\pi e^\pi\)
2.4.2.20.(✳).
Answer.
(a) \(\alpha=1\text{,}\)\(\beta=\gamma\)
(b) \(e^e-\beta(e+1)\)
2.4.2.21.(✳).
Answer.
(a) \(\vZero\)
(b) Yes. In fact \(\vF=\vnabla f\) with \(f=\sin x +2y-\cos y +e^z \text{.}\)
(c) \(-4\)
2.4.2.22.(✳).
Answer.
(a) \(\vnabla\times\vF=\vZero\text{.}\)\(\vF\) is conservative.
(b) \(\int_C \vF \cdot \dee{\vr}=2\pi^2\)
2.4.2.23.(✳).
Answer.
(a) \(a=-1\text{,}\)\(b=3\)
(b) \(f(x,y,z) = xye^x + yz^3 + C\) works for any constant \(C\)
(c) \(\pi e^\pi-2\)
(d) \(\pi e^\pi -\frac{32}{15}\)
2.4.2.24.(✳).
Answer.
(a) \(A=2\text{,}\)\(B=3\)
(b) \(\varphi(x,y,z)=xy^2e^{3z}+x^2y^3\) is one allowed scalar potential.
(c) \(6+e-2[e-1]=8-e\approx5.2817\)
2.4.2.25.(✳).
Answer.
(a) \(a=\pi,\ b=3\)
(b) \(\varphi(x,y,z)=x^2\sin(\pi y)-xe^z-3ye^z+C\) for any constant \(C\)
(c) \(-8\)
(d) \(-\frac{13}{2}\)
2.4.2.26.(✳).
Answer.
(a) \(f(x,y,z) = ye^{yz} + y\cos^2 x + C\) works for any constant \(C\)
(b) \(2e^\pi-e^{-\pi^2}-1\)
2.4.2.27.(✳).
Answer.
(a) \(\vZero\text{.}\)
(b) \(\vF\) is conservative with potential \(\varphi(x,y,z) = x^2 + y^2 +z^2\text{.}\) So the integral is \(\varphi(a_1,a_2,a_3) - \varphi(0,0,0) = \va\cdot\va\text{.}\)
2.4.2.28.(✳).
Answer.
(a) \(\vnabla\times\vF=\vZero\)
(b) \(\frac{\pi e}{2} - 1\)
2.4.2.29.(✳).
Answer.
(a), (b) \(f(x,y) = y\sin(x^2)+\cos(y) + C\) is a potential for any constant \(C\text{.}\) Because \(\vF\) has a potential, it is conservative.
(c) \(-1 -\frac{\pi}{2}\sin(1)\)
2.4.2.30.(✳).
Answer.
(a) \(p=2\text{,}\)\(m=2\text{,}\)\(n=2\text{,}\) but \(q\in\bbbr\) is completely free
(b) One possibility is the path consisting of the line segment from \((2,2)\) to \((2,-3)\text{,}\) followed by the line segment from \((2,-3)\) to \((1,-3)\text{,}\) followed by the line segment from \((1,-3)\) to \((1,1)\text{.}\)
Another possibility is the path from \((2,2)\) to \((1,1)\) along the parabola \(27x^2-80x+54\text{.}\)
2.4.2.37.(✳).
Answer.
One possibility is the path consisting of the line segment from \((0,0)\) to \((0,1)\text{,}\) followed by the line segment from \((0,1)\) to \((2,1)\text{,}\) followed by the line segment from \((2,1)\) to \((2,0)\text{.}\)
Another possibility is the path tracing out the half ellipse \(\left(\cos t+1\, , \, \frac{4}{\pi}\sin t\right)\text{,}\) with \(t\) running from \(\pi\) to \(0\text{.}\)
The normal plane is \(\vn\cdot\left( x-x_0\,,\,y-y_0\,,\,z-z_0\right) =0\text{,}\) where the normal vector \(\vn = \vnabla F(x_0,y_0,z_0)\times \vnabla G(x_0,y_0,z_0)\text{.}\)
\(z = -\frac{3}{4} x- \frac{3}{2} y + \frac{11}{4}\)
3.2.9.(✳).
Answer.
(a) \(2ax -2ay +z = -a^2\)
(b) \(a=\frac{1}{2}\text{.}\)
3.2.10.(✳).
Answer.
\(x+3y-2z = 1\)
3.2.11.(✳).
Answer.
\(y=2x-2\)
3.2.12.(✳).
Answer.
The tangent plane is \(\frac{8}{25}x-\frac{6}{25}y-z=-\frac{8}{5}\text{.}\)
The normal line is \(( x,y,z) = ( -1,2,\frac{4}{5})
+t ( \frac{8}{25}\,,\,-\frac{6}{25}\,,\,-1)\text{.}\)
3.2.13.(✳).
Answer.
\(\pm(1,0,-2)\)
3.2.14.(✳).
Answer.
\(\big(\frac{1}{\sqrt{2}}\,,\,-1\,,\,-\frac{1}{2}\big)\) and \(\big(-\frac{1}{\sqrt{2}}\,,\,-1\,,\,-\frac{1}{2}\big)\)
3.2.15.(✳).
Answer.
\(\pm \big(\half,-1,-1\big)\)
3.2.16.(✳).
Answer.
(a) \(( 1,0,3)\)
(b) \(( 3,3,-1)\)
(c) \(\vr(t)=( 1,1,3)+t( 3,3,-1)\)
3.2.17.(✳).
Answer.
\(49.11^\circ\) (to two decimal places)
3.2.18.
Answer.
The horizontal tangent planes are \(z=0\text{,}\)\(z=e^{-1}\) and \(z=-e^{-1}\text{.}\) The largest and smallest values of \(z\) are \(e^{-1}\) and \(-e^{-1}\text{,}\) respectively.
(a) \(\vnabla\cdot\vF=0\) if \((x,y,z)\ne\vZero\) and is not defined if \((x,y,z)=\vZero\text{.}\)
(b) \(4\pi\)
(c) \(0\)
(d) The flux integrals \(\dblInt_{S_1} \vF\cdot\hn\,\dee{S}\) and \(\dblInt_{S_2} \vF\cdot\hn\,\dee{S}\) are different, because the one point, \((0,0,0)\text{,}\) where \(\vnabla\cdot\vF\) fails to be well-defined and zero, is contained inside \(S_1\) but is not contained inside \(S_2\text{.}\)
4.2.6.24.(✳).
Answer.
\(72\)
4.2.6.25.(✳).
Answer.
(a) \(3\)
(b) \(-14\pi\)
4.2.6.26.(✳).
Answer.
\(-\frac{4\,\pi}{5}\ 3^{5/2}\)
4.2.6.27.(✳).
Answer.
(a) \(0\)
(b) \(\frac {625}{2}\pi\)
4.2.6.28.(✳).
Answer.
\(4\pi\)
4.2.6.29.(✳).
Answer.
\(\pi\)
4.2.6.30.(✳).
Answer.
\(12\pi\)
4.2.6.31.(✳).
Answer.
\(\frac{188}{15}\pi\approx39.37\)
4.2.6.32.(✳).
Answer.
\(5\pi\)
4.2.6.33.
Answer.
\(\left[\frac{\pi}{6}-\frac{1}{3}\right]a^3\)
4.2.6.34.
Answer.
(a) \(-8\sqrt{2}\pi\)
(b) \(8\sqrt{2}\pi\)
(c) \(16\sqrt{2}\pi\)
4.2.6.35.
Answer.
See the solution.
4.2.6.36.
Answer.
See the solution.
4.2.6.37.(✳).
Answer.
(a), (b) \(36\pi\)
4.2.6.38.(✳).
Answer.
\(\frac{e}{4}\)
4.2.6.39.(✳).
Answer.
(a) \(\frac{1}{\sqrt{3}}(-1,-1,1)\)
(b) \(-\tfrac{27\pi}{2}\)
(c) \(-\tfrac{81\pi}{2}\)
4.2.6.40.(✳).
Answer.
(a) \(\vnabla\cdot\vF =2+2z\)
(b) \(\pi\frac{23}{6} 5^3 =\frac{2875}{6}\pi\)
(c) Let \(S\) be an oriented surface that encloses a solid \(V\) and has outward pointing normal. If \(\bar z = -\frac{9}{2|V|}-1\text{,}\) where \(|V|\) is the volume of \(V\) and \(\bar z\) is the \(z\)-component of the centroid (i.e. centre of mass with constant density) of \(V\text{,}\) then \(\dblInt_{S} \vF\cdot\hn\,\dee{S}=-9\text{.}\) One surface which obeys this condition is the unit cube (with outward normal) centred on \(\big(0,0, -\frac{11}{2}\big)\text{.}\)
4.2.6.41.(✳).
Answer.
(a) \(\frac{\sqrt{5}\pi}{8}\)
(b) \(20\)
(c) \(18\)
4.2.6.42.(✳).
Answer.
(a) \(\frac{\pi}{4} +\frac{\pi(b+d)}{6}\)
(b) \(\dblInt_{\sigma_1\cup\sigma_3} \vF\cdot\hn\,\dee{S}\) is zero if and only if \(d=-b\text{.}\)
(c) \(\dblInt_{\sigma_1\cup\sigma_3} \vF\cdot\hn\,\dee{S}\) is zero for all \(a\text{,}\)\(b\text{,}\)\(c\text{,}\)\(d\text{.}\)
4.2.6.43.(✳).
Answer.
(a) \(4\pi\)
(b) \(0\text{.}\)
4.2.6.44.(✳).
Answer.
See the solution.
4.2.6.45.(✳).
Answer.
See the solution.
4.2.6.46.(✳).
Answer.
\(\frac{3}{4}\)
4.2.6.47.(✳).
Answer.
\(9\pi a^3+9\pi a^2\)
4.2.6.48.(✳).
Answer.
See the solution.
4.2.6.49.(✳).
Answer.
(a) \(0\)
(b) \(\frac{15}{2}\pi\)
4.2.6.50.(✳).
Answer.
\(30+24\pi\)
4.2.6.51.(✳).
Answer.
(a) \(\frac{64}{3}\pi\)
(b) \(\frac{128}{3}\pi\)
4.3Green’s Theorem
Exercises
4.3.1.
Answer.
See the solution.
4.3.2.
Answer.
See the solution.
4.3.3.
Answer.
(a) \(1\)
(b) \(1\)
(c) \(0\)
4.3.4.
Answer.
See the solution.
4.3.5.
Answer.
\(-54\)
4.3.6.
Answer.
\(9\)
4.3.7.(✳).
Answer.
\(\frac{32}{3}\pi\)
4.3.8.(✳).
Answer.
\(-6\)
4.3.9.(✳).
Answer.
(a)
(b) \(-\frac{8}{3}\)
4.3.10.(✳).
Answer.
\(-\frac{1}{3}\)
4.3.11.(✳).
Answer.
\(54\)
4.3.12.(✳).
Answer.
\(\frac{10}{3}\)
4.3.13.(✳).
Answer.
(a) \(-\frac{\pi}{2}\)
(b) \(\frac{3\pi}{2}\)
(c) No.
4.3.14.(✳).
Answer.
\(54\)
4.3.15.(✳).
Answer.
(a) \(I_2=0\)
(b) \(I_3=\pi\)
(c) \(I_4=\pi\)
4.3.16.(✳).
Answer.
(a) \(Q_x-P_y=0\) except at \((0,0)\) where it is not defined.
(b) \(-2\pi\)
(c) No.
(d) \(0\)
(e) \(-2\pi\)
4.3.17.(✳).
Answer.
(a) \(\frac{\pi}{2}\)
(b) \(\frac{\pi}{4}+\frac{2}{3}\)
4.3.18.(✳).
Answer.
\(\frac{3\pi}{2}\)
4.3.19.(✳).
Answer.
\(\frac{3\pi}{8}\)
4.3.20.(✳).
Answer.
\(A=-2\)
4.3.21.(✳).
Answer.
\(-\pi\)
4.3.22.(✳).
Answer.
\(\oint_{\cC_1}\vF\cdot \dee{\vr}=0\) and \(\oint_{\cC_2}\vF\cdot \dee{\vr}=2\pi\)
Rewrite \(\oint_C\vE\cdot\dee{\vr}\) as a surface integral. For the details, see the solution.
4.4.3.37.(✳).
Answer.
\(\sqrt{2}\pi\)
4.4.3.38.(✳).
Answer.
\(\frac{2\pi}{3\sqrt{3}}\)
4.4.3.39.(✳).
Answer.
(a) One possible parametrization is \(\vr(r, \theta) = r\cos \theta\, \hi +r\sin \theta\, \hj + r\,\hk\) with \(0\le r\le 1\text{,}\)\(0\le \theta \le \pi\text{.}\)
(b) \(\pi\)
4.4.3.40.
Answer.
\(-\frac{4}{\sqrt{3}}\pi\)
5True/False and Other Short Questions 5.2Exercises
5.2.1.(✳).
Answer.
(a) True
(b) True
(c) True
(d) False
(e) True
(f) That depends. If \(\ka=0\text{,}\) the curve is part of a straight line. If \(\ka \gt 0\) it is part of a circle of radius \(\frac{1}{\ka}\text{.}\)
