So there is a cancelation. Hence the limit is
\begin{align*} \lim_{t\to1}\frac{3t-3}{2 - \sqrt{5-t}} &=\lim_{t\to1} \left(2 + \sqrt{5-t}\right) \cdot 3\\ &= 12 \end{align*}Since we cancelled out the term that was causing the numerator and denominator to be zero when \(t=\frac{1}{2}\text{,}\) now \(t=\frac{1}{2}\) is in the domain of our function, so we simply plug it in:
\begin{align*} &=\frac{1+1}{\frac{3}{4}\left(\frac{1}{4}-1\right)}\\ &=\frac{2}{\frac{3}{4}\left(-\frac{3}{4}\right)}\\ &=-\frac{32}{9} \end{align*}So,
\begin{align*} \frac{|x|}{x} &= \left\{\begin{array}{rcl} \frac{x}{x}&,&x \gt 0\\ \frac{-x}{x}&,&x \lt 0 \end{array}\right.\\ &= \left\{\begin{array}{rcl} 1&,&x \gt 0\\ -1&,&x \lt 0 \end{array}\right.\\ \end{align*}Therefore,
\begin{align*} 3+\frac{|x|}{x}&=\left\{\begin{array}{rcl} 4&,&x \gt 0\\ 2&,&x \lt 0 \end{array}\right. \end{align*}
So, with \(X=d+4\text{,}\)
\begin{align*} 3\frac{|d+4|}{d+4} &= \left\{\begin{array}{lcl} 3\frac{d+4}{d+4}&,&d+4 \gt 0\\ &\\ 3\frac{-(d+4)}{d+4}&,&d+4 \lt 0 \end{array}\right.\\ &= \left\{\begin{array}{rcl} 3&,&d \gt -4\\ -3&,&d \lt -4 \end{array}\right. \end{align*}
| \(x\) | \(f(x)\) | \(g(x)\) | \(\dfrac{f(x)}{g(x)}\) |
| \(-3\) | \(-3\) | \(-1.5\) | \(2\) |
| \(-2\) | \(0\) | \(0\) | UND |
| \(-1\) | \(3\) | \(1.5\) | \(2\) |
| \(-0\) | \(3\) | \(1.5\) | \(2\) |
| \(1\) | \(1.5\) | \(.75\) | \(2\) |
| \(2\) | \(0\) | \(0\) | UND |
| \(3\) | \(1\) | \(.5\) | \(2\) |
because
\begin{align*} \lim\limits_{x\rightarrow\infty} \left(1 -\frac{1}{3x^4} -\frac{100}{3 x^3}\right) &=1-0-0=1. \end{align*}Now we divide the numerator and denominator by \(x\text{.}\) In the case of the denominator, since \(x \gt 0\text{,}\) \(x=\sqrt{x^2}\text{.}\)
\begin{align*} &\hskip0.25in= \lim_{x\rightarrow \infty}\dfrac{6(x)}{\sqrt{x^2}\sqrt{1+\frac{5}{x}}+\sqrt{x^2}\sqrt{1-\frac{1}{x}}}\\ &\hskip0.25in= \lim_{x\rightarrow \infty}\dfrac{6(x)}{(x)\sqrt{1+\frac{5}{x}}+(x)\sqrt{1-\frac{1}{x}}}\\ & \hskip0.25in= \lim_{x\rightarrow \infty}\dfrac{6}{\sqrt{1+\frac{5}{x}}+\sqrt{1-\frac{1}{x}}}\\ &\hskip0.25in=\frac{6}{\sqrt{1+0}+\sqrt{1-0}}= 3 \end{align*}Since \(n \gt 0\text{,}\) we can simplify \(\sqrt{n^2}=n\text{.}\)
\begin{align*} &=\lim_{n\rightarrow\infty}\frac{5\cdot n}{n\sqrt{1+\frac{5}{n}}+n}\\ &=\lim_{n\rightarrow\infty}\frac{5}{\sqrt{1+\frac{5}{n}}+1}\\ &=\frac{5}{\sqrt{1+0}+1}=\frac{5}{2} \end{align*}So,
\begin{align*} \lim_{a \to 0^+}\frac{a^2-\frac{1}{a}}{a-1}&= \lim_{a \to 0^+}\frac{(a-1)(a^2+a+1)}{a(a-1)}\\ &=\lim_{a \to 0^+}\frac{a^2+a+1}{a}\\ &=\lim_{a \to 0^+}a+1+\frac{1}{a}=\infty \end{align*}
In particular, this limit does not exist. Since the one-sided limit does not exist,
\begin{align*} \lim_{h \to 0}\frac{f(0+h)-f(0)}{h}&=DNE \end{align*}
So, if \((1,-3)\) is on the tangent line, then
\begin{align*} -3&=(2\alpha)(1)-\alpha^2\\ \iff\qquad 0&=\alpha^2-2\alpha-3\\ \iff\qquad 0&=(\alpha-3)(\alpha+1)\\ \iff\qquad \alpha&=3, \quad\mbox{or}\qquad \alpha=-1.\\ \end{align*}So, the tangent lines \(y=(2\alpha)x-\alpha^2\) are
\begin{align*} y&=6x-9 \quad\mbox{and}\quad y=-2x-1. \end{align*}exists. In particular, this means \(f\) is differentiable at \(0\) if and only if both one-sided limits exist and are equal to each other.
\begin{align*} \\ \end{align*}When \(h \lt 0\text{,}\) \(f(h)=0\text{,}\) so
\begin{align*} \lim_{h \to 0^-}\frac{f(h)-f(0)}{h}&=\lim_{h \to 0^-}\frac{0-0}{h}=0\\ \end{align*}So, \(f\) is differentiable at \(x=0\) if and only if
\begin{align*} \lim_{h \to 0^+}\frac{f(h)-f(0)}{h}&=0.\\ \end{align*}To evaluate the limit above, we note \(f(0)=0\) and, when \(h \gt 0\text{,}\) \(f(h)=h^a\sin\left(\frac{1}{h}\right)\text{,}\) so
\begin{align*} \lim_{h \to 0^+}\frac{f(h)-f(0)}{h}&=\lim_{h \to 0^+}\frac{h^a\sin\left(\frac{1}{h}\right)}{h}\\ &=\lim_{h \to 0^+}h^{a-1}\sin\left(\frac{1}{h}\right) \end{align*}
So, plugging in \(x=1\text{:}\)
\begin{align*} y'(1)&=\dfrac{-18(3+1)}{(3-2)^3}=-72 \end{align*}Now, product rule:
\begin{align*} f'(x)&=(2x\!+\!5)(x^{-1/2}\!+\!x^{-2/3})+(x^2\!+\!5x\!+\!1) \left(\frac{-1}{2}x^{-3/2}\!-\!\frac{2}{3}x^{-5/3}\right) \end{align*}
Since \(e^a\) is just a constant,
\begin{align*} \diff{}{x}\{e^{a}e^{x}\}&=e^a\diff{}{x}\{e^x\}=e^ae^x=e^{a+x} \end{align*}Using the rule for differentiating the reciprocal:
\begin{align*} \diff{}{x}\{e^{-x}\}&=\frac{-e^x}{(e^x)^2}=\frac{-1}{e^x}=-e^{-x} \end{align*}In particular, we need the one-sided limits to exist and be equal:
\begin{align*} \textcolor{red}{\lim_{h \to 0^-}\frac{f(1+h)-f(1)}{h}}&=\textcolor{blue}{\lim_{h \to 0^+}\frac{f(1+h)-f(1)}{h}}\\ \end{align*}If \(h \lt 0\text{,}\) then \(1+h \lt 1\text{,}\) so \(f(1+h)=a(1+h)^2+b\text{.}\) If \(h \gt 0\text{,}\) then \(1+h \gt 1\text{,}\) so \(f(1+h)=e^{1+h}\text{.}\) With this in mind, we begin to evaluate the one-sided limits:
\begin{align*} \color{red}{\lim_{h \to 0^-}\frac{f(1+h)-f(1)}{h}}&\color{red}{= \lim_{h \to 0^-}\frac{[a(1+h)^2+b]-[a+b]}{h}}\\ &\color{red}={\lim_{h \to 0^-}\frac{ah^2+2ah}{h}=2a}\\ \color{blue}{\lim_{h \to 0^+}\frac{f(1+h)-f(1)}{h}}&\color{blue}{= \lim_{h \to 0^+}\frac{e^{1+h}-(a+b)}{h}}\\ \end{align*}Since we take \(a+b\) to be equal to \(e\) (to ensure continuity):
\begin{align*} &\color{blue}{= \lim_{h \to 0^+}\frac{e^{1+h}-e^1}{h}}\\ &\color{blue}={\left.\diff{}{x}\{e^x\}\right|_{x=1}=e^1=e} \end{align*}
and
\begin{align*} \lim_{h\to 0^+}\frac{f(0+h)-f(0)}{h} \amp =\lim_{h\to 0^+}\frac{(ah+b)-\cos(0)}{h} =a\qquad\text{since $b=1$}\\ \end{align*}exist and are equal. Because \(\cos(x)\) is differentiable at \(x=0\) we have
\begin{align*} \lim_{h\to 0^-}\frac{\cos(h)-\cos(0)}{h} \amp = \diff{}{x}\cos(x)\bigg|_{x=0} = -\sin(x)\Big|_{x=0}=0 \amp \end{align*}So, using the quotient rule,
\begin{align*} \diff{}{\theta}\{\tan \theta\}&=\frac{\cos\theta\cos\theta-\sin\theta(-\sin\theta)}{\cos^2\theta} =\frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta}\\ &=\left(\frac{1}{\cos \theta}\right)^2=\sec^2\theta \end{align*}
So,
\begin{align*} \sin|x|&=\left\{\begin{array}{rl} \sin x&x\ge 0\\ \sin(-x)&x \lt 0 \end{array}\right. =\left\{\begin{array}{rl} \sin x&x\ge 0\\ -\sin x&x \lt 0 \end{array}\right.\\ \end{align*}where we used the identity \(\sin(-x)=-\sin x\text{.}\) From here, it’s easy to see \(h'(x)\) when \(x\) is anything other than zero.
\begin{align*} \diff{}{x}\{\sin|x|\}&=\left\{\begin{array}{rl} \cos x&x \gt 0\\ ??&x=0\\ -\cos x&x \lt 0 \end{array}\right.\\ \end{align*}To decide whether \(h(x)\) is differentiable at \(x=0\text{,}\) we use the definition of the derivative. One word of explanation: usually in the definition of the derivative, \(h\) is the tiny “change in \(x\)” that is going to zero. Since \(h\) is the name of our function, we need another letter to stand for the tiny change in \(x\text{,}\) the size of which is tending to zero. We chose \(t\text{.}\)
\begin{align*} \lim_{t \to 0} \frac{h(t+0)-h(0)}{t}&=\lim_{t \to 0}\frac{\sin|t|}{t}\\ \end{align*}We consider the behaviour of this function to the left and right of \(t=0\text{:}\)
\begin{align*} \frac{\sin |t|}{t}&=\left\{\begin{array}{ll} \frac{\sin t}{t} & t\ge 0\\ \frac{\sin (-t)}{t} & t \lt 0 \end{array}\right. =\left\{\begin{array}{ll} \frac{\sin t}{t} & t\ge 0\\ -\frac{\sin t}{t} & t \lt 0 \end{array}\right.\\ \end{align*}Since we’re evaluating the limit as \(t\) goes to zero, we need the fact that \(\ds\lim_{t \to 0} \dfrac{\sin t}{t}=1\text{.}\) We saw this in Section 2.7, but also we know enough now to evaluate it another way. Using the definition of the derivative:
\begin{align*} \lim_{t \to 0}\frac{\sin t}{t}&=\lim_{t \to 0}\frac{\sin (t+0)-\sin (0)}{t}=\left.\diff{}{x}\{\sin x\}\right|_{t=0}=\cos 0=1 \end{align*}So, since the one-sided limits disagree,
\begin{align*} \lim_{t \to 0} \frac{h(t+0)-h(0)}{t}&=DNE\\ \end{align*}so \(h(x)\) is not differentiable at \(x=0\text{.}\) Therefore,
\begin{align*} h'(x)&=\left\{\begin{array}{rl} \cos x&x \gt 0\\ -\cos x&x \lt 0 \end{array}\right. \end{align*}So \(f\) is continuous at \(x=0\text{,}\) and so Statement 2.8.8.29.ii does not hold. Now let’s consider \(f'(x)\)
\begin{align*} \lim_{x\rightarrow 0+}\frac{f(x)-f(0)}{x} &=\lim_{x\rightarrow 0+}\frac{\frac{\sin x}{\sqrt{x}}-0}{x}\\ &=\lim_{x\rightarrow 0+}\frac{1}{\sqrt{x}}\frac{\sin x}{x}=+\infty\\ \end{align*}Therefore, using the definition of the derivative,
\begin{align*} f'(0)&=\lim_{x \to 0}\frac{f(x)-f(0)}{x}\quad\mbox{ if it exists, but}\\ \lim_{x \to 0}\frac{f(x)-f(0)}{x}&=DNE \end{align*}And now, the quotient rule:
\begin{align*} &=\frac{1}{2}\sqrt{\frac{x^2-1}{x^2+1}}\cdot\left(\frac{(x^2-1)(2x)-(x^2+1)2x}{(x^2-1)^2}\right)\\ &=\frac{1}{2}\sqrt{\frac{x^2-1}{x^2+1}}\cdot\left(\frac{-4x}{(x^2-1)^2}\right)\\ &=\sqrt{\frac{x^2-1}{x^2+1}}\cdot\left(\frac{-2x}{(x^2-1)^2}\right)\\ &=\frac{-2x}{({x^2-1})\sqrt{x^4-1}} \end{align*}In order to evaluate \(\diff{}{x}\{{\cos(x^2)}\}\text{,}\) we’ll need the chain rule again.
\begin{align*} &=e^{\cos(x^2)}\cdot [-\sin(\textcolor{orange}{x^2})]\cdot\diff{}{x}\{\textcolor{orange}{x^2}\}\\ &=-e^{\cos(x^2)}\cdot \sin(x^2)\cdot 2x \end{align*}When \(x=2\text{:}\)
\begin{align*} f'(2) &= g'\left(\frac{2}{h(2)}\right)\frac{h(2)-2h'(2)}{h(2)^2}\\ &= 4\frac{2-2\times3}{2^2} =-4 \end{align*}Now, we can use the power rule to differentiate \(\dfrac{1}{x^2}\text{.}\) This will be easier than differentiating \(\dfrac{1}{x^2}\) using quotient rule, but if you prefer, quotient rule will also work.
\begin{align*} f'(x)&=-2x^{-3}+\frac{1}{2}(\textcolor{red}{x^2-1})^{-1/2}\cdot\diff{}{x}\{\textcolor{red}{x^2-1}\}\\ &=-2x^{-3}+\frac{1}{2}({x^2-1})^{-1/2}(2x)\\ &=\frac{-2}{x^3}+\frac{x}{\sqrt{x^2-1}} \end{align*}Here, we need the chain rule again:
\begin{align*} &=\sec(e^{2x+7})\tan(e^{2x+7}) \cdot \left[e^{\textcolor{red}{2x+7}}\cdot \diff{}{x}\{\textcolor{red}{2x+7}\}\right]\\ &=\sec(e^{2x+7})\tan(e^{2x+7}) \cdot \left[e^{2x+7}\cdot2\right]\\ &=2e^{2x+7}\sec(e^{2x+7})\tan(e^{2x+7}) \end{align*}To determine where this function is zero, we factor:
\begin{align*} &=e^{{t^3-7t^2+8t}}\cdot(3t-2)(t-4) \end{align*}We find ourselves once more in need of the chain rule:
\begin{align*} &=\sec^2({e^{x^2}})\cdot{e^{\textcolor{red}{x^2}}}\diff{}{x}\{\textcolor{red}{x^2}\}\\ &=\sec^2(e^{x^2})\cdot e^{x^2}\cdot 2x\\ \end{align*}Finally, we evaluate this derivative at the point \(x=1\text{:}\)
\begin{align*} y'(1)&=\sec^2(e)\cdot e \cdot 2\\ &=2e\sec^2e \end{align*}and the chain rule:
\begin{align*} &=e^{\textcolor{red}{4x}}\cdot\diff{}{x}\{\textcolor{red}{4x}\}\cdot\tan x+e^{4x}\sec^2 x\\ &=4e^{4x}\tan x+e^{4x}\sec^2 x \end{align*}Now, the chain rule:
\begin{align*} &=\frac{(3x^2)(1+e^{3x})-(x^3)(3e^{3x})}{{(1+e^{3x})}^2}\\ \end{align*}So, when \(x=1\text{:}\)
\begin{align*} f'(1)&=\frac{3(1+e^{3})-3e^{3}}{{(1+e^{3})}^2}=\frac{3}{{(1+e^{3})}^2} \end{align*}and find ourselves in need of chain rule a second time:
\begin{align*} &=-\sin\big(x^2+\sqrt{x^2+1}\big)\cdot\left(2x+\dfrac{1}{2\sqrt{\textcolor{red}{x^2+1}}}\cdot\diff{}{x}\left\{\textcolor{red}{x^2+1}\right\}\right)\\ &=-\sin\big(x^2+\sqrt{x^2+1}\big)\cdot\left(2x+\dfrac{2x}{2\sqrt{x^2+1}}\right) \end{align*}Using the product rule,
\begin{align*} y'&=(2x)\cos^2 x + (1+x^2)\diff{}{x}\{\cos^2 x\}\\ \end{align*}Here, we’ll need to use the chain rule. Remember \(\cos^2 x = [\cos x]^2\text{.}\)
\begin{align*} &=2x\cos^2x+(1+x^2) 2\textcolor{red}{\cos x} \cdot \diff{}{x}\{\textcolor{red}{\cos x}\}\\ &=2x\cos^2x+(1+x^2) 2\cos x \cdot (-\sin x)\\ &=2x\cos^2x-2(1+x^2) \sin x\cos x \end{align*}Plugging in \(x=2\text{:}\)
\begin{align*} g'(2)&=3(2^2)h(2^2)+2^3h'(2^2)2\times 2\cr &=12h(4)+32h'(4)=12\times 2-32\times 2\cr &=-40 \end{align*}Here, we need the chain rule:
\begin{align*} &=e^{(1-x^2)/2}+x\cdot e^{\textcolor{red}{(1-x^2)/2}}\diff{}{x}\left\{\textcolor{red}{\frac{1}{2}(1-x^2)}\right\}\\ &=e^{(1-x^2)/2}+x\cdot e^{{(1-x^2)/2}}\cdot (-x)\\ &=(1-x^2)e^{(1-x^2)/2} \end{align*}Now, we use the chain rule. Since \(\cos^3(5x-7)=[\cos(5x-7)]^3\text{,}\) our “outside” function is \(g(x)=x^3\text{,}\) and our “inside” function is \(h(x)=\cos(5x-1)\text{.}\)
\begin{align*} &= \frac{\cos^3(5x-7)e^x-e^x\cdot3\textcolor{red}{\cos}^2\textcolor{red}{(5x-7)} \cdot \diff{}{x}\{\textcolor{red}{\cos(5x-7)}\}}{\cos^6(5x-7)}\\ \end{align*}We need the chain rule again!
\begin{align*} &= \frac{\cos^3(5x\!-\!7)e^x\!-\!e^x\cdot3{\cos}^2{(5x\!-\!7)} \cdot[{-\sin(\textcolor{red}{5x\!-\!7})\cdot \diff{}{x}\{\textcolor{red}{5x\!-\!7}\}}]}{\cos^6(5x-7)}\\ &= \frac{\cos^3(5x-7)e^x-e^x\cdot3{\cos}^2{(5x-7)} \cdot[{-\sin({5x-7})\cdot5}]}{\cos^6(5x-7)}\\ \end{align*}We finish by simplifying:
\begin{align*} &= \frac{e^x\cos^2(5x-7)\left(\cos(5x-7)+15\sin(5x-7)\right)}{\cos^6(5x-7)}\\ &=e^x \frac{\cos(5x-7)+15\sin(5x-7)}{\cos^4(5x-7)}\\ &=e^x(\sec^3(5x-7)+15\tan(5x-7)\sec^3(5x-7))\\ &=e^x\sec^3(5x-7)(1+15\tan(5x-7)) \end{align*}Now we use the product rule to differentiate:
\begin{align*} f'(x)&=e^x\sec^3(5x-7)+e^x\diff{}{x}\{\sec^3(5x-7)\}\\ \end{align*}Here, we’ll need the chain rule. Since \(\sec^3(5x-7)=[\sec (5x-7)]^3\text{,}\) our “outside” function is \(g(x)=x^3\) and our “inside” function is \(h(x)=\sec(5x-7)\text{,}\) so that \(g(h(x))=[\sec(5x-7)]^3=\sec^3(5x-7)\text{.}\)
\begin{align*} &=e^x\sec^3(5x-7)+e^x\cdot3\;\textcolor{red}{\sec}^2\textcolor{red}{(5x-7)} \cdot \diff{}{x}\{\textcolor{red}{\sec(5x-7)}\}\\ \end{align*}We need the chain rule again! Recall \(\diff{}{x}\{\sec x\}=\sec x \tan x\text{.}\)
\begin{align*} &=e^x\sec^3(5x-7)\\ \amp\hskip0.2in+e^x\cdot3\;{\sec}^2{(5x\!-\!7)} \cdot {\sec(\textcolor{orange}{5x\!-\!7})\tan(\textcolor{orange}{5x\!-\!7})\cdot\diff{}{x}\{\textcolor{orange}{5x\!-\!7}\}}\\ &=e^x\sec^3(5x\!-\!7)\\ \amp\hskip0.2in+e^x\cdot3\;{\sec}^2{(5x\!-\!7)} \cdot {\sec({5x\!-\!7})\tan({5x\!-\!7})\cdot 5}\\ \end{align*}We finish by simplifying:
\begin{align*} &=e^x\sec^3(5x-7)(1+15\tan({5x-7})) \end{align*}| \(t\) | \((\sin t,\cos^2 t)\) |
| \(0\) | \((0,1)\) |
| \(\pi/4\) | \((\frac{1}{\sqrt{2}},\frac{1}{2})\) |
| \(\pi/2\) | \((1,0)\) |
| \(3\pi/4\) | \((\frac{1}{\sqrt{2}},\frac{1}{2})\) |
| \(\pi\) | \((0,1)\) |
| \(5\pi/4\) | \((-\frac{1}{\sqrt{2}},\frac{1}{2})\) |
| \(3\pi/2\) | \((-1,0)\) |
| \(7\pi/4\) | \((-\frac{1}{\sqrt{2}},\frac{1}{2})\) |
| \(2\pi\) | \((0,1)\) |
Using \(y(t)=\cos^2 t\) and \(x(t)=\sin t\text{:}\)
\begin{align*} \diff{f}{x}&=\left(-2\cos t \sin t \right)\div\left( \cos t \right)=-2\sin t=-2x\\ \end{align*}So, when \(t=\dfrac{10\pi}{3}\) and \(x=-\dfrac{\sqrt{3}}{2},\)
\begin{align*} \diff{f}{x}\left(\dfrac{-\sqrt{3}}{2}\right)&=-2\cdot\frac{-\sqrt{3}}{2}=\sqrt{3}. \end{align*}The slope of the curve is \(\ds\diff{y}{x}\text{.}\) To find \(\ds\diff{y}{x}\text{,}\) we use the chain rule:\begin{align*} \diff{y}{t}&=\diff{y}{x}\cdot\diff{x}{t}\\ \diff{}{t}\left\{\cos^2 t\right\}&=\diff{y}{x}\cdot\diff{}{t}\{\sin t\}\\ -2\cos t \sin t &= \diff{y}{x} \cdot \cos t\\ \diff{y}{x}&=-2\sin t\\ \end{align*}So, when \(t=\dfrac{10\pi}{3}\text{,}\)
\begin{align*} \diff{y}{x}&=-2\sin\left(\frac{10\pi}{3}\right)=-2\left(-\frac{\sqrt{3}}{2}\right)=\sqrt{3}. \end{align*}
So, for ten speakers:
\begin{align*} V(10P)&=10\log_{10}\left(\frac{10P}{S}\right)=10\log_{10}\left(\frac{P}{S}\right)+10\log_{10}\left(10\right)\\ &=3+10(1)=13 \mathrm{dB}\\ \end{align*}and for one hundred speakers:
\begin{align*} V(100P)&=10\log_{10}\left(\frac{100P}{S}\right)=10\log_{10}\left(\frac{P}{S}\right)+10\log_{10}\left(100\right)\\ &=3+10(2)=23 \mathrm{dB} \end{align*}Since \(\log 10\) is a constant:
\begin{align*} f'(x)&=\frac{1}{x\log 10}. \end{align*}We’ll need the chain rule again:
\begin{align*} &= e^{\cos\left(\log x\right)} (-\sin(\textcolor{orange}{\log x})) \cdot \diff{}{x}\{\textcolor{orange}{ \log x} \}\\ &= e^{\cos\left(\log x\right)} (-\sin(\log x)) \cdot \frac{1}{x}\\ &=\frac{-e^{\cos(\log x)}\sin(\log x)}{x} \end{align*}So, we’ll need the chain rule:
\begin{align*} y'&=\frac{\diff{}{x}\left\{\textcolor{red}{x^2+\sqrt{x^4+1}}\right\}}{\textcolor{red}{x^2+\sqrt{x^4+1}}}\\ &=\frac{2x+\diff{}{x}\left\{\sqrt{x^4+1}\right\}}{x^2+\sqrt{x^4+1}}\\ \end{align*}We need the chain rule again:
\begin{align*} &=\frac{2x+\frac{\diff{}{x}\left\{\textcolor{red}{x^4+1}\right\}}{2\sqrt{\textcolor{red}{x^4+1}}}}{x^2+\sqrt{x^4+1}}\\ &=\frac{2x+\frac{4x^3}{2\sqrt{x^4+1}}}{x^2+\sqrt{x^4+1}}. \end{align*}Now, we differentiate using the chain rule:
\begin{align*} f'(x)&=\frac{1}{2}\left[ 3\frac{2x}{x^2+5}-\frac{4x^3}{x^4+10} \right]\\ &=\frac{3x}{x^2+5}-\frac{2x^3}{x^4+10} \end{align*}In particular, when \(x=2\text{:}\)
\begin{align*} f'(2) &= \dfrac{1}{g\big(2h(2)\big)}\cdot g'\big(2h(2)\big)\cdot \big[h(2)+2h'(2)\big]\\ &= \dfrac{g'(4)}{g(4)}\big[2+2\times 3\big] = \dfrac{5}{3}\big[2+2\times 3\big]\\ &=\dfrac{40}{3} \end{align*}Now, we can use the product rule to differentiate the right side, and the chain rule to differentiate \(\log(f(x))\text{:}\)
\begin{align*} g'(x)=\frac{f'(x)}{f(x)}&=\log x +x\frac{1}{x}=\log x +1\\ \end{align*}Finally, we solve for \(f'(x)\text{:}\)
\begin{align*} f'(x)&=f(x)(\log x + 1) = x^x(\log x + 1) \end{align*}Now that we’ve simplified, we can efficiently differentiate both sides. It is important to remember that we aren’t differentiating \(f(x)\) directly--we’re differentiating \(\log(f(x))\text{.}\)
\begin{align*} \frac{f'(x)}{f(x)}&=\frac{1}{4}\left(\frac{4x^3}{x^4+12}+\frac{4x^3-2x}{x^4-x^2+2}-\frac{3}{x}\right) \end{align*}Now we can use logarithm rules to change \(g(x)\) into a form that is friendlier to differentiate:
\begin{align*} &=\log(x+1)+\log(x^2+1)^2+\log(x^3+1)^3+\log(x^4+1)^4\\ \amp\hskip2in+\log(x^5+1)^5\\ &=\log(x+1)+2\log(x^2+1)+3\log(x^3+1)+4\log(x^4+1)\\ \amp\hskip2in+5\log(x^5+1)\\ \end{align*}Now, we differentiate \(g(x)\) using the chain rule:
\begin{align*} g'(x)=\frac{f'(x)}{f(x)}&=\frac{1}{x+1}+\frac{4x}{x^2+1} +\frac{9x^2}{x^3+1}+\frac{16x^3}{x^4+1}+\frac{25x^4}{x^5+1} \end{align*}Logarithm rules allowed us to simplify. Now, we differentiate both sides of this equation:
\begin{align*} \frac{f'(x)}{f(x)}&=(\cos x ) \log(\cos x)+ \sin x \cdot \frac{-\sin x}{\cos x}\\ &=(\cos x) \log (\cos x) - \sin x \tan x\\ \end{align*}Finally, we solve for \(f'(x)\text{:}\)
\begin{align*} f'(x)&=f(x)\left[(\cos x) \log (\cos x) - \sin x \tan x\right]\\ &= (\cos x)^{\sin x}\left[(\cos x) \log (\cos x) - \sin x \tan x\right] \end{align*}Now, we can differentiate:
\begin{align*} \frac{\diff{}{x}\left\{(\tan x)^x\right\}}{(\tan x)^x}&=\log(\tan x) + x\cdot\frac{\sec^2 x}{\tan x}\\ &=\log(\tan x) + \frac{x}{\sin x \cos x}\\ \end{align*}Finally, we solve for the derivative we want, \(\ds\diff{}{x}\{(\tan x)^x\}\text{:}\)
\begin{align*} {\diff{}{x}\left\{(\tan x)^x\right\}}&={(\tan x)^x}\left(\log(\tan x) + \frac{x}{\sin x \cos x}\right) \end{align*}We differentiate both sides to obtain:
\begin{align*} \dfrac{f'(x)}{f(x)} &= \diff{}{x} \left\{ \log(x^2+1) \cdot (x^2+1) \right\}\\ &= \frac{2x}{x^2+1}(x^2+1)+2x\log(x^2+1)\\ &=2x(1+\log(x^2+1))\\ \end{align*}Now, we solve for \(f'(x)\text{:}\)
\begin{align*} f'(x) &= f(x) \cdot 2x(1+\log(x^2+1))\\ &= (x^2+1)^{x^2+1} \cdot 2x(1+\log(x^2+1)) \end{align*}We differentiate using the product and chain rules:
\begin{align*} \dfrac{f'(x)}{f(x)} &= \diff{}{x} \left\{ \log(x^2+1) \cdot \sin x \right\} = \cos x \cdot \log(x^2+1) + \frac{2x\sin x}{x^2+1}\\ \end{align*}Finally, we solve for \(f'(x)\)
\begin{align*} f'(x) &= f(x) \cdot \left( \cos x \cdot \log(x^2+1) + \frac{2x\sin x}{x^2+1} \right)\\ &= (x^2+1)^{\sin(x)} \cdot \left( \cos x \cdot \log(x^2+1) + \frac{2x\sin x}{x^2+1} \right) \end{align*}Differentiating, we find:
\begin{align*} \dfrac{f'(x)}{f(x)} &= \diff{}{x} \left\{ \log(x) \cdot \cos^3(x) \right\}\\ \amp= 3\cos^2(x)\cdot (-\sin(x)) \cdot \log(x) + \frac{\cos^3(x)}{x}\\ \end{align*}Finally, we solve for \(f'(x)\text{:}\)
\begin{align*} f'(x) &= f(x) \cdot \left( -3\cos^2(x)\sin(x) \log(x) + \frac{\cos^3(x)}{x} \right)\\ &= x^{\cos^3(x)} \cdot \left( -3\cos^2(x)\sin(x) \log(x) + \frac{\cos^3(x)}{x} \right) \end{align*}We differentiate:
\begin{align*} \frac{f'(x)}{f(x)} &= \diff{}{x} \left\{(x^2-3)\cdot \log(3+\sin(x)) \right\}\\ &=2x\log(3+\sin(x)) + (x^2-3)\frac{\cos(x)}{3+\sin(x)}\\ \end{align*}Finally, we solve for \(f'(x)\text{:}\)
\begin{align*} f'(x)&= f(x)\cdot \left[ 2x\log(3+\sin(x)) + \frac{(x^2-3)\cos(x)}{3+\sin(x)}\right]\\ &= (3+\sin(x))^{x^2-3}\cdot \left[ 2x\log(3+\sin(x)) + \frac{(x^2-3)\cos(x)}{3+\sin(x)}\right] \end{align*}Now, we differentiate. On the left side we use the chain rule, and on the right side we use product and chain rules.
\begin{align*} \ds\diff{}{x}\left\{\log\left([f(x)]^{g(x)}\right)\right\}&=\ds\diff{}{x}\left\{g(x)\log(f(x))\right\}\\ \frac{\diff{}{x}\{[f(x)]^{g(x)}\}}{[f(x)]^{g(x)}}&= g'(x)\log(f(x))+g(x)\cdot\frac{f'(x)}{f(x)}\\ \end{align*}Finally, we solve for the derivative of our original function.
\begin{align*} {\diff{}{x}\{[f(x)]^{g(x)}\}}&={[f(x)]^{g(x)}}\left( g'(x)\log(f(x))+g(x)\cdot\frac{f'(x)}{f(x)}\right) \end{align*}
and solve for \(\ds\diff{y}{x}\)
\begin{align*} \diff{y}{x}&=-\frac{x}{y} \end{align*}Now, we solve for \(\ds\diff{y}{x}\text{.}\)
\begin{align*} x\diff{y}{x}+e^y\diff{y}{x}&=-(e^x+y)\\ (x+e^y)\diff{y}{x}&=-(e^x+y)\\ \diff{y}{x}&=-\frac{e^x+y}{e^y+x} \end{align*}Now, get the derivative on one side and solve
\begin{align*} e^y\diff{y}{x}-2xy\diff{y}{x}&=y^2+1\\ \diff{y}{x}\left(e^y-2xy\right)&=y^2+1\\ \diff{y}{x}&=\frac{y^2+1}{e^y-2xy} \end{align*}We could solve for \(\ds\diff{y}{x}\) at this point, but it’s not necessary. We want to know when \(\ds\diff{y}{x}\) is equal to one:
\begin{align*} 2x+6y(1)&=0\\ x&=-3y\\ \end{align*}That is, \(\ds\diff{y}{x}=1\) at those points along the ellipse where \(x=-3y\text{.}\) We plug this into the equation of the ellipse to find the coordinates of these points.
\begin{align*} \left(-3y\right)^2+3y^2&=1\\ 12y^2&=1\\ y=\pm\frac{1}{\sqrt{12}}=\pm\frac{1}{2\sqrt{3}} \end{align*}Now, we plug in \(x=1\text{,}\) \(y=4\text{,}\) and solve for \(\ds\diff{y}{x}\text{:}\)
\begin{align*} \frac{4+\diff{y}{x}}{4}&=8+\diff{y}{x}\\ \diff{y}{x}&=-\frac{28}{3} \end{align*}Then we gather the terms containing \(y'\) on one side, so we can solve for \(y'\text{:}\)
\begin{align*} 2x^2yy'&+xy'\cos y=-2xy^2-\sin y\\ y'(2x^2y&+x\cos y)=-2xy^2-\sin y\\ y'&=-\frac{2xy^2+\sin y}{2x^2y+x\cos y} \end{align*}For the circle, we differentiate implicitly
\begin{align*} 2x+2y\diff{y}{x}&=0\\ \end{align*}and solve for \(\ds\diff{y}{x}\)
\begin{align*} \diff{y}{x}&=-\frac{x}{y}\\ \end{align*}For the ellipse, we also differentiate implicitly:
\begin{align*} 2x+6y\diff{y}{x}&=0\\ \end{align*}and solve for \(\ds\diff{y}{x}\)
\begin{align*} \diff{y}{x}&=-\frac{x}{3y}\\ \end{align*}What we want is a value of \(x\) where both derivatives are equal. However, they might have different values of \(y\text{,}\) so let’s let \(y_1\) be the \(y\)-values associated with \(x\) on the circle, and let \(y_2\) be the \(y\)-values associated with \(x\) on the ellipse. That is, \(x^2+ y_1^2=1\) and \(x^2+3y_2^2=1\text{.}\) For the slopes at \((x,y_1)\) on the circle and \((x,y_2)\) on the ellipse to be equal, we need:
\begin{align*} -\frac{x}{y_1}&=-\frac{x}{3y_2}\\ x\left(\frac{1}{y_1}-\frac{1}{3y_2}\right)&=0\\ \end{align*}So \(x=0\) or \(y_1=3y_2\text{.}\) Let’s think about which \(x\)-values will have a \(y\)-coordinate of the circle be three times as large as a \(y\)-coordinate of the ellipse. If \(y_1=3y_2\text{,}\) \((x,y_1)\) is on the circle, and \((x,y_2)\) is on the ellipse, then \(x^2+y_1^2 =x^2+(3y_2)^2=1\) and \(x^2+3y_2^2=1\text{.}\) In this case:
\begin{align*} x^2+9y_2^2&=x^2+3y_2^2\\ 9y_2^2&=3y_2^2\\ y_2&=0\\ x&=\pm 1 \end{align*}
where \(x\) and \(\theta\) are the same in both expressions, and \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\text{.}\) Then
\begin{align*} \arcsin x &=\theta & \arccos x &= \frac{\pi}{2}-\theta \end{align*}
Now, we can differentiate with respect to \(y\) using the chain rule.
\begin{align*} \diff{}{y}\left\{f(g(y))\right\}&=\diff{}{y}\{y\}\\ f'(g(y))\cdot g'(y)&=1\\ g'(y)&=\frac{1}{f'(g(y))}=\frac{1}{1-\sin g(y)} \end{align*}What we want is \(g'(\pi-1)\text{,}\) so we need to figure out which value of \(x\) gives \(f(x)=\pi-1\text{.}\) A little trial and error leads us to \(x=\frac{\pi}{2}\text{.}\)
\begin{align*} g'(\pi-1)\cdot f'\left(\frac{\pi}{2}\right)&=1\\ \end{align*}Since \(f'(x)=2-\cos(x)\text{,}\) \(f'\left(\frac{\pi}{2}\right)=2-0=2\text{:}\)
\begin{align*} g'(\pi-1)\cdot 2&=1\\ g'(\pi-1)=\frac{1}{2} \end{align*}What we want is \(g'(e+1)\text{,}\) so we need to figure out which value of \(x\) gives \(f(x)=e+1\text{.}\) A little trial and error leads us to \(x=1\text{.}\)
\begin{align*} g'(f(1))f'(1)&=1\\ g'(e+1)\cdot f'(1)&=1\\ g'(e+1) &= \frac{1}{f'(1)}\\ \end{align*}It remains only to note that \(f'(x)=e^x+1\text{,}\) so \(f'(1)=e+1\)
\begin{align*} g'(e+1)&=\frac{1}{e+1} \end{align*}and so
\begin{align*} 7(3x+7)&=5x-9\\ x&=-\frac{29}{8}\\ \end{align*}So, when \(x=-\dfrac{29}{8}\text{,}\) our equation \(f(2x+1)=\dfrac{5x-9}{3x+7}\) becomes:
\begin{align*} f\left(2\cdot\frac{-29}{8}+1\right)&=\dfrac{5\cdot\frac{-29}{8}-9}{3\cdot\frac{-29}{8}+7}\\ \end{align*}Or, equivalently:
\begin{align*} f\left(-\frac{25}{4}\right)&=7 \end{align*}Now, the equation \(f^{-1}(4x-1)=\dfrac{2x+3}{x+1}\) with \(x=\dfrac{-3}{2}\) tells us:
\begin{align*} f^{-1}\left(4\cdot\frac{-3}{2}-1\right)&=\frac{2\cdot\frac{-3}{2}+3}{\frac{-3}{2}+1}\\ \end{align*}Or, equivalently:
\begin{align*} f^{-1}(-7)&=0 \end{align*}Now, we differentiate implicitly.
\begin{align*} 1+2y'&=\cos(x^2+y^2)\cdot(2x+2yy')\\ 1+2y'&=2x\cos(x^2+y^2)+2yy'\cos(x^2+y^2)\\ 2y'-2yy'\cos(x^2+y^2)&=2x\cos(x^2+y^2)-1\\ y'\left(2-2y\cos(x^2+y^2)\right)&=2x\cos(x^2+y^2)-1\\ y'&=\frac{2x\cos(x^2+y^2)-1}{2-2y\cos(x^2+y^2)} \end{align*}
The problem asks us to find this value of \(c\text{.}\) Solving:
\begin{align*} e^c&=\frac{e^T-e^0}{T}\\ e^c&=\frac{e^T-1}{T}\\ c&=\log\left(\frac{e^T-1}{T}\right) \end{align*}
Plugging in \(x=1\text{,}\) \(y=3\text{:}\)
\begin{align*} -28+6+2y'+6y'&=0\\ y'&=\frac{11}{4} \quad\mbox{\textcolor{red}{ at the point $(1,3)$}}\\ \end{align*}Differentiating the equation \(-28x+2y+2xy'+2yy'=0\):
\begin{align*} -28+2y'+2y'+2xy''+2y'y'+2yy''&=0\\ 4y'+2(y')^2+2xy''+2yy''&=28\\ \end{align*}At the point \((1,3)\text{,}\) \(y'=\dfrac{11}{4}\text{.}\) Plugging in:
\begin{align*} 4\left(\frac{11}{4}\right)+2\left(\frac{11}{4}\right)^2+2(1)y''+2(3)y''&=28\\ y''&=\frac{15}{64} \end{align*}So, we need an expression for \(y'\text{.}\) We use the equation \(2x+2yy'=0\) to conclude \(y'=-\dfrac{x}{y}\text{:}\)
\begin{align*} y''&=-\frac{\left(-\frac{x}{y}\right)^2+1}{y}\\ &=-\frac{\frac{x^2}{y^2}+1}{y}\\ &=-\frac{x^2+y^2}{y^3}\\ &=-\frac{1}{y^3} \end{align*}Using the quotient rule:
\begin{align*} \ddiff{2}{}{x}\{\log(5x^2-12)\}&=\diff{}{x}\left\{\frac{10x}{5x^2-12}\right\}\\ &=\frac{(5x^2-12)(10)-10x(10x)}{(5x^2-12)^2}\\ &=\frac{-10(5x^2+12)}{(5x^2-12)^2} \end{align*}We differentiate implicitly. For ease of notation, we write \(y'\) for \(\ds\diff{y}{x}\text{.}\)
\begin{align*} 2x+1+y'&=\cos(xy)(y+xy')\\ \end{align*}We’re interested in \(y''\text{,}\) so we implicitly differentiate again.
\begin{align*} 2+y''&=-\sin(xy)(y+xy')^2+\cos(xy)(2y'+xy'')\\ \end{align*}We want to know what \(y''\) is when \(x=y=0\text{.}\) Plugging these in yields the following:
\begin{align*} 2+y''&=2y'\\ \end{align*}So, we need to know what \(y'\) is when \(x=y=0\text{.}\) We can get this from the equation \(2x+1+y'=\cos(xy)(y+xy')\text{,}\) which becomes \(1+y'=0\) when \(x=y=0\text{.}\) So, at the origin, \(y'=-1\text{,}\) and
\begin{align*} 2+y''&=2(-1)\\ y''&=-4 \end{align*}The fourth derivative is \(\sin x\) is \(\sin x\text{,}\) and the fourth derivative of \(\cos x\) is \(\cos x\text{,}\) so (a) and (b) are true.
\begin{align*} \diff{}{x}\tan x &=\sec^2 x\\ \diff{}{x}\sec^2 x &=2\sec x (\sec x \tan x)=2\sec^2x\tan x\\ \diff{}{x}\{2\sec^2x\tan x\}&=(4\sec x \cdot \sec x \tan x)\tan x+2\sec^2x\sec^2x\\ &=4\sec^2x\tan^2x+2\sec^4x \end{align*}Every time we differentiate, we multiply the original function by another factor of \(\log 2\text{.}\) So, the \(n\)th derivative is given by:
\begin{align*} \ddiff{n}{}{x}\{2^x\}&=2^x(\log2)^n \end{align*}Subbing in \(x=1\) and \(y(1)=2\) gives
\begin{align*} (3)(1)( 2)+(1)y'(1)+(3)(4) y'(1)&=10\\ 13y'(1)&=4\\ y'(1)&=\frac{4}{13} \end{align*}
Since \(f\) looks different to the left and right of 0, in order to evaluate this limit, we look at the corresponding one-sided limits. Note that when \(h\) approaches 0 from the right, \(h \gt 0\) so \(f(h)=h^2\text{.}\) By contrast, when \(h\) approaches 0 from the left, \(h \lt 0\) so \(f(h)=-h^2\text{.}\)
\begin{align*} &\;\; \lim_{h \to 0^+} \frac{f(h)}{h}=\lim_{h \to 0^+}\frac{h^2}{h}=\lim_{h \to 0^+}h=0\\ &\;\; \lim_{h \to 0^-} \frac{f(h)}{h}=\lim_{h \to 0^-}\frac{-h^2}{h}=\lim_{h \to 0^-}-h=0\\ \end{align*}Since both one-sided limits exist and are equal to 0,
\begin{align*} &\;\; \lim_{h \to 0} \frac{f(0+h)-f(0)}{h}=0 \end{align*}exists, then \(f''(0)\) exists.
\begin{align*} \lim_{h \to 0}\frac{f'(0+h)-f'(0)}{h}&=\lim_{h \to 0} \frac{f'(h)-0}{h}=\lim_{h \to 0}\frac{f'(h)}{h}\\ \end{align*}Since \(f(h)\) behaves differently when \(h\) is greater than or less than zero, we look at the one-sided limits.
\begin{align*} \lim_{h \to 0^+}\frac{f'(h)}{h}&=\lim_{h \to 0^+}\frac{2h}{h}=2\\ \lim_{h \to 0^-}\frac{f'(h)}{h}&=\lim_{h \to 0^-}\frac{-2h}{h}=-2\\ \end{align*}Since the one-sided limits do not agree,
\begin{align*} \lim_{h \to 0}\frac{f'(0+h)-f'(0)}{h}&=DNE \end{align*}The velocity of the pot at this time is
\begin{align*} h'\left(\sqrt{\frac{20}{9.8}}\right)&=-9.8\left(\sqrt{\frac{20}{9.8}}\right)=-\sqrt{20\cdot 9.8} =-14 \frac{\mathrm{m}}{\mathrm{s}} \end{align*}Multiplying both sides by \(d\text{:}\)
\begin{align*} \frac{d}{10}&=120^2-\frac{120^2}{2}\\ d&=5\cdot 120^2 \end{align*}https://www.guinnessworldrecords.com/world-records/fastest-baseball-pitch-(male)
We differentiate implicitly with respect to a third variable, \(t\text{:}\)
\begin{align*} \diff{P}{t}&=3Q^2\cdot\diff{Q}{t} \end{align*}
Differentiating with respect to \(t\text{,}\)
\begin{align*} 2x(t)x'(t)+2y(t)y'(t)&=2z(t)z'(t)\\ x(t)x'(t)+y(t)y'(t)&=z(t)z'(t)\\ \end{align*}At the specified time, \(x(t)\) is decreasing, so \(x'(t)\) is negative, and \(y(t)\) is increasing, so \(y'(t)\) is positive.
\begin{align*} (300)(-15)+(400)(20) &= \sqrt{300^2+400^2}z'(t)\\ 500 z'(t)&=3500\\ z'(t)&=7 \mbox{ mph} \end{align*}
so
\begin{align*} V&=100h(60+w)\\ &=100h(120+\frac{4}{5}h)\\ &=80h^2+12000h\\ \end{align*}This is the equation we need, relating \(V\) and \(h\text{.}\) Differentiating implicitly with respect to \(t\text{:}\)
\begin{align*} \diff{V}{t} &=2\cdot80h\cdot\diff{h}{t}+12000\diff{h}{t}\\ &=\left(160h+12000\right)\diff{h}{t}\\ \end{align*}We are given that \(h=25\) and \(\ds\diff{V}{t}=3\) litres per minute. Converting to cubic centimetres, \(\ds\diff{V}{t} = -3000\) cubic centimetres per minute. So:
\begin{align*} -3000&=\left(160\cdot25+12000\right)\diff{h}{t}\\ \diff{h}{t}&=-\frac{3}{16}=-.1875\;\frac{\mathrm{cm}}{\mathrm{min}} \end{align*}
So, the area of the triangle on the left is
\begin{align*} \frac{1}{2}ah&=\frac{2}{5}h^2 \end{align*}The little triangle (of base \(b\) and height \(1.25-h\)) is formed by the air on the right side of the tank. It is a similar triangle to the triangle of base 3 and height 1.25, so
\begin{align*} \frac{b}{1.25-h}&=\frac{3}{1.25}\\ b&=\frac{3}{1.25}(1.25-h)\\ \end{align*}So, the area of the trapezoid on the right is
\begin{align*} &\frac{1}{2}(3)(1.25)-\frac{1}{2}\left(\frac{3}{1.25}\right)\left(1.25-h\right)(1.25-h)\\ &=3h-\frac{6}{5}h^2 \end{align*}So, the volume of water is
\begin{align*} V&=5\left(6h-\frac{4}{5}h^2\right)=30h-4h^2\\ \end{align*}Differentiating with respect to time, \(t\text{:}\)
\begin{align*} \diff{V}{t}&=30\diff{h}{t}-8h\diff{h}{t}\\ \end{align*}When \(h=\dfrac{1}{10}\) metre, and \(\ds\diff{V}{t}=\frac{1}{10^3}\) cubic metres per second,
\begin{align*} \frac{1}{10^3}&=30\diff{h}{t}-8\left(\frac{1}{10}\right)\diff{h}{t}\\ \diff{h}{t}&=\frac{1}{29200} \mbox{ metres per second} \end{align*}
Differentiating with respect to \(t\text{:}\)
\begin{align*} \sec^2\theta \cdot \diff{\theta}{t}&=\frac{1}{2}\diff{h}{t}\\ \diff{\theta}{t}&=\frac{1}{2}\cos^2\theta\cdot\diff{h}{t} \end{align*}
So, the quantities we need to know one minute after liftoff (that is, when \(t = \dfrac{1}{60}\)) are \(h\left(\dfrac{1}{60}\right)\) and \(\ds\diff{h}{t}\left(\frac{1}{60}\right)\text{.}\) Recall \(h(t)=61750t^2\text{.}\)
\begin{align*} h\left(\frac{1}{60}\right)&=\frac{61750}{3600}=\frac{1235}{72}\\ \diff{h}{t}&=2(61750)t\\ \diff{h}{t}\left(\frac{1}{60}\right)&=\frac{2(61750)}{60}=\frac{6175}{3}\\ \end{align*}Returning to the equation \(\ds\diff{\theta}{t}=\left(\dfrac{2}{h^2+4}\right)\ds\diff{h}{t} \text{:}\)
\begin{align*} \diff{\theta}{t}\left(\frac{1}{60}\right)&=\left(\frac{2}{\left(\frac{1235}{72}\right)^2+4}\right)\left(\frac{6175}{3} \right)\approx 13.8\;\frac{\mathrm{rad}}{\mathrm{hour}}\approx 0.0038\;\frac{\mathrm{rad}}{\mathrm{sec}} \end{align*}
Differentiating with respect to \(t\text{:}\)
\begin{align*} \theta'(t)\cos\left(\theta(t)\right) &=\frac{x'(t)z(t)-x(t)z'(t)}{z(t)^2}\\ \theta'(t)&=\frac{x'(t)z(t)-x(t)z'(t)}{z(t)^2\cos\left(\theta(t)\right)}\\ \end{align*}From our diagram, we see \(\cos\left(\theta(t)\right)=\dfrac{0.5}{z(t)}\text{,}\) so:
\begin{align*} &=2\frac{x'(t)z(t)-x(t)z'(t)}{z(t)}\\ \end{align*}Substituting in \(x'(t)=2\text{,}\) \(z(t)=1.3\text{,}\) \(x(t)=1.2\text{,}\) and \(z'(t)=\dfrac{2\times1.2}{1.3}\text{:}\)
\begin{align*} \theta'(t) &=2\frac{2\times 1.3-1.2\times\frac{2\times 1.2}{1.3}}{1.3} \approx .592 \mbox{ radians/min} \end{align*}
Differentiating with respect to time \(t\text{,}\)
\begin{align*} 2D\diff{D}{t}&=100\sin\theta\cdot\diff{\theta}{t} \end{align*}Plugging in the given data,
\begin{align*} A'&=2\pi\big(3\cdot 2-1\cdot 7\big)=-2\pi \end{align*}Note, \(b\) will change with time as well as \(h\text{.}\) So, differentiating with respect to time, \(t\text{:}\)
\begin{align*} \diff{A}{t}&=\frac{1}{2}\left(\diff{b}{t}\cdot h+b\cdot\diff{h}{t}\right) \end{align*}
Differentiating with respect to \(t\text{:}\)
\begin{align*} \diff{b}{t}&=\frac{-2h\diff{h}{t}}{2\sqrt{150^2-h^2}}+ \frac{-2h\diff{h}{t}}{2\sqrt{200^2-h^2}}\\ &=\frac{-h\diff{h}{t}}{\sqrt{150^2-h^2}}+ \frac{-h\diff{h}{t}}{\sqrt{200^2-h^2}}\\ \end{align*}Using \(\ds\diff{h}{t}=-3\) centimetres per minute:
\begin{align*} \diff{b}{t}&=\frac{3h}{\sqrt{150^2-h^2}}+ \frac{3h}{\sqrt{200^2-h^2}}\\ \end{align*}When \(h=120\text{,}\) \(\sqrt{150^2-h^2}=90\) and \(\sqrt{200^2-h^2}=160\text{.}\) So, at this moment in time:
\begin{align*} b&=90+160=250\\ \diff{b}{t}&=\frac{3(120)}{90}+\frac{3(120)}{160} = 4+\frac{9}{4}=\frac{25}{4}\\ \end{align*}We return to our equation relating the derivatives of \(A\text{,}\) \(b\text{,}\) and \(h\text{.}\)
\begin{align*} \diff{A}{t}&=\frac{1}{2}\left(\diff{b}{t}\cdot h + b \cdot \diff{h}{t}\right)\\ \end{align*}When \(h=120\) cm, \(b=250\text{,}\) \(\ds\diff{h}{t}=-3\text{,}\) and \(\ds\diff{b}{t}=\dfrac{25}{4}\text{:}\)
\begin{align*} \diff{A}{t}&=\frac{1}{2}\left(\frac{25}{4}(120)+250(-3)\right)\\ &=0 \end{align*}
This is the formula we wanted, relating \(A\) and \(\theta\text{.}\) Differentiating with respect to \(t\text{,}\)
\begin{align*} \diff{A}{t}&=\frac{35}{2}\diff{\theta}{t}=\frac{35}{2}\left(-\frac{\pi}{6}\right)=-\frac{35\pi}{12}\\ \end{align*}Since \(\ds\diff{S}{t}=\frac{1}{5}\ds\diff{A}{t}\text{,}\)
\begin{align*} \diff{S}{t}&=-\frac{1}{5}\frac{35\pi}{12}=-\frac{7\pi}{12}\approx -1.8\;\frac{\mathrm{cm}^3}{\mathrm{sec}^2} \end{align*}
We want to differentiate with respect to \(t\text{.}\) Using the chain rule:
\begin{align*} \diff{A}{t}&=\diff{A}{h}\cdot\diff{h}{t}\\ \diff{A}{t}&=\left((1-h)\frac{2-2h}{2\sqrt{2h-h^2}}+(-1)\sqrt{2h-h^2} +\frac{-1}{\sqrt{1-(1-h)^2}}\right)\diff{h}{t}\\ &=\left(\frac{(1-h)^2}{\sqrt{2h-h^2}}-\sqrt{2h-h^2}-\frac{1}{\sqrt{2h-h^2}}\right)\diff{h}{t}\\ &=\left(\frac{(1-h)^2-1}{\sqrt{2h-h^2}}-\sqrt{2h-h^2}\right)\diff{h}{t}\\ &=\left(\frac{-(2h-h^2)}{\sqrt{2h-h^2}}-\sqrt{2h-h^2}\right)\diff{h}{t}\\ &=\left(-\sqrt{2h-h^2}-\sqrt{2h-h^2}\right)\diff{h}{t}\\ &=-2\sqrt{2h-h^2}\diff{h}{t}\\ \end{align*}We note here that the negative sign makes sense: as the door lowers, \(h\) increases and \(A\) decreases, so \(\ds\diff{h}{t}\) and \(\ds\diff{A}{t}\) should have opposite signs.
\begin{align*} \\ \end{align*}When \(h=\dfrac{1}{4}\) metres, and \(\ds\diff{h}{t}=\frac{1}{100}\) metres per second:
\begin{align*} \diff{A}{t}&=-2\sqrt{\frac{2}{4}-\frac{1}{4^2}}\left(\frac{1}{100}\right)=-\frac{\sqrt{7}}{200}\;\frac{\mathrm{cm}^2}{\mathrm{s}}\\ \end{align*}Since \(\ds\diff{F}{t}=\frac{1}{5}\ds\diff{A}{t}\text{:}\)
\begin{align*} \diff{F}{t}&=-\frac{\sqrt{7}}{1000}\;\frac{\mathrm{m}^3}{\mathrm{sec^2}} \end{align*}
Differentiating with respect to \(t\text{:}\)
\begin{align*} \diff{V}{t}&=\frac{\pi}{3}h^2\diff{h}{t}\\ \end{align*}When \(h=7\) cm and \(\ds\diff{V}{t}=-5\) mL per minute,
\begin{align*} -5&=\frac{\pi}{3}(49)\diff{h}{t}\\ \diff{h}{t}&=\frac{-15}{49\pi}\approx -0.097 \mbox{ cm per minute} \end{align*}
Using the chain rule, we differentiate both sides with respect to time, \(t\text{.}\)
\begin{align*} \diff{D}{t}&=2\cos\theta\cdot\diff{\theta}{t}\\ \end{align*}So, if \(\ds\diff{\theta}{t}=0.25\) radians per hour and \(\theta = \dfrac{\pi}{4}\) radians, then
\begin{align*} (a)\qquad\diff{D}{t}&=2\cos\left(\frac{\pi}{4}\right)\cdot 0.25=2\left(\frac{1}{\sqrt{2}}\right)\frac{1}{4}=\frac{1}{2\sqrt{2}}\;\mbox{metres per hour}. \end{align*}
So, since \(\tan(-\theta)=-\tan(\theta)\text{:}\)
\begin{align*} \tan\theta&=\frac{x}{2} \end{align*}When the point is at \((0,-2)\text{,}\) since the observer is turning at one radian per second, also \(\ds\diff{\theta}{t}=1\text{.}\) Also, looking at the diagram, \(\theta=0\text{.}\) Plugging in these values:
\begin{align*} \sec^2\left(0\right)\cdot(1)&=\frac{1}{2}\cdot\diff{x}{t}\\ 1&=\frac{1}{2}\cdot\diff{x}{t}\\ \diff{x}{t}&=2 \end{align*}Differentiating with respect to \(t\text{:}\)
\begin{align*} 2D\cdot\diff{D}{t}&=2(x-2)\cdot\diff{x}{t}+2(y-1)\cdot\diff{y}{t}\\ \end{align*}We plug in \(x=0\text{,}\) \(y=2\text{,}\) \(\diff{x}{t}=1\text{,}\) \(\diff{y}{t}=-\frac{1}{2}\text{,}\) and \(D=-\sqrt{(0-2)^2+(2-1)^2}=-\sqrt{5}\) (negative because the point is to the left of \((2,1)\)):
\begin{align*} -2\sqrt{5}\cdot\diff{D}{t}&=2(-2)(1)+2(1)\left(-\frac{1}{2}\right)\\ \diff{D}{t}&=\frac{\sqrt{5}}{2}\;\mbox{units per second} \end{align*}where we use the fact that \(b\) is a positive number, so \(\sqrt{b^2}=|b|=b\text{.}\)
\begin{align*} \\ \end{align*}Since \(a=2b\text{,}\)
\begin{align*} \color{red}a&\color{red}{=\frac{20}{9-\sqrt{35}}}\\ \end{align*}Now we know \(a\) and \(b\) at the moment when \(a=2b\text{.}\) We still need to know \(\ds\diff{a}{t}\) and \(\ds\diff{b}{t}\) at that moment. Since \(a=5+t\text{,}\) always \(\textcolor{red}{\ds\diff{a}{t}=1}\text{.}\) The equation from (a) relates \(a\) and \(b\text{,}\) so differentiating both sides with respect to \(t\) will give us an equation relating \(\ds\diff{a}{t}\) and \(\ds\diff{b}{t}\text{.}\) When differentiating the portion with a square root, be careful not to forget the chain rule.
\begin{align*} 0&=3\left(\diff{a}{t}+\diff{b}{t}\right)-\frac{\left(\diff{a}{t}+3\diff{b}{t}\right)(3a+b)+(a+3b)\left(3\diff{a}{t}+\diff{b}{t}\right)}{2\sqrt{(a+3b)(3a+b)}}\\ \end{align*}Since \(\ds\diff{a}{t}=1\text{:}\)
\begin{align*} 0&=3\left(1+\diff{b}{t}\right)-\frac{\left(1+3\diff{b}{t}\right)(3a+b)+(a+3b)\left(3+\diff{b}{t}\right)}{2\sqrt{(a+3b)(3a+b)}} \end{align*}This will probably be useful information. Since we’re also given the value of a derivative, let’s differentiate the equation relating the variables implicitly with respect to \(t\text{.}\) For ease of notation, we will write \(\ds\diff{A}{t}=A'\text{,}\) etc.
\begin{align*} A'B+AB'&=\frac{2CC'+2DD'}{C^2+D^2+1}\\ \end{align*}At \(t=10\text{,}\) \(A=C=D=0\text{:}\)
\begin{align*} A'B+0&=\frac{0+0}{0+0+1}\\ A'B&=0\\ \end{align*}at \(t=10\text{,}\) \(A'=2\) units per second:
\begin{align*} 2B&=0\\ B&=0. \end{align*}That is,
\begin{align*} Ce^{-kt}&=0 \end{align*}So:
\begin{align*} Q(t)&=5\left(e^{-k}\right)^t\\ &=5\cdot2^{-\tfrac{t}{5730}}\\ \end{align*}Now, we can evaluate:
\begin{align*} Q(10000)&=5\cdot 2^{-\tfrac{10000}{5730}}\approx 1.5\;\mu g \end{align*}for some positive constant \(k\text{.}\) When \(t=100\text{,}\) the amount of Radium-226 left is 0.9576 grams, so
\begin{align*} 0.9576=Q(100)&=e^{-k\cdot 100}=\left(e^{-k}\right)^{100}\\ e^{-k}&=0.9576^{\tfrac{1}{100}}\\ \end{align*}This tells us
\begin{align*} Q(t)&=0.9576^{\tfrac{t}{100}}\\ \end{align*}So, if half the original amount of Radium-226 is left,
\begin{align*} \frac{1}{2}&=0.9576^{\tfrac{t}{100}}\\ \log\left(\frac{1}{2}\right)&=\log\left(0.9576^{\tfrac{t}{100}}\right)\\ -\log 2&=\frac{t}{100}\log(0.9576)\\ t&=-100\frac{\log 2}{\log 0.9576}\approx 1600 \end{align*}So, all together,
\begin{align*} Q(t)&=6\left(e^{-k}\right)^t=6\cdot \left(6^{-1}\right)^t=6^{1-t} \end{align*}where \(C\) is the amount in the sample at time \(t=0\text{,}\) and \(k\) is some positive constant. We know the half-life of the isotope, so we can find \(e^{-k}\text{:}\)
\begin{align*} \frac{C}{2}=Q(30)&=Ce^{-k\cdot 30}\\ \frac{1}{2}&=\left(e^{-k}\right)^{30}\\ 2^{-\tfrac{1}{30}}&=e^{-k}\\ \end{align*}So,
\begin{align*} Q(t)&=C\left(e^{-k}\right)^t=C\cdot2^{-\tfrac{t}{30}}\\ \end{align*}When only 0.01% of the original sample is left, \(Q(t)=0.0001C\text{:}\)
\begin{align*} 0.0001C=Q(t)&=C\cdot 2^{-\tfrac{t}{30}}\\ 0.0001&=2^{-\tfrac{t}{30}}\\ \log(0.0001)&=\log\left(2^{-\tfrac{t}{30}}\right)\\ \log\left(10^{-4}\right)&=-\frac{t}{30}\log2\\ -4\log 10&=-\frac{t}{30}\log2\\ t&=120\cdot\frac{\log10}{\log2}\approx 398.6 \end{align*}Now, we have a better formula for \(Q(t)\text{:}\)
\begin{align*} Q(t)&=C\left(e^{-k}\right)^t\\ Q(t)&=C\cdot2^{-\tfrac{t}{138}}\\ \end{align*}Finally, we can evaluate what percentage of the sample decays in a day.
\begin{align*} 100\frac{Q(t)-Q(t+1)}{Q(t)}&=100\frac{C\cdot2^{-\tfrac{t}{138}}-C\cdot2^{-\tfrac{t+1}{138}}}{C\cdot2^{-\tfrac{t}{138}}}\left(\frac{\frac{1}{C}}{\frac{1}{C}}\right)\\ &=100\frac{2^{-\tfrac{t}{138}}-2^{-\tfrac{t+1}{138}}}{2^{-\tfrac{t}{138}}}\\ &=100\left(2^{-\tfrac{t}{138}}-2^{-\tfrac{t+1}{138}}\right)2^{\tfrac{t}{138}}\\ &=100\left(1-2^{-\tfrac{1}{138}}\right)\approx 0.5 \end{align*}where \(6.9 \leq Q(0) \leq 7.5\text{.}\) We don’t exactly know \(Q(0)\text{,}\) and we don’t exactly know the half-life, so we also won’t exactly know \(Q(10)\text{:}\) we can only say that is it between two numbers. Our strategy is to find the highest and lowest possible values of \(Q(10)\text{,}\) given the information in the problem. In order for the most possible Uranium-232 to be in the sample after 10 years, we should start with the most and have the longest half-life (since this represents the slowest decay). So, we take \(Q(0)=7.5\) and \(Q(70)=\frac{1}{2}(7.5)\text{.}\)
\begin{align*} Q(t)&=7.5e^{-kt}\\ \frac{1}{2}(7.5)=Q(70)&=7.5e^{-k(70)}\\ \frac{1}{2}&=\left(e^{-k}\right)^{70}\\ 2^{-\tfrac{1}{70}}&=e^{-k}\\ \end{align*}So, in this secenario,
\begin{align*} Q(t)&=7.5\cdot 2^{-\tfrac{t}{70}}\\ \end{align*}After ten years,
\begin{align*} Q(10)&=7.5\cdot 2^{-\tfrac{10}{70}}\approx 6.79\\ \end{align*}So after ten years, the sample contains at most 6.8 \(\mu\)g. Now, let’s think about the least possible amount of Uranium-232 that could be left after 10 years. We should start with as little as possible, so take \(Q(0)=6.9\text{,}\) and the sample should decay quickly, so take the half-life to be 68.8 years.
\begin{align*} Q(t)&=6.9e^{-kt}\\ \frac{1}{2}6.9=Q(68.8)&=6.9e^{-k(68.8)}\\ \frac{1}{2}&=\left(e^{-k}\right)^{68.8}\\ 2^{-\tfrac{1}{68.8}}&=e^{-k}\\ \end{align*}In this scenario,
\begin{align*} Q(t)&=6.9\cdot 2^{-\tfrac{t}{68.8}}\\ \end{align*}After ten years,
\begin{align*} Q(10)&=6.9\cdot 2^{-\tfrac{10}{68.8}}\approx 6.24 \end{align*}Since the object is warmer than the room, \(T(0)-A\) is a nonzero constant. So,
\begin{align*} \lim_{t \to \infty}e^{Kt}&=0 \end{align*}That is, when
\begin{align*} [T(0)-A]e^{kt}&=0\\ \end{align*}Since \(e^{kt} \gt 0\) for all values of \(k\) and \(t\text{,}\) this happens exactly when
\begin{align*} T(0)-A=0 \end{align*}The information given tells us that \(T(10)=90\text{,}\) so
\begin{align*} 90&=-75e^{10K}+100\\ 75\left(e^K\right)^{10}&=10\\ \left(e^K\right)^{10}&=\frac{2}{15}\\ e^K&=\left(\frac{2}{15}\right)^{\tfrac{1}{10}}\\ \end{align*}This lets us describe \(T(t)\) without any unknown constants.
\begin{align*} T(t)&=-75\left(e^K\right)^{t}+100\\ &=-75\left(\frac{2}{15}\right)^{\tfrac{t}{10}}+100\\ \end{align*}The question asks what value of \(t\) gives \(T(t)=99.9\text{.}\)
\begin{align*} 99.9&=-75\left(\frac{2}{15}\right)^{\tfrac{t}{10}}+100\\ 75\left(\frac{2}{15}\right)^{\tfrac{t}{10}}&=0.1\\ \left(\frac{2}{15}\right)^{\tfrac{t}{10}}&=\frac{1}{750}\\ \log\left(\left(\frac{2}{15}\right)^{\tfrac{t}{10}}\right)&=\log\left(\frac{1}{750}\right)\\ \frac{t}{10}\log\left(\frac{2}{15}\right)&=-\log(750)\\ t&=\frac{-10\log(750)}{\log\left(\tfrac{2}{15}\right)}\approx 32.9 \end{align*}for some constant \(K\text{.}\) We are given that \(T(10)=100\text{.}\)
\begin{align*} 100=T(10)&=500e^{10K}\\ e^{10K}&=\frac{1}{5}\\ e^K&=5^{-\tfrac{1}{10}}\\ \end{align*}This gives us the more complete picture for the temperature of the stone.
\begin{align*} T(t)&=500\left(e^K\right)^t=500\cdot 5^{-\tfrac{t}{10}}\\ \end{align*}If \(T(t)=50:\)
\begin{align*} 50=T(t)&=500\cdot 5^{-\tfrac{t}{10}}\\ \frac{1}{10}=10^{-1}&=5^{-\tfrac{t}{10}}\\ 10&=5^{\tfrac{t}{10}}\\ \log(10)&=\frac{t}{10}\log(5)\\ t&=10\frac{\log(10)}{\log(5)}\approx 14.3 \end{align*}After 10 minutes,
\begin{align*} \textcolor{red}{54}=T(10)&=22+64e^{-10 k}\\ e^{-10 k}&=\frac{54-22}{64}=\frac{1}{2} \end{align*}We are told \(T(5)=10\text{:}\)
\begin{align*} 10&=-25e^{5k}+30\\ 25e^{5k}&=20\\ e^{5k}&=\frac{4}{5}\\ 5k&=\log(4/5)\\ k&=\tfrac{1}{5}\log(4/5)\\ \end{align*}So, the differential equation is
\begin{align*} \diff{T}{t}(t)&=\frac{1}{5}\log(4/5)[T(t)-30] \end{align*}This gives us a better idea of \(P(t)\text{:}\)
\begin{align*} P(t)&=121e^{bt}=121\left(\frac{136}{121}\right)^t\\ \end{align*}2020 is 5 years after 2015, so in 2020 (assuming the population keeps growing according to the Malthusian model) the population will be
\begin{align*} P(5)&=121\left(\frac{136}{121}\right)^5\approx217 \end{align*}So, the population at time \(t\) is
\begin{align*} P(t)&=1000\cdot 2^t\\ \end{align*}We want to know at what time the population triples, to 3,000 individuals.
\begin{align*} 3000&=1000\cdot 2^t\\ 3&=2^t\\ \log(3)&=\log\left(2^t\right)=t\log(2)\\ t&=\frac{\log(3)}{\log(2)}\approx 1.6 \end{align*}for some constant \(b\text{.}\) We want to get rid of this extraneous variable \(b\text{,}\) so we use the given information. If 1928 is \(a\) years after the wreck:
\begin{align*} 1000=P(a)&=2e^{ba}\\ 1500=P(a+1)&=2e^{b(a+1)}=2e^{ba}e^b\\ \end{align*}So,
\begin{align*} (1000)\left(e^b\right)&=\left(2e^{ba}\right)\left(e^b\right)=1500\\ \end{align*}Which tells us
\begin{align*} e^b&=\frac{1500}{1000}=1.5\\ \end{align*}Now, our model is complete:
\begin{align*} P(t)&=2\left(e^b\right)^t=2\cdot1.5^t\\ \end{align*}Since \(P(a)=1000\text{,}\) we can find \(a\text{:}\)
\begin{align*} 1000=P(a)&=2\cdot1.5^a\\ 500&=1.5^a\\ \log(500)&=\log\left(1.5^a\right)=a\log(1.5)\\ a&=\frac{\log(500)}{\log(1.5)}\approx 15.3 \end{align*}for some constant \(b\text{.}\) The constant \(b\) is the net birthrate per population member per unit time. Assuming that the net birthrate for a larger population will be the same as the test population, we can use the data from the test to find \(e^b\text{.}\) Let \(Q(t)\) be the number of individuals in the test population at time \(t\text{.}\)
\begin{align*} Q(t)&=Q(0)e^{bt}=200e^{bt}\\ 1000&=Q(3)=200e^{3b}\\ 5&=e^{3b}\\ e^b&=5^{1/3}\\ \end{align*}Now that we have an idea of the birthrate, we predict
\begin{align*} P(t)&=P(0)\left(e^{b}\right)^t=P(0)\cdot5^{\tfrac{t}{3}}\\ \end{align*}We want \(P(12)=10^6+P(0)\text{.}\)
\begin{align*} 10^6+P(0)=P(12)&=P(0)\cdot 5^{\tfrac{12}{3}}=P(0)\cdot 5^4\\ 10^6&=P(0)\cdot 5^4-P(0)=P(0)\left[5^4-1\right]\\ P(0)&=\frac{10^6}{5^4-1}\approx 1603 \end{align*}for some constant \(C\text{.}\) The second piece of given information tells us \(f(0)=2\text{.}\) Using this, we can solve for \(C\text{:}\)
\begin{align*} 2&=f(0)=Ce^{0}=C\\ \end{align*}Now, we know \(f(x)\) entirely:
\begin{align*} f(x)&=2e^{\pi x}\\ \end{align*}So, we can evaluate \(f(2)\)
\begin{align*} f(2)&=2e^{2\pi} \end{align*}We can check that this is reasonable: if
\begin{align*} T(t)&=\left[T(0)+\frac{9}{7}\right]e^{7t}-\frac{9}{7}\\ \end{align*}then
\begin{align*} \diff{T}{t}&=7\left[T(0)+\frac{9}{7}\right]e^{7t}\\ &=7\left[T+\frac{9}{7}\right]\\ &=7T+9. \end{align*}So,
\begin{align*} Q(0)e^{8k}&=0.8\,Q(0)\\ e^{8k}&=0.8\\ e^k&=0.8^{\tfrac{1}{8}}\\ \end{align*}If \(Q(0)=100\text{,}\) the time \(t\) at which \(Q(t)=40\) is determined by
\begin{align*} 40=Q(t)&=Q(0)e^{kt}=100e^{kt}=100\left(0.8^{\tfrac{1}{8}}\right)^t=100\cdot 0.8^{\tfrac{t}{8}}\\ \end{align*}Solving for \(t\text{:}\)
\begin{align*} \frac{40}{100}&=0.8^{\tfrac{t}{8}}\\ \log\left(0.4\right)&=\log\left(0.8^{\tfrac{t}{8}}\right)=\frac{t}{8}\log(0.8)\\ t&=\frac{8\log(0.4)}{\log(0.8)}\approx 32.85\mbox{ days} \end{align*}where \(A\) is the room temperature, and \(K\) is some constant. We are told that \(T(15)=85\) and \(T(30)=73\text{,}\) so:
\begin{align*} 85=T(15)&=[100-A]e^{15K}+A\\ 73=T(30)&=[100-A]e^{30K}+A\\ \end{align*}Rearranging both equations:
\begin{align*} \frac{85-A}{100-A}&=e^{15K}\\ \frac{73-A}{100-A}&=e^{30K}=\left(e^{15K}\right)^2\\ \end{align*}Using these equations:
\begin{align*} \left(\frac{85-A}{100-A}\right)^2=\left(e^{15K}\right)^2&=e^{30K}=\frac{73-A}{100-A}\\ \frac{(85-A)^2}{100-A}&=73-A\\ (85-A)^2&=(73-A)(100-A)\\ 85^2-170A+A^2&=7300-173A+A^2\\ 173A-170A&=7300-85^2\\ 3A&=75\\ A&=25 \end{align*}Using Corollary 3.3.8
\begin{align*} A(t)&=[A(0)+40,\!000]e^{0.05t}-40,\!000\\ &=90,\!000\cdot e^{0.05t}-40,\!000\\ \end{align*}where \(t=0\) corresponds to the year when the graduate is 25.
\begin{align*} \\ \end{align*}When the graduate is 65 years old, \(t=40\text{,}\) so
\begin{align*} A(40)&=90,\!000\, e^{0.05 \times 40}-40,000\approx \$625,015.05 \end{align*}Using Corollary 3.3.8 and assuming that the interest rate remains 5%
\begin{align*} A(t)&=[A(0)-20W]e^{0.05t}+20W\\ &=[625,015.05-20W]e^{0.05t}+20W\\ \end{align*}Note that, for part (b), we only care about what happens when the graduate starts withdrawing money. We take \(t=0\) to correspond to the year when the graduate is 65--so we’re using a different \(t\) from part (a). Then from part (a), \(A(0)=625,025.05\text{.}\)
\begin{align*} \\ \end{align*}We want the account to be depleted when the graduate is 85. So, we want
\begin{align*} 0&=A(20)\\ 0&=20W+ e^{0.05\times 20}(625,015.05-20W)\\ 0&=20W+ e(625,015.05-20W)\\ 20(e-1)W&= 625,015.05e\\ W&=\frac{625,015.05e}{20(e-1)}\approx\$49,437.96 \end{align*}Using Corollary 3.3.8
\begin{align*} A(t)&=[A(0)-150,\!000]e^{0.06t}+150,\!000\\ &=[120,\!000-150,\!000]e^{0.06t}+150,\!000\\ &=-30,\!000e^{0.06t}+150,\!000 \end{align*}The doubling time \(t\) obeys:
\begin{align*} Q(t)&=2Q(0)\\ \mbox{So,}\qquad Q(0)e^{kt}&=2Q(0)\\ e^{kt}&=2\\ 3^{\tfrac{t}{9}}&=2\\ \frac{t}{9}\log 3 &=\log 2\\ t&=9\frac{\log 2}{\log 3}\approx5.68 \mbox{ hr} \end{align*}So, we can use the corollary, with \(K=-k\text{,}\) \(T=v\text{,}\) and \(A=-\dfrac{g}{k}\text{.}\)
\begin{align*} v(t)&=\left(v(0)-\left(-\frac{g}{k}\right)\right)e^{-kt}-\frac{g}{k}\\ &=\left(v_0+\frac{g}{k}\right)e^{-kt}-\frac{g}{k} \end{align*}Since \(k\) is positive:
\begin{align*} &=\left(v_0+\frac{g}{k}\right)\left(0\right)-\frac{g}{k}\\ &=-\dfrac{g}{k} \end{align*}
So, we calculate:
\begin{align*} f(1)&=\log(1)=0\\ f'(x)&=\frac{1}{x}\\ f'(1)&=\frac{1}{1}=1\\ \end{align*}Therefore,
\begin{align*} f(x) &\approx 0+1(x-1)=x-1\\ \end{align*}When \(x=0.93\text{:}\)
\begin{align*} f(0.93) &\approx 0.93-1=-0.07 \end{align*}
\(a\) needs to be a number whose fifth root we know. Since \(\sqrt[5]{32}=2\text{,}\) and 32 is reasonably close to 30, \(a=32\) is a great choice.
\begin{align*} f(32)&=\sqrt[5]{32}=2\\ f'(32)&=\frac{1}{5\cdot 2^4}=\frac{1}{80}\\ \end{align*}The linear approximation of \(f(x)\) about \(x=32\) is
\begin{align*} f(x) &\approx 2+\frac{1}{80}(x-32)\\ \end{align*}When \(x=30\text{:}\)
\begin{align*} f(30) &\approx 2+\frac{1}{80}(30-32)=2-\frac{1}{40}=\frac{79}{40} \end{align*}
We subsitute \(f(a)=2a+5\text{,}\) \(f'(a)=2\text{,}\) and \(f''(a)=0\text{:}\)
\begin{align*} f(x) &\approx (2a+5)+2(x-a)=2x+5 \end{align*}So,
\begin{align*} f(x) &\approx f(1)+f'(1)(x-1)+\frac{1}{2}f''(1)(x-1)^2\\ &=0+(x-1)-\frac{1}{2}(x-1)^2 \end{align*}Using the quadratic approximation \(f(x) \approx f(0)+f'(0)(x-0)+\frac{1}{2}f''(0)(x-0)^2\text{:}\)
\begin{align*} f(x)&\approx 1 -\frac{1}{2}x^2\\ f\left(\frac{1}{15}\right)&\approx 1-\frac{1}{2\cdot 15^2}=\frac{449}{450} \end{align*}We compute derivatives.
\begin{align*} f(0)&=e^0=1\\ f'(x)&=2e^{2x}\\ f'(0)&=2e^0=2\\ f''(x)&=4e^{2x}\\ f''(0)&=4e^0=4\\ \end{align*}Substituting:
\begin{align*} f(x) &\approx 1+2(x-0)+\frac{4}{2}(x-0)^2\\ f(x) &\approx 1+2x+2x^2 \end{align*}Using the quadratic approximation \(f(x) \approx f(a)+f'(a)(x-a)+\frac{1}{2}f''(a)(x-a)^2\text{:}\)
\begin{alignat*}{3} f(x)&\approx 10+\frac{5}{12}(x-8)-\frac{5}{288}(x-8)^2\\ f(5)&\approx 10+\frac{5}{12}(-3)-\frac{5}{288}(9)=\frac{275}{32} \end{alignat*}Most of these cancel!
\begin{align*} &=\frac{1}{1}\underbrace{-\frac{1}{2}+\frac{1}{2}}_0 \underbrace{-\frac{1}{3}+\frac{1}{3}}_0 \underbrace{-\frac{1}{4}+\frac{1}{4}}_0 \underbrace{-\frac{1}{5}+\frac{1}{5}}_0 \underbrace{-\frac{1}{6}+\frac{1}{6}}_0\\ &\underbrace{-\frac{1}{7}+\frac{1}{7}}_0 \underbrace{-\frac{1}{8}+\frac{1}{8}}_0 \underbrace{-\frac{1}{9}+\frac{1}{9}}_0 \underbrace{-\frac{1}{10}+\frac{1}{10}}_0 -\frac{1}{11}\\ &=1-\frac{1}{11}=\frac{10}{11} \end{align*}Now, we can find the quadratic approximation about \(x=0\text{.}\)
\begin{align*} f(x)&\approx f(0)+f'(0)x+\frac{1}{2}f''(0)x^2\\ &=2x\\ f(1)& \approx 2 \end{align*}So 3.4.11.11.a and 3.4.11.11.d are equivalent.
\begin{align*} \knowl{./knowl/xref/s3_4_3equiv3.html}{\text{3.4.11.11.e}}&=2\ds\sum_{n=2}^{11} (n-1)=2(1+2+\cdots + 10) = \knowl{./knowl/xref/s3_4_3equiv2.html}{\text{3.4.11.11.d}} \end{align*}So,
\begin{align*} \sum_{k=0}^{n} \frac{f^{(k)}(5)}{k!}(x-5)^k&=\sum_{k=0}^{n}\frac{2k+1}{3k-9}(x-5)^k\\ \end{align*}For any \(k\) from 0 to \(n\text{,}\)
\begin{align*} \frac{f^{(k)}(5)}{k!}&=\frac{2k+1}{3k-9}\\ \end{align*}In particular, when \(k=10\text{,}\)
\begin{align*} \frac{f^{(10)}(5)}{10!}&=\frac{20+1}{30-9}=1\\ f^{(10)}(5)&=10! \end{align*}while the third order Maclaurin polynomial for \(f(x)\) is
\begin{align*} T_3(x)&=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+\frac{1}{3!}f'''(0)x^3\\ \end{align*}So, we simply “chop off” the part of \(T_4(x)\) that includes \(x^4\text{:}\)
\begin{align*} T_3(x)&=-x^3+x^2-x+1 \end{align*}while the third order Taylor polynomial for \(f(x)\) about \(x=1\) is
\begin{align*} T_3(x)&=f(1)+f'(1)(x-1)+\frac{1}{2}f''(1)(x-1)^2+\frac{1}{3!}f'''(1)(x-1)^3 \end{align*}Recall that \(T_4(x)\) and \(f(x)\) have the same values at \(x=1\) (although maybe not anywhere else!), and they also have the same first, second, third, and fourth derivatives at \(x=1\) (but again, maybe not anywhere else, and maybe their fifth derivatives don’t agree). This tells us the following:
\begin{align*} T_4(x)&=x^4+x^3-9&&\Rightarrow& f(1)=T_4(1)&=-7\\ T_4'(x)&=4x^3+3x^2&&\Rightarrow& f'(1)=T_4'(1)&=7\\ T_4''(x)&=12x^2+6x&&\Rightarrow& f''(1)=T_4''(1)&=18\\ T_4'''(x)&=24x+6&&\Rightarrow& f'''(1)=T_4'''(1)&=30 \end{align*}Now, let’s move to the Taylor polynomial. Remember that \(e\) is a constant.
\begin{align*} T_3(x)&=-\frac{2}{3}\sqrt{e^3}+3ex-6\sqrt{e}x^2+x^3\\ T_3'(x)&=3e-12\sqrt{e}x+3x^2\\ T_3''(x)&=-12\sqrt{e}+6x\\ T_3'''(x)&=6\\ \color{red}{T_3'''(a)}&\color{red}{=6}\\ \end{align*}The final bullet point gives us the equation:
\begin{align*} \color{blue}{f'''(a)}&=\color{red}{T_3'''(a)}\\ \color{blue}{12\log a} &= \color{red}{6}\\ \log a &= \frac{1}{2}\\ a&=e^{\tfrac{1}{2}} \end{align*}So, in general,
\begin{align*} f^{(k)}(x)&=2^x\left(\log 2\right)^k\\ \end{align*}We notice that this formula works even when \(k=0\) and \(k=1\text{.}\) When \(x=1\text{,}\)
\begin{align*} f^{(k)}(1)&=2\left(\log 2\right)^k\\ \end{align*}The \(n\)th order Taylor polynomial of \(f(x)\) about \(x=1\) is
\begin{align*} T_{n}(x)&=\sum_{k=0}^n \frac{f^{(k)}(1)}{k!}(x-1)^k\\ &=\sum_{k=0}^n \frac{2(\log 2)^k}{k!}(x-1)^k \end{align*}Using the chain rule,
\begin{align*} f'(x)&=(-1)(1-x)^{-2}(-1)=(1-x)^{-2}\\ f''(x)&=(-2)(1-x)^{-3}(-1)=2(1-x)^{-3}\\ f^{(3)}(x)&=(-3)(2)(1-x)^{-4}(-1)=2(3)(1-x)^{-4}\\ f^{(4)}(x)&=(-4)(2)(3)(1-x)^{-5}(-1)=2(3)(4)(1-x)^{-5}\\ f^{(5)}(x)&=(-5)(2)(3)(4)(1-x)^{-6}(-1)=2(3)(4)(5)(1-x)^{-6}\\ \end{align*}Recognizing the pattern,
\begin{align*} f^{(k)}(x)&=k!(1-x)^{-(k+1)}\\ f^{(k)}(0)&=k!(1)^{-(k+1)}=k!\\ \end{align*}The \(n\)th order Maclaurin polynomial for \(f(x)\) is
\begin{align*} T_n(x)&=\sum_{k=0}^n \frac{f^{(k)}(0)}{k!}x^k\\ &=\sum_{k=0}^n \frac{k!}{k!}x^k\\ &=\sum_{k=0}^n x^k \end{align*}Now, if we want to approximate \(f\left(\dfrac{1}{\sqrt{3}}\right)=6\arctan\left(\dfrac{1}{\sqrt{3}}\right)=\pi\text{:}\)
\begin{align*} \pi&=f\left(\dfrac{1}{\sqrt{3}}\right) \approx T_5\left(\dfrac{1}{\sqrt{3}}\right)= \frac{6}{\sqrt{3}}-\frac{2}{\sqrt{3}^3}+\frac{6}{5\sqrt{3}^5}\\ &=2\sqrt{3}\left(1-\frac{1}{3\cdot3}+\frac{1}{5\cdot 9}\right)\approx 3.156 \end{align*}When \(k \geq 2\text{,}\) making use of the fact that \(0!=1\) and \((-1)^{k-2}=(-1)^k\text{:}\)
\begin{align*} f^{(k)}(x)&=(-1)^{k-2}(k-2)!x^{-(k-1)}&f^{(k)}(1)&=(-1)^k(k-2)!\\ \end{align*}Now we use the standard form of a Taylor polynomial. Since the first two terms don’t fit the pattern, we add those outside of the sigma.
\begin{align*} T_{100}(x)&=\sum_{k=0}^{100}\frac{f^{(k)}(1)}{k!}(x-1)^k\\ &=f(1)+f'(1)(x-1)+\sum_{k=2}^{100}\frac{f^{(k)}(1)}{k!}(x-1)^k\\ &=-1+0(x-1)+\sum_{k=2}^{100}\frac{(-1)^k(k-2)!}{k!}(x-1)^k\\ &=-1+\sum_{k=2}^{100}\frac{(-1)^k}{k(k-1)}(x-1)^k \end{align*}If \(n=157\) and \(x=1\text{,}\)
\begin{align*} T_{157}(1)&=\sum_{k=0}^{157}\frac{1}{k!}=1+1+\frac{1}{2}+\frac{1}{3!}+\frac{1}{4!}+\cdots +\frac{1}{157!} \end{align*}If \(n=100\) and \(x=\dfrac{5\pi}{4}\text{,}\) this equation becomes
\begin{align*} T_{200}\left(\frac{5\pi}{4}\right)&=\sum_{k=0}^{100} \frac{(-1)^k}{(2k)! }\cdot \left(\frac{5\pi}{4}\right)^{2k} \end{align*}
The linear approximation for \(\Delta y\) when \(\Delta x = \dfrac{1}{10}\) is
\begin{align*} \Delta y &\approx f'(5)\Delta x = \frac{1}{26}\cdot \frac{1}{10}=\frac{1}{260} \approx 0.003846 \end{align*}From the problem, \(\Delta x= 5-4=1\text{.}\) Differentiating,
\begin{align*} s'(t)&=\sqrt{19.6}\cdot\frac{1}{2\sqrt{x}}=\sqrt{\frac{4.9}{x}},\mbox{ so }\\ s'(4)&=\sqrt{\frac{4.9}{4}}\\ \end{align*}The linear approximation gives us
\begin{align*} \Delta y& \approx \sqrt{\frac{4.9}{4}}(1)\approx1.1 \end{align*}while the change in terminal speed between the next two jumps is
\begin{align*} \Delta y &\approx \sqrt{\frac{4.9}{x+\Delta x}}\cdot\Delta x \end{align*}Differentiating with respect to \(r\text{,}\)
\begin{align*} A'(r)&=2\pi r\\ \end{align*}The exact area desired is \(A_0\text{.}\) Let the corresponding exact radius desired be \(r_0\text{.}\)
\begin{align*} \\ \end{align*}Using the linear approximation formula, where \(\Delta A\) is the change in \(A\) corresponding to a change in \(r\) of \(\Delta r\text{,}\)
\begin{align*} \Delta A &\approx A'(r_0)\Delta r = 2\pi r_0 \Delta r\\ \Delta r &\approx \dfrac{\Delta A}{2\pi r_0}\\ \end{align*}What we’re interested in is the percent error \(r\) can have. The percent error is:
\begin{align*} 100\frac{\Delta r}{r_0}&\approx100\frac{\Delta A}{2\pi r_0\cdot r_0}\\ &=100\frac{\Delta A}{2(\pi r_0^2)}\\ &=100\frac{\Delta A}{2\cdot A_0}\\ &=\left( 100\frac{\Delta A}{A_0}\right)\frac{1}{2}\\ &\leq\left( 2\right)\frac{1}{2}=1 \end{align*}
Since the question tells us the sector is no more than a quarter of the circle, we know \(0 \leq \theta \leq \dfrac{\pi}{2}\text{,}\) so \(0 \leq \dfrac{\theta}{2} \leq \dfrac{\pi}{4}\text{.}\) This puts \(\dfrac{\theta}{2}\) well within the range of arcsine.
\begin{align*} \theta&=2\arcsin\left(\frac{d}{6}\right) \end{align*}Differentiating,
\begin{align*} A'(d)&=\frac{9}{\sqrt{1-\left(\frac{d}{6}\right)^2}}\cdot\frac{1}{6}\\ &=\frac{9}{\sqrt{36-d^2}}\\ \end{align*}Let \(\Delta A\) is the error in \(A\) corresponding to an error of \(\Delta d\) in \(d\text{.}\) Since we measured \(d\) to be 0.7 instead of \(0.68\text{,}\) in the linear approximation we take \(\Delta d = 0.02\) and \(d_0=0.68\text{.}\)
\begin{align*} \Delta A &\approx A'(d_0)\cdot\Delta d\\ &=A'\left(0.68\right)\cdot 0.02\\ &=\frac{9}{\sqrt{36-0.68^2}}\cdot0.02\\ &\approx 0.03 \end{align*}
Using this, we find our equation for the volume of the water in terms of \(h\text{.}\)
\begin{align*} V(h)&=\frac{1}{3}\pi r^2 h = \frac{\pi}{3}\left(\frac{h}{4}\right)^2h=\frac{\pi h^3}{48}\\ \end{align*}Differentiating,
\begin{align*} V'(h)&=\frac{\pi h^2}{16}\\ V'(0.5)&=\frac{0.25\pi}{16}=\frac{\pi}{64}\\ \end{align*}Finally, using the approximation \(\Delta V \approx -0.05V'(0.5)\text{,}\)
\begin{align*} \Delta V &\approx \frac{-0.05\pi}{64}=-\frac{\pi}{1280}\approx -0.00245\;\mathrm{m}^3 \end{align*}So,
\begin{align*} \left|\Delta h\right|&\approx\left| h'(0.9)\right|\left|\Delta q\right|= h'(0.9)\cdot 0.05 \end{align*}for some constant \(k\text{.}\) Let’s take \(t=0\) to be the time when precisely one \(\mu\)g was present. Then
\begin{align*} Q(t)&=e^{-kt}\\ \end{align*}After three years, \(q\) is the amount of the isotope remaining, so
\begin{align*} q&=e^{-k\cdot 3}\\ q^{\frac{1}{3}}&=e^{-k}\\ Q(t)&=\left(e^{-k}\right)^t=q^{\frac{t}{3}}\\ \end{align*}The half-life is the value of \(t\) for which \(Q(t)=\frac{1}{2}Q(0)=\frac{1}{2}\text{.}\)
\begin{align*} \frac{1}{2}&=Q(t)=q^{\frac{t}{3}}\\ \log\left(\frac{1}{2}\right)&=\log\left(q^{\frac{t}{3}}\right)\\ -\log 2 &=\frac{t}{3}\log q\\ t&=\frac{-3\log 2}{\log q} \end{align*}Following our plan, we find \(h'(0.9)\text{.}\)
\begin{align*} h'(q)&=\diff{}{q}\left\{\dfrac{-3\log2}{\log q}\right\}\\ &=-3\log 2\cdot\diff{}{q}\left\{\left(\log q\right)^{-1}\right\}\\ &=-3\log 2\cdot(-1)\left(\log q\right)^{-2}\cdot\frac{1}{q}\\ &=\frac{3\log 2}{q\log^2q}\\ h'(0.9)&=\frac{3\log2}{0.9\log^2(0.9)}\approx 208\\ \end{align*}Finally, as outlined in our plan,
\begin{align*} \left|\Delta h\right|&=h'(0.9)\cdot0.05\\ &=\frac{3\log2}{18\log^2(0.9)}\approx 10.4 \end{align*}for some \(c\) strictly between \(x\) and \(a\text{.}\) In our case, \(n=3\text{,}\) \(a=0\text{,}\) \(x=2\text{,}\) and \(f^{(4)}(c)=e^c\text{,}\) so
\begin{align*} |f(2)-T_3(2)| &= \left|\frac{f^{(4)}(c)}{4!}(2-0)^4\right|\\ &=\frac{2^4}{4!}e^c=\frac{2}{3}e^c\\ \end{align*}Since \(c\) is strictly between 0 and 2, \(e^c \lt e^2\text{:}\)
\begin{align*} &\leq \frac{2}{3}e^2\\ \end{align*}but this isn’t a number we really know. Indeed: \(e^2\) is the very number we’re trying to approximate. So, we use the estimation \(e \lt 3\text{:}\)
\begin{align*} & \lt \frac{2}{3}\cdot 3^2 =6 \end{align*}while the linear approximation gives
\begin{align*} f(x)&\approx f(0)+f'(0)x\\ \sin(x)& \approx \sin (0)+\cos(0) x\\ &= x\\ \sin(33)&\approx 33 \end{align*}
for some \(c\) strictly between \(x\) and \(a\text{.}\) In our case, \(n=5\text{,}\) \(a=11\text{,}\) \(x=11.5\text{,}\) and \(f^{(6)}(c)=\dfrac{6!(2c-5)}{c+3}\text{.}\)
\begin{align*} \left|f(11.5)-T_5(11.5)\right|& = \left|\dfrac{1}{6!}\left(\dfrac{6!(2c-5)}{c+3}\right)(11.5-11)^{6}\right|\\ & = \left|\dfrac{2c-5}{c+3}\right|\cdot\frac{1}{2^6} \end{align*}for some \(c\) strictly between \(x\) and \(a\text{.}\) In our case, \(n=2\text{,}\) \(a=0\text{,}\) and \(x=0.1\text{,}\) so
\begin{align*} \left|f(0.1)-T_2(0.1)\right| &= \left|\frac{f^{(3)}(c)}{3!}(0.1-0)^{3}\right|\\ &=\frac{\left|f'''(c)\right|}{6000} \end{align*}for some \(c\) strictly between \(x\) and \(a\text{.}\) In our case, \(n=5\text{,}\) \(a=0\text{,}\) and \(x=-\dfrac{1}{4}\text{,}\) so
\begin{align*} \left|f\left(-\frac{1}{4}\right)-T_5\left(-\frac{1}{4}\right)\right| &= \left|\frac{f^{(6)}(c)}{6!}\left(-\frac{1}{4}-0\right)^6\right|\\ &= \frac{\left|f^{(6)}(c)\right|}{6!\cdot 4^6}\\ \end{align*}for some \(c\) in \(\left(-\frac{1}{4},0\right)\text{.}\) We’ll need to know the sixth derivative of \(f(x)\text{.}\)
\begin{align*} f(x)&=\log (1-x)\\ f'(x)&=-(1-x)^{-1}\\ f''(x)&=-(1-x)^{-2}\\ f'''(x)&=-2(1-x)^{-3}\\ f^{(4)}(x)&=-3!(1-x)^{-4}\\ f^{(5)}(x)&=-4!(1-x)^{-5}\\ f^{(6)}(x)&=-5!(1-x)^{-6}\\ \end{align*}Plugging in \(\left|f^{(6)}(c)\right|=\dfrac{5!}{(1-c)^{6}}\text{:}\)
\begin{align*} \left|f\left(-\frac{1}{4}\right)-T_5\left(-\frac{1}{4}\right)\right| &= \frac{5!}{6!\cdot 4^6\cdot(1-c)^6}=\frac{1}{6\cdot 4^6 \cdot (1-c)^6} \end{align*}for some \(c\) strictly between \(x\) and \(a\text{.}\) In our case, \(n=3\text{,}\) \(a=30\text{,}\) and \(x=32\text{,}\) so
\begin{align*} |f(30)-T_3(30)| &= \left|\frac{f^{(4)}(c)}{4!}(30-32)^{4}\right|\\ &=\frac{2}{3}\left|f^{(4)}(c)\right| \end{align*}Since \(30 \lt c \lt 32\text{,}\)
\begin{align*} & \lt \frac{336}{5^4\cdot 30^{\tfrac{19}{5}}} =\frac{336}{5^4\cdot 30^3 \cdot 30^{\tfrac{4}{5}}}\\ &=\frac{14}{5^7\cdot 9 \cdot 30^{\tfrac{4}{5}}}\\ \end{align*}This isn’t a number we know. We’re trying to find the error in our estimation of \(\sqrt[5]{30}\text{,}\) but \(\sqrt[5]{30}\) shows up in our error. From here, we have to be a little creative to get a bound that actually makes sense to us. There are different ways to go about it. You could simply use \(30^{\tfrac{4}{5}} \gt 1\text{.}\) We will be a little more careful, and use the following estimation:
\begin{align*} \frac{14}{5^7 \cdot 9 \cdot \textcolor{red}{30^{\tfrac{4}{5}}}}&= \frac{14\cdot \textcolor{red}{30^{\tfrac{1}{5}}}}{5^7 \cdot 9 \cdot \textcolor{red}{30}}\\ & \lt \frac{14\cdot \textcolor{red}{32^{\tfrac{1}{5}}}}{5^7 \cdot 9 \cdot \textcolor{red}{30}}\\ & \lt \frac{14\cdot \textcolor{red}{2}}{5^7 \cdot 9 \cdot \textcolor{red}{30}}\\ & \lt \frac{14}{5^7\cdot 9 \cdot \textcolor{red}{15}}\\ & \lt 0.000002 \end{align*}for some \(c\) strictly between \(x\) and \(a\text{.}\) In our case, \(n=1\text{,}\) \(a=\dfrac{1}{\pi}\text{,}\) and \(x=0.01\text{,}\) so
\begin{align*} |f(0.01)-T_1(0.01)| &= \left|\frac{f''(c)}{2}\left(0.01-\frac{1}{\pi}\right)^{2}\right|\\ &=\frac{1}{2}\left(\frac{100-\pi}{100\pi}\right)^2\cdot\left|f''(c)\right| \end{align*}for some \(c\) strictly between \(x\) and \(a\text{.}\) In our case, \(n=2\text{,}\) \(a=0\text{,}\) and \(x=\dfrac{1}{2}\text{,}\) so
\begin{align*} \left|f\left(\frac{1}{2}\right)-T_2\left(\frac{1}{2}\right)\right| &= \left|\frac{f^{(3)}(c)}{3!}\left(\frac{1}{2}-0\right)^{3}\right|\\ &= \frac{\left|f^{(3)}(c)\right|}{3!\cdot2^3} \end{align*}Since \(\left|f\left(\frac{1}{2}\right)-T_2\left(\frac{1}{2}\right)\right| =\dfrac{\left|f^{(3)}(c)\right|}{3!\cdot2^3}\) for some \(c\) in \(\left(0,\frac{1}{2}\right)\text{,}\)
\begin{align*} \left|f\left(\frac{1}{2}\right)-T_2\left(\frac{1}{2}\right)\right| &=\dfrac{\left|\dfrac{1+2c^2} {\left(\sqrt{1-c^2}\right)^5} \right|}{3!\cdot2^3}= \frac{1+2c^2}{48\left(\sqrt{1-c^2}\right)^5} \end{align*}Conveniently, we’ve already found the first few derivatives of \(f(x)\text{.}\)
\begin{align*} T_2(x)&=\arcsin(0)+\left(\frac{1}{\sqrt{1-0^2}}\right)x+\frac{1}{2}\left(\frac{0}{\left(\sqrt{1-0^2}\right)^3}\right)x^2\\ &=0+x+0\\ &=x\\ T_2\left(\frac{1}{2}\right)&=\frac{1}{2}\\ \end{align*}So, the actual error is
\begin{align*} \left|f\left(\frac{1}{2}\right)-T_2\left(\frac{1}{2}\right)\right|&=\left| \frac{\pi}{6}-\frac{1}{2}\right|= \frac{\pi}{6}-\frac{1}{2} \end{align*}So, when \(n \geq 1\text{,}\)
\begin{align*} f^{(n)}(x)&=(-1)^{n-1}(n-1)!\cdot x^{-n} \end{align*}for some \(c\) in \((1,1.1)\)
\begin{align*} & \lt \frac{1}{(n+1)10^{n+1}\cdot 1^{n+1}}\\ &=\frac{1}{(n+1)10^{n+1}}\\ \end{align*}What we’ve shown so far is
\begin{align*} |f(1.1)-T_n(1.1)|& \lt \dfrac{1}{(n+1)10^{n+1}}\\ \end{align*}If we can show that \(\dfrac{1}{(n+1)10^{n+1}}\le10^{-4}\text{,}\) then we’ll be able to conclude
\begin{align*} |f(1.1)-T_n(1.1)|& \lt \dfrac{1}{(n+1)10^{n+1}}\le10^{-4} \end{align*}for some \(c\) in \((2187,2200)\text{.}\) In order for this to be less than 0.001, we need
\begin{align*} \left|f^{(n+1)}(c)\right|\cdot\frac{13^{n+1}}{(n+1)!}& \lt 0.001\\ \left|f^{(n+1)}(c)\right|& \lt \frac{ (n+1)!}{1000\cdot13^{n+1}} \end{align*}for some \(c\) between 0 and 1. Since \(-1 \leq \cos c \leq 1\text{,}\)
\begin{align*} \frac{4242-1}{7!}\leq \sin(1)&\leq \frac{4242+1}{7!}\\ \frac{4241}{7!}\leq \sin(1)&\leq \frac{4243}{7!}\\ \frac{4241}{5040}\leq \sin(1)&\leq \frac{4243}{5040} \end{align*}Since \(e^x\) is a strictly increasing function, and \(0 \lt c \lt 1\text{,}\) we conclude \(e^0 \lt e^c \lt e^1\text{:}\)
\begin{align*} \frac{65}{24} +\frac{1}{120}& \lt e \lt \frac{65}{24} +\frac{e}{120}\\ \end{align*}Simplifying the left inequality, we see
\begin{align*} \frac{326}{120}& \lt e\\ \end{align*}From the right inequality, we see
\begin{align*} e& \lt \frac{65}{24}+\frac{e}{120}\\ e-\frac{e}{120}& \lt \frac{65}{24}\\ e\cdot \frac{119}{120}& \lt \frac{65}{24}\\ e& \lt \frac{65}{24}\cdot \frac{120}{119}=\frac{325}{119}\\ \end{align*}So, we conclude
\begin{align*} \frac{326}{120}& \lt e \lt \frac{325}{119}, \end{align*}In particular
\begin{align*} f(2.98)&\approx 2+4(2.98-3)-5(2.98-3)^2\\ &=2-0.08-0.002={1.918} \end{align*}Using the error formula:
\begin{align*} g(10)&=g(8)+g'(8)(10-8)+\half g''(c)(10-8)^2 \end{align*}So, the linear approximation of \(\sin\left(\frac{101\pi}{100}\right) \) is
\begin{align*} f\left(\frac{101\pi}{100}\right)&\approx T_1\left(\frac{101\pi}{100}\right)\\ &= f(\pi) + f'(\pi)\cdot \left(\frac{101\pi}{100}-\pi\right)\\ &=0+(-1)\cdot\frac{\pi}{100}=-\frac{\pi}{100}. \end{align*}while the third order Maclaurin polynomial for \(f(x)\) is
\begin{align*} T_3(x)&=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+\frac{1}{3!}f'''(0)x^3\\ \end{align*}So, we simply “chop off” the part of \(T_4(x)\) that includes \(x^4\text{.}\) Since that’s already 0, in this case \(T_3(x)=T_4(x)\text{.}\)
\begin{align*} T_3(x)&=5x^2-9 \end{align*}By implicit differentiation (and then subbing in \(x=2\text{,}\) \(y(2)=1\))
\begin{align*} 4y(x)^3y'(x)+y(x)+xy'(x)&=2x\\ 4y'(2)+1+2y'(2)&=4\\ y'(2)&=\frac{1}{2} \end{align*}In particular,
\begin{align*} y(2.1)&\approx y(2)+y'(2)(2.1-2)=1+\frac{1}{2}(.1)=\boxed{1.05}\\ \end{align*}The quadratic approximation to \(y(x)\) at \(x=2\) is
\begin{align*} y(x)&\approx y(2)+y'(2)(x-2)+\frac{1}{2} y''(2)(x-2)^2\\ &=1+\frac{1}{2}(x-2)-\frac{1}{6}(x-2)^2\\ \end{align*}In particular,
\begin{align*} y(2.1)&\approx y(2)+y'(2)(2.1-2)+\frac{1}{2} y''(2)(2.1-2)^2\\ &=1+\frac{1}{2}(.1)-\frac{1}{6}(.1)^2=\boxed{1.0483} \end{align*}
By implicit differentiation (and then subbing in \(x=-1\text{,}\) \(y(-1)=1\))
\begin{align*} 0&=4x^3+y'(x)+y(x)^4+4xy(x)^3y'(x)\\ 0&= -4+y'(-1)+1-4y'(-1)\\ -1&= y'(-1)\\ \end{align*}Differentiating with respect to \(x\) a second time and then subbing in \(x=-1\text{,}\) \(y(-1)=1\text{,}\) and \(y'(-1)=-1\text{:}\)
\begin{align*} 0&=12x^2+y''(x)+4y(x)^3y'(x)+4y(x)^3y'(x)+12xy(x)^2y'(x)^2\\ \amp\hskip3in+4xy(x)^3y''(x)\\ 0&= 12+y''(-1)-4-4-12-4y''(-1)\\ -8&= 3y''(-1)\\ y''(-1)&=-\frac{8}{3} \end{align*}In particular,
\begin{align*} y(-0.9)&\approx 0.9 \end{align*}In particular,
\begin{align*} y(-0.9)&\approx 1-(.1)-\frac{4}{3}(.1)^2\approx{0.8867} \end{align*}
The product \((x-5)(x+7)\) is positive when \((x-5)\) and \((x+7)\) have the same sign, and negative when they have opposite signs, so
\begin{align*} f(x)&=\left\{\begin{array}{ll} \sqrt{(x-5)(x+7)}&\mbox{ if } x\in (-\infty,-7] \cup [5,\infty)\\ \sqrt{-(x-5)(x+7)}&\mbox{ if } x\in(-7,5) \end{array}\right.\\ \end{align*}Now, when \(x \neq -7,5\text{,}\) we can differentiate, using the chain rule.
\begin{align*} f'(x)&=\left\{\begin{array}{ll} \frac{\diff{}{x}\left\{(x-5)(x+7)\right\}}{2\sqrt{(x-5)(x+7)}}&\mbox{ if } x\in (-\infty,-7) \cup (5,\infty)\\ \frac{\diff{}{x}\left\{-(x-5)(x+7)\right\}}{2\sqrt{-(x-5)(x+7)}}&\mbox{ if } x\in(-7,5)\\ ?? & \mbox{ if } x=-7,\,x=5 \end{array}\right.\\ &=\left\{\begin{array}{ll} \frac{2x+2}{2\sqrt{(x-5)(x+7)}}&\mbox{ if } x\in (-\infty,-7) \cup (5,\infty)\\ \frac{-2x-2}{2\sqrt{-(x-5)(x+7)}}&\mbox{ if } x\in(-7,5)\\ ?? & \mbox{ if } x=-7,\,x=5 \end{array}\right. \end{align*}Let’s first consider the case \(h \gt 0\text{.}\)
\begin{align*} \lim_{h \to 0^+} \frac{\sqrt{|(-13+h)(h)|}}{h}&=\lim_{h \to 0^+}\frac{\sqrt{(13-h)(h)}}{h}\\ &=\lim_{h \to 0^+}\frac{\sqrt{13h-h^2}}{\sqrt{h^2}}\\ &=\lim_{h \to 0^+}\sqrt{\frac{13h-h^2}{h^2}}\\ &=\lim_{h \to 0^+}\sqrt{\frac{13}{h}-1}\\ &=\infty\\ \end{align*}Since one side of the limit doesn’t exist,
\begin{align*} \lim_{h \to 0} \frac{f(-7+h)-(-7)}{h}&=DNE \end{align*}
| \(c\) | \(-3\) | \(-5\) | \(5\) |
| type | critical point | endpoint | endpoint |
| \(f(c)\) | \(-19\) | \(-15\) | \(45\) |
| \(c\) | \(-3\) | \(-4\) | \(0\) |
| type | critical point | endpoint | endpoint |
| \(f(c)\) | \(61\) | \(\frac{157}{3}=52+\frac{1}{3}\) | \(7\) |
| \(c\) | \(-2\) | \(0\) | \(-1\) |
| type | endpoint | endpoint | critical point |
| \(f(c)\) | \(-20\) | \(2\) | \(6\) |
| \(c\) | \(2\) | \(0\) | \(1\) |
| type | endpoint | endpoint | critical point |
| \(f(c)\) | \(12\) | \(-10\) | \(-14\) |
| \(c\) | \(1\) | \(4\) | \(2\) |
| type | endpoint | endpoint | critical point |
| \(f(c)\) | \(-6\) | \(30\) | \(-10\) |
| \(x\) | \((-\infty,-2)\) | \((-2,2)\) | \((2,\infty)\) |
| \(h'(x)\) | \(\gt 0\) | \(\lt 0\) | \(\gt 0\) |
| \(h(x)\) | increasing | decreasing | increasing |
| \(x\) | \((-\infty,-2)\) | \((-2,2)\) | \((2,\infty)\) |
| \(h'(x)\) | \(\gt 0\) | \(\lt 0\) | \(\gt 0\) |
| \(h(x)\) | increasing | decreasing | increasing |
Therefore,
\begin{align*} \ell&=3w=30\\ h&=\frac{1500}{w^2}=15. \end{align*}
So, this is what we optimize. The endpoints of the domain for this function are \(x=0\) and \(x=R\text{.}\) To find the critical points, we differentiate:
\begin{align*} P'(x)&=4-\dfrac{2x}{\sqrt{R^2-x^2}}\\ P'(x)=0\iff 4&=\dfrac{2x}{\sqrt{R^2-x^2}}\\ x&=2\sqrt{R^2-x^2}\\ x^2&=4(R^2-x^2)\\ 5x^2&=4R^2\\ x&=\dfrac{2}{\sqrt{5}}R \end{align*}| \(c\) | \(0\) | \(R\) | \(\frac{2}{\sqrt{5}}R\) |
| type | endpoint | endpoint | critical point |
| \(P(c)\) | \(2R\) | \(4R\) | \(2\sqrt{5}R\) |
This is the function we want to maximize. Let’s find its critical points.
\begin{align*} V'(r)&=\dfrac{1}{2}\left(A-6\pi r^2\right)\\ V'(r)=0 &\iff A=6\pi r^2 \iff r=\sqrt{\dfrac{A}{6\pi}}\\ \end{align*}since negative values of \(r\) don’t make sense. At this critical point,
\begin{align*} V\left(\sqrt{\dfrac{A}{6\pi}}\right)&= \dfrac{1}{2}\left[A\left(\sqrt{\dfrac{A}{6\pi}}\right)-2\pi \left(\sqrt{\dfrac{A}{6\pi}}\right)^3\right]\\ &=\dfrac{1}{2}\left[\dfrac{A^{3/2}}{\sqrt{6\pi}}-\dfrac{2\pi A^{3/2}}{6\pi\sqrt{6\pi}}\right]\\ &=\dfrac{1}{2}\left[\dfrac{A^{3/2}}{\sqrt{6\pi}}-\dfrac{A^{3/2}}{3\sqrt{6\pi}}\right]\\ &=\dfrac{A^{3/2}}{3\sqrt{6\pi}}. \end{align*}
Finding all critical points:
\begin{align*} 0=A'(r)&=P-(\pi+4)r\\ r&=\dfrac{P}{\pi+4} \end{align*}
Using the product rule,
\begin{align*} V'(x)&=\dfrac{p}{2(1+p)}\left[x(-2px)+(A-px^2)\right]\\ &=\dfrac{p}{2(1+p)}\big[A-3px^2\big] \end{align*}We want to find the global max and min for this function, given the constraint \(0 \leq \ell \leq L\text{,}\) so we find its derivative:
\begin{align*} A'(\ell)&=\dfrac{\ell}{8} -\dfrac{L-\ell}{2\pi} =\dfrac{\pi+4}{8\pi}\ell-\dfrac{L}{2\pi}\\ \end{align*}Now, we find the critical point.
\begin{align*} A'(\ell)&=0\\ \frac{\pi+4}{8\pi}\ell&=\frac{L}{2\pi}\\ \ell&=\frac{4L}{\pi+4} \end{align*}| \(\ell\) | \(0\) | \(L\) | \(\textcolor{white}{\dfrac{.^.}{.}}\frac{4L}{\pi+4}\) |
| type | endpoint | endpoint | critical point |
| \(A(\ell)\) | \(\textcolor{white}{\dfrac{.^.}{.}}\frac{L^2}{4\pi}\) | \(\frac{L^2}{16}_{}\) | \(A\left(\frac{4L}{\pi+4}\right)\) |
where \(a\text{,}\) \(b\text{,}\) ad \(c\) are some constants. Remember, for rational functions, you can figure out the end behaviour by looking only at the terms with the highest degree--the others won’t matter, so we don’t bother finding them. From here, we divide the numerator and denominator by the highest power of \(x\) in the denominator, \(x^3\text{.}\)
\begin{align*} &=\lim_{x \to \pm \infty}\dfrac{2x^3+ax^2+bx+c}{3x^3-81}\left(\frac{\tfrac{1}{x^3}}{\tfrac{1}{x^3}}\right)\\ &=\lim_{x \to \pm \infty}\dfrac{2+\tfrac{a}{x}+\tfrac{b}{x^2}+\tfrac{c}{x^3}}{3-\tfrac{81}{x^3}}\\ &=\dfrac{2+0+0+0}{3-0}=\frac{2}{3} \end{align*}So, there’s no horizontal asymptote as \(x \to \infty\text{.}\)
\begin{align*} \lim_{x \to -\infty}10^{3x-7}&=\underbrace{\lim_{X \to -\infty}10^{X}}_{\mbox{let }X=3x-7}\\ &=\underbrace{\lim_{X' \to \infty}10^{-X'}}_{\mbox{let }X'=-X}\\ &=\lim_{X' \to \infty}\frac{1}{10^{X'}}\\ &=0 \end{align*}
Replacing \(X\) with \(\pi x\text{:}\)
\begin{align*} \tan(\pi x + \pi)&=\tan(\pi x)&&\mbox{for any $x$ in the domain of $\tan(\pi x)$}\\ \tan(\pi(x+1))&=\tan(\pi x)&&\mbox{for any $x$ in the domain of $\tan(\pi x)$}\\ f(x+1)&=f(x)&&\mbox{for any $x$ in the domain of $\tan(\pi x)$} \end{align*}Replacing \(X\) with \(3x\text{:}\)
\begin{align*} \tan(3x)&=\tan(3x+\pi)&&\mbox{for every $x$ in the domain of $\tan 3x$}\\ \tan(3x)&=\tan\left(3\left(x+\frac{\pi}{3}\right)\right)&&\mbox{for every $x$ in the domain of $\tan 3x$}\\ g(x)&=g\left(x+\frac{\pi}{3}\right)&&\mbox{for every $x$ in the domain of $\tan 3x$} \end{align*}Replacing \(X\) with \(4x\text{:}\)
\begin{align*} \sin(4x)&=\sin(4x+2\pi)&&\mbox{for every $x$ in the domain of $\sin(4x)$}\\ \sin(4x)&=\sin\left(4\left(x+\frac{\pi}{2}\right)\right)&&\mbox{for every $x$ in the domain of $\sin(4x)$}\\ h(x)&=h\left(x+\frac{\pi}{2}\right)&&\mbox{for every $x$ in the domain of $\sin(4x)$} \end{align*}| \(x\) | \((-\infty,2)\) | \(2\) | \((2,3)\) | \(3\) |
| \(f'(x)\) | positive | 0 | negative | DNE |
| \(f(x)\) | increasing | maximum | decreasing | endpoint |
This is undefined at \(x=3\text{.}\) Indeed,
\begin{align*} \ds\lim_{x \rightarrow 3^-} 3\dfrac{2-x}{2\sqrt{3-x}}&=-\infty, \end{align*}
| \(x\) | \((-\infty,0)\) | \(0\) | \((0,2)\) | \(2\) | \((2,\infty)\) |
| \(f'(x)\) | negative | DNE | positive | 0 | negative |
| \(f(x)\) | decreasing | vertical asymptote | increasing | local max | decreasing |
| \(x\) | \((-\infty,-\sqrt[3]{4})\) | \(-\sqrt[3]{4}\) | \((-\sqrt[3]{4},-1)\) | \(-1\) | \((-1,0)\) | \(0\) | \((0,\infty)\) |
| \(f'(x)\) | positive | 0 | negative | DNE | negative | 0 | positive |
| \(f(x)\) | increasing | l. max | decreasing | VA | decreasing | l. min | increasing |
| \(x\) | \((-\infty,-1)\) | \(-1\) | \((-1,0)\) | \(0\) | \((0,\sqrt[3]{2})\) | \(\sqrt[3]{2}\) | \((\sqrt[3]{2},\infty)\) |
| \(f''(x)\) | negative | DNE | positive | 0 | positive | 0 | negative |
| \(f(x)\) | concave down | VA | concave up | concave up | IP | concave down |
| \(x\) | \((-\infty,-\sqrt{3})\) | \(-\sqrt{3}\) | \((-\sqrt 3,-1)\) | \(-1\) | \((-1,0)\) |
| \(f'(x)\) | negative | 0 | positive | DNE | positive |
| \(f(x)\) | decreasing | local min | increasing | VA | increasing |
| \(x\) | \(0\) | \((0,1)\) | \(1\) | \((1,\sqrt 3)\) | \(\sqrt{3}\) | \((\sqrt{3},\infty)\) |
| \(f'(x)\) | 0 | positive | DNE | positive | 0 | negative |
| \(f(x)\) | increasing | VA | increasing | local max | decreasing |
| \(x\) | \((-\infty,-1)\) | \((-1,0)\) | 0 | \((0,1)\) | \((1,\infty)\) |
| \(f''(x)\) | positive | negative | 0 | positive | negative |
| \(f(x)\) | concave up | concave down | inflection point | concave up | concave down |
Differentiating again,
\begin{align*} \diff{^2}{x^2}\left\{\dfrac{x^2+3}{3(x+1)}\right\}&= \diff{}{x}\left\{\frac{x^2+2x-3}{3(x+1)^2} \right\}\\ &=\frac{3(x+1)^2(2x+2)-(x^2+2x-3)(6)(x+1)}{9(x+1)^4}\\ \amp\hskip3in\left(\frac{\div3(x+1)}{\div3(x+1)}\right)\\ &=\frac{(x+1)(2x+2)-2(x^2+2x-3)}{3(x+1)^3}\\ &=\frac{8}{3(x+1)^3}\qquad\mbox{where $x \neq -1$}\\ \mbox{so}\qquad f''(x)&=\left\{\begin{array}{ll} e^x&x \lt 0\\ DNE&x=0\\ \frac{8}{3(x+1)^3}&x \gt 0 \end{array}\right. \end{align*}| \(x\) | \((-\infty,0)\) | \(0\) | \((0,1)\) | 1 | \((1,\infty)\) |
| \(f'(x)\) | positive | DNE | negative | 0 | positive |
| \(f(x)\) | increasing | local max | decreasing | local min | increasing |
So, there is no horizontal asymptote to the right.
\begin{align*} \lim_{x \to -\infty} f(x)&=\lim_{x \to -\infty}e^x=0 \end{align*}
| \(x\) | \((-\infty,-1)\) | \(-1\) | \((-1,\frac{1}{2})\) | \(\frac{1}{2}\) | \((\frac{1}{2},\infty)\) |
| \(f'(x)\) | negative | 0 | positive | 0 | negative |
| \(f(x)\) | decreasing | local min | increasing | local max | decreasing |
| \(x\) | \((-\infty,-1)\) | \(-1\) | \((-1,1)\) | \(1\) | \((1,\infty)\) |
| \(f'(x)\) | negative | 0 | positive | 0 | negative |
| \(f(x)\) | decreasing | local min | increasing | local max | decreasing |
| \(x\) | \((-\infty,-\sqrt{3})\) | \(-\sqrt{3}\) | \((-\sqrt{3},0)\) | 0 | \((0,\sqrt{3})\) | \(\sqrt{3}\) | \((\sqrt{3},\infty)\) |
| \(f''(x)\) | negative | 0 | positive | 0 | negative | 0 | positive |
| \(f(x)\) | concave down | IP | concave up | IP | concave down | IP | concave up |
So, the critical points of \(f(x)\) occur when
\begin{align*} \cos x &= -\frac{1}{2}\\ x&=2\pi n \pm \frac{2\pi}{3}\mbox{ for any integer } n \end{align*}| \(x\) | \(\!\left(-\frac{2\pi}{3},\frac{2\pi}{3}\right)\!\) | \(\frac{2\pi}{3}\) | \(\left(\frac{2\pi}{3},\frac{4\pi}{3}\right)\) | \(\frac{4\pi}{3}\) | \(\left(\frac{4\pi}{3},\frac{8\pi}{3}\right)\) | \(\frac{8\pi}{3}\) | \(\left(\frac{8\pi}{3},\frac{10\pi}{3}\right)\) |
| \(f'(x)\) | positive | 0 | negative | 0 | positive | 0 | negative |
| \(f(x)\) | increasing\(\!\) | l. max | decreasing | l. min | increasing | l. max | decreasing |
| \(x\) | \(\left(0,\pi\right)\) | \(\pi\) | \(\left(\pi,2\pi\right)\) | \(2\pi\) | \(\left(2\pi,3\pi\right)\) | \(3\pi\!\) | \(\left(3\pi,4\pi\right)\) |
| \(f''(x)\) | negative | 0 | positive | 0 | negative | 0 | positive |
| \(f(x)\) | concave down | IP | concave up | IP\(\!\) | concave down | IP | concave up |
| \(x\) | \((-\pi,-\frac{5\pi}{6})\) | \(\left(-\frac{5\pi}{6},-\frac{\pi}{2}\right)\) | \(\left(-\frac{\pi}{2},-\frac{\pi}{6}\right)\) | \(\left(-\frac{\pi}{6},\frac{\pi}{2}\right)\) | \(\left(\frac{\pi}{2},\pi\right)\) |
| \(f'(x)\) | negative | positive | negative | positive | negative |
| \(f(x)\) | decreasing | increasing | decreasing | increasing | decreasing |
| \(x\) | \(-\dfrac{5}{6}\pi\) | \(-\dfrac{\pi}{2}\) | \(\dfrac{\pi}{2}\) | \(\dfrac{5}{6}\pi\) |
| \(\sin(x)\) | \(-\frac{1}{2}\) | \(-1\) | \(-\frac{1}{2}\) | \(1\) |
| \(\cos(2x)\) | \(\frac{1}{2}\) | \(-1\) | \(\frac{1}{2}\) | \(-1\) |
| \(f(x)\) | \(-3\) | \(-2\) | \(-3\) | \(6\) |
| \(x\) | \((-\infty,-2)\) | \(-2\) | \((-2,-1)\) | \(-1\) | \((-1,0)\) | 0 | \((0,\infty)\) |
| \(f'(x)\) | negative | 0 | positive | DNE | positive | DNE | negative |
| \(f(x)\) | decreasing | l. min | increasing | vertical | increasing | VA | decreasing |
| \(x\) | \(\left(-\infty,-2-\sqrt{1.5}\right)\) | \(-2-\sqrt{1.5}\) | \((-2-\sqrt{1.5},-1)\) | \(-1\) |
| \(f''(x)\) | negative | 0 | positive | DNE |
| \(f(x)\) | concave down | IP | concave up | IP |
| \(x\) | \((-1,-2+\sqrt{1.5})\) | \(-2+\sqrt{1.5}\) | \((-2+\sqrt{1.5},0)\) | \((0,\infty)\) |
| \(f''(x)\) | negative | 0 | positive | positive |
| \(f(x)\) | concave down | IP | concave up | concave up |
So,
\begin{align*} &\lim_{x \to -\infty} K(2x-x^2)e^{-x}=-\infty \end{align*}
| \(x\) | \((-\infty,0)\) | 0 | \((0,2)\) | 2 | \((2,\infty)\) |
| \(f'(x)\) | negative | 0 | positive | 0 | negative |
| \(f(x)\) | decreasing | local min | increasing | local max | decreasing |
| \(x\) | \((-\infty,2-\sqrt{2})\) | \(2-\sqrt{2}\) | \((2-\sqrt{2},2+\sqrt{2})\) | \(2+\sqrt 2\) | \((2+\sqrt 2,\infty)\) |
| \(f'(x)\) | positive | 0 | negative | 0 | positive |
| \(f(x)\) | concave up | IP | concave down | IP | concave up |
\((y,x)\) is on the graph of \(g\) if and only if \((x,y)\) is on the graph of \(f\text{.}\)
Using the chain rule,
\begin{align*} g'(f(x))\cdot f'(x)&=1\\ \end{align*}Since \(f'(x)=-e^{-x}=-f(x)\text{:}\)
\begin{align*} g'(f(x))\cdot f(x)&=-1\\ \end{align*}We plug in \(f(x)=\frac{1}{2}\text{.}\)
\begin{align*} g'\left(\frac{1}{2}\right)\cdot \frac{1}{2}&=-1\\ g'\left(\frac{1}{2}\right)&=-2 \end{align*}
We differentiate with respect to \(x\) using the chain rule.
\begin{align*} \diff{}{x}\left\{\cosh y(x)\right\}&=\diff{}{x}\{x\}\\ y'(x) \sinh y(x)&=1\\ \end{align*}We solve for \(y'(x)\text{.}\)
\begin{align*} y'(x) &=\dfrac{1}{\sinh y(x)}\\ \end{align*}We want to have our answer in terms of \(x\text{,}\) not \(y\text{.}\) We know that \(\cosh y=x\text{,}\) so if we can convert hyperbolic sine into hyperbolic cosine, we can get rid of \(y\text{.}\) Our tool for this is the identity, given in the question statement, \(\cosh^2(x)-\sinh^2(x)=1\text{.}\) This tells us \(\sinh^2(y)=1-\cosh^2(y)\text{.}\) Now we have to decide whether \(\sinh(y)\) is the positive or negative square root of \(1-\cosh^2(y)\) in our context. Looking at the graph of \(y(x)=\cosh^{-1}(x)\text{,}\) we see \(y'(x) \gt 0\text{.}\) So we use the positive square root:
\begin{align*} y'(x) & =\dfrac{1}{\sqrt{\cosh^2 y(x)-1}} =\dfrac{1}{\sqrt{x^2-1}} \end{align*}Since plugging in \(x=0\) makes both the numerator and the denominator equal to zero, this is a candidate for l’Hôspital’s Rule. However, a much easier way is to simplify the trig first.
\begin{align*} \lim_{x\to0}\frac{\tan x\cdot (x^2+5)}{\sin x \cdot e^x}&= \lim_{x\to0}\frac{\sin x\cdot (x^2+5)}{\cos x \cdot\sin x \cdot e^x}\\ &= \lim_{x\to0}\frac{x^2+5}{\cos x \cdot e^x}\\ &=\frac{0^2+5}{\cos(0)\cdot e^0}=5 \end{align*}Here, we have the indeterminate form \(\frac{\infty}{\infty}\text{,}\) so l’Hôpital’s Rule applies. However, if we try to use it here, we quickly get a huge mess. Instead, remember how we dealt with these kinds of limits in the past: factor out the highest power of the denominator, which is \(x\text{.}\)
\begin{align*} &=\lim_{x \to \infty} \frac{x\left(x-1+\frac{1}{x}\right)}{\sqrt{x^2(2+\frac{1}{x^2})}+\sqrt{x^2(1+\frac{1}{x})}}\\ &=\lim_{x \to \infty} \frac{x\left(x-1+\frac{1}{x}\right)}{x\left(\sqrt{2+\frac{1}{x^2}}+\sqrt{1+\frac{1}{x}}\right)}\\ &=\lim_{x \to \infty} \underbrace{\frac{x-1+\frac{1}{x}}{\sqrt{2+\frac{1}{x^2}}+\sqrt{1+\frac{1}{x}}}}_{\atp {\mathrm{num}\to \infty} {\mathrm{den}\to \sqrt{2}+1}}\\ &=\infty \end{align*}We have the indeterminate form \(0^0\text{.}\) We want to use l’Hôpital, but we need a different indeterminate form. So, we’ll use logarithms.
\begin{align*} \lim_{x \to 0^+}\log y &=\lim_{x \to 0^+}\log(x^x)=\lim_{x \to 0^+}x\log x\\ \end{align*}Now we have the indeterminate form \(0 \cdot \infty\text{,}\) so we need one last adjustment before we can use l’Hôpital’s Rule.
\begin{align*} \lim_{x \to 0^+} y &=\lim_{x \to 0^+}\underbrace{\frac{\log x}{\frac{1}{x}}}_{\atp {\mathrm{num}\to 0} {\mathrm{den} \to -\infty}} =\lim_{x \to 0^+}\frac{\frac{1}{x}}{\frac{-1}{x^2}} =\lim_{x \to 0^+}-x=0\\ \end{align*}Now, we can figure out what happens to our original function, \(y\text{:}\)
\begin{align*} \lim_{x \to 0^+} y &=\lim_{x \to 0^+} e^{\log y} = e^0=1 \end{align*}Now, we’re ready to figure out our original limit.
\begin{align*} \lim_{x\to 0} y &= \lim_{x \to 0} e^{\log y}=e^0=1 \end{align*}This isn’t exactly right, so we modify it by multiplying by a constant.
\begin{align*} \diff{}{x}\left\{\frac{1}{5}e^{5x+11}\right\}&=e^{5x+11} \end{align*}So, an antiderivative of \(7\cos(13x)\) is \(\dfrac{7}{13}\sin(13x)\text{.}\)
\begin{align*} \diff{}{x}\{\cos x\}&=-\sin x\\ \diff{}{x}\{-\cos x\}&=\sin x\\ \diff{}{x}\{-\cos (5x)\}&=5\sin (5x)\\ \diff{}{x}\left\{-\frac{3}{5}\cos (5x)\right\}&=\left(\frac{3}{5}\right)5\sin (5x)=3\sin(5x) \end{align*}This gives us a first guess for our antiderivative: perhaps \(\arctan(5x)\) will work. We test it by differentiating, making sure we don’t forget the chain rule.
\begin{align*} \diff{}{x}\left\{\arctan(\textcolor{red}{5x})\right\}&=\frac{1}{1+\left(\textcolor{red}{5x}\right)^2}\cdot\textcolor{red}{5}\\ \end{align*}We’re close to \(f(x)\text{,}\) but we’ve multiplied by 5. That’s easy to take care of: we can divide our guess by 5.
\begin{align*} \diff{}{x}\left\{\frac{1}{5}\arctan(\textcolor{red}{5x})\right\}&= \frac{1}{5}\left(\frac{1}{1+\left(\textcolor{red}{5x}\right)^2}\right)\cdot\textcolor{red}{5}\\ &=\frac{1}{1+25x^2}=f(x) \end{align*}Use the fact that \(f(1)=10\) to solve for \(C\text{.}\)
\begin{align*} 10&=1-\frac{9}{2}+4+C\\ C&=\frac{19}{2}\\ \end{align*}All together,
\begin{align*} f(x)&=x^3-\frac{9}{2}x^2+4x+\frac{19}{2}. \end{align*}This is close to \(f'(x)\text{,}\) but we need to divide by 2.
\begin{align*} \diff{}{x}\left\{\frac{1}{2}\sin(2x)\right\}&=\cos(2x)\\ \end{align*}So, \(f(x)=\dfrac{1}{2}\sin(2x)+C\) for some constant \(C\text{.}\) We can find \(C\) using the given information \(f(\pi)=\pi\text{.}\)
\begin{align*} \pi=f(\pi)&=\frac{1}{2}\sin(2\pi)+C\\ \pi&=C \end{align*}Antidifferentiating,
\begin{align*} A(t)&=75000e^{\tfrac{t}{50}}+C \end{align*}Antidifferentiating,
\begin{align*} P(t)&=-\frac{12}{\pi}\cos\left(\frac{\pi}{24}t\right)+0.25t+C\\ \end{align*}Since \(P(0)\) is the amount of energy consumed after 0 hours, \(P(0)=0\text{,}\) so
\begin{align*} 0=P(0)&=-\frac{12}{\pi}+C\\ C&=\frac{12}{\pi}\\ P(t)&=\frac{12}{\pi}\left[1-\cos\left(\frac{\pi}{24}t\right)\right]+0.25t\\ \end{align*}After 24 hours, your energy consumed is
\begin{align*} P(24)&=\frac{12}{\pi}\left[1-\cos\left(\pi\right)\right]+0.25(24)\\ &=\frac{24}{\pi}+6\approx13.6\; \mathrm{kWh} \end{align*}Then, since the denominator of \(f(x)\) is \((x^2+1)^2\text{,}\) we might guess \(v(x)=(x^2+1)\text{.}\) That leaves \(u(x)=e^x\text{.}\)
\begin{align*} \diff{}{x}\left\{\frac{e^x}{x^2+1}\right\}&=\frac{(x^2+1)e^x-e^x(2x)}{(x^2+1)^2}=f(x) \end{align*}This differs from \(f(x)\) only by a constant multiple.
\begin{align*} \diff{}{x}\left\{-\frac{5}{2}\cos\left(x^2\right)\right\}&=\frac{5}{2}\sin\left(x^2\right)\cdot2x=5x\sin(x^2)=f(x) \end{align*}What we really want under that square root, instead of \(\dfrac{x^2}{3}\text{,}\) is simply \(x^2\text{.}\) We can get close: we can get something squared.
\begin{align*} f(x)&=\frac{7}{\sqrt{3}}\left(\frac{1}{\sqrt{1-\left(\frac{x}{\sqrt{3}}\right)^2}}\right)\\ \end{align*}Now, the thing that’s squared isn’t \(x\text{,}\) it’s \(\dfrac{x}{\sqrt{3}}\text{.}\) This gives us a first guess for an antiderivative: perhaps \(F(x)=\dfrac{7}{\sqrt{3}}\arcsin\left(\dfrac{x}{\sqrt{3}}\right)\) will work. Let’s try it! Remember to use the chain rule when you differentiate.
\begin{align*} \diff{}{x}\left\{\frac{7}{\sqrt{3}}\arcsin\left(\textcolor{red}{\frac{x}{\sqrt{3}}}\right)\right\}&= \frac{7}{\sqrt{3}}\left(\frac{1}{\sqrt{1-\left(\textcolor{red}{\frac{x}{\sqrt{3}}}\right)^2}}\right)\cdot\textcolor{red}{\frac{1}{\sqrt{3}}}\\ \end{align*}We’re very close! We’re only off by a constant, and those are easy to fix. We’re dividing by \(\sqrt{3}\) when we differentiate, so let’s multiply our function by \(\sqrt{3}\text{.}\)
\begin{align*} \diff{}{x}\left\{7\arcsin\left(\textcolor{red}{\frac{x}{\sqrt{3}}}\right)\right\}&= 7\left(\frac{1}{\sqrt{1-\left(\textcolor{red}{\frac{x}{\sqrt{3}}}\right)^2}}\right)\cdot\textcolor{red}{\frac{1}{\sqrt{3}}}\\ &=\frac{7}{\sqrt{3}\sqrt{1-\left(\frac{x}{\sqrt{3}}\right)^2}}\\ &=\frac{7}{\sqrt{3-x^2}}=f(x) \end{align*}
When \(H=0\text{,}\) there is no solid, so \(V(0)=0\text{.}\) Therefore,
\begin{align*} V(H)&=2\pi\left(\frac{1}{5}H^5+\frac{2}{3}H^3+H\right). \end{align*}
