Chapter 6

Use l'Hôpital's Rule to evaluate the following limits. \(\lim _{x \rightarrow 1^{-}} \frac{\tanh ^{-1} x}{\tan (\pi x / 2)}\)

Use l'Hôpital's Rule to evaluate the following limits. \(\lim _{x \rightarrow 0^{+}}(\tanh x)^{x}\)

Evaluate the following integrals. \(\int \frac{\cosh z}{\sinh ^{2} z} d z\)

Evaluate the following integrals. \(\int \frac{\cos \theta}{9-\sin ^{2} \theta} d \theta\)

Evaluate the following integrals. \(\int_{5 / 12}^{3 / 4} \frac{\sinh ^{-1} x}{\sqrt{x^{2}+1}} d x\)

Evaluate the following integrals. \(\int_{25}^{225} \frac{d x}{\sqrt{x^{2}+25 x}}(\text { Hint: } \sqrt{x^{2}+25 x}=\sqrt{x} \sqrt{x+25}\)

Use Newton's method to find all local extreme values of \(f(x)=x \operatorname{sech} x\).

When an object falling from rest encounters air resistance proportional to the square of its velocity, the distance it falls (in meters) after \(t\) seconds is given by \(d(t)=\frac{m}{k} \ln [\cosh (\sqrt{\frac{k g}{m}} t)],\) where \(m\) is the mass of the object in kilograms, \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, and \(k\) is a physical constant. a. A BASE jumper \((m=75 \mathrm{kg})\) leaps from a tall cliff and performs a ten-second delay (she free-falls for 10 s and then opens her chute). How far does she fall in \(10 \mathrm{s} ?\) Assume \(k=0.2\) b. How long does it take for her to fall the first \(100 \mathrm{m} ?\) The second 100 \(\mathrm{m} ?\) What is her average velocity over each of these intervals?

Refer to Exercise \(95,\) which gives the position function for a falling body. Use \(m=75 \mathrm{kg}\) and \(k=0.2\) a. Confirm that the base jumper's velocity \(t\) seconds after $$\text { jumping is } v(t)=d^{\prime}(t)=\sqrt{\frac{m g}{k}} \tanh (\sqrt{\frac{k g}{m}} t)$$ b. How fast is the BASE jumper falling at the end of a 10 s delay? c. How long does it take for the BASE jumper to reach a speed of \(45 \mathrm{m} / \mathrm{s} \text { (roughly } 100 \mathrm{mi} / \mathrm{hr}) ?\)

Refer to Exercises 95 and 96. a. Compute a jumper's terminal velocity, which is defined as \(\lim _{t \rightarrow \infty} v(t)=\lim _{t \rightarrow \infty} \sqrt{\frac{m g}{k}} \tanh (\sqrt{\frac{k g}{m}} t)\) b. Find the terminal velocity for the jumper in Exercise 96 \((m=75 \mathrm{kg} \text { and } k=0.2)\) c. How long does it take for any falling object to reach a speed equal to \(95 \%\) of its terminal velocity? Leave your answer in terms of \(k, g,\) and \(m\) d. How tall must a cliff be so that the BASE jumper \((m=75 \mathrm{kg}\) and \(k=0.2\) ) reaches \(95 \%\) of terminal velocity? Assume that the jumper needs at least \(300 \mathrm{m}\) at the end of free fall to deploy the chute and land safely.

Hyperbolic functions are useful in solving differential equations. Show that the functions \(y=A \sinh k x\) and \(y=B \cosh k x,\) where \(A, B,\) and \(k\) are constants, satisfy the equation \(y^{\prime \prime}(x)-k^{2} y(x)=0\).

When the catenary \(y=a \cosh (x / a)\) is rotated around the \(x\) -axis, it sweeps out a surface of revolution called a catenoid. Find the area of the surface generated when \(y=\cosh x\) on \([-\ln 2, \ln 2]\) is rotated around the \(x\) -axis.

Verify the following identities. \(\sinh \left(\cosh ^{-1} x\right)=\sqrt{x^{2}-1},\) for \(x \geq 1\)

Verify the following identities. \(\cosh \left(\sinh ^{-1} x\right)=\sqrt{x^{2}+1},\) for all \(x\)

Verify the following identities. \(\cosh (x+y)=\cosh x \cosh y+\sinh x \sinh y\)

Verify the following identities. \(\sinh (x+y)=\sinh x \cosh y+\cosh x \sinh y\)

Show that \(\cosh ^{-1}(\cosh x)=|x|\) by using the formula \(\cosh ^{-1} t=\ln (t+\sqrt{t^{2}-1})\) and by considering the cases \(x \geq 0\) and \(x<0\).

a. The definition of the inverse hyperbolic cosine is \(y=\cosh ^{-1} x \Leftrightarrow x=\cosh y,\) for \(x \geq 1,0 \leq y<\infty .\) Use implicit differentiation to show that \(\frac{d}{d x}\left(\cosh ^{-1} x\right)=\) \(1 / \sqrt{x^{2}-1}\). b. Differentiate \(\sinh ^{-1} x=\ln (x+\sqrt{x^{2}+1})\) to show that \(\frac{d}{d x}\left(\sinh ^{-1} x\right)=1 / \sqrt{x^{2}+1}\).

There are several ways to express the indefinite integral of sech \(x\). a. Show that \(\int \operatorname{sech} x d x=\tan ^{-1}(\sinh x)+C\) (Theorem 9 ). (Hint: Write sech \(x=\frac{1}{\cosh x}=\frac{\cosh x}{\cosh ^{2} x}=\frac{\cosh x}{1+\sinh ^{2} x},\) and then make a change of variables.) b. Show that \(\int \operatorname{sech} x d x=\sin ^{-1}(\tanh x)+C .\) (Hint: Show that sech \(x=\frac{\operatorname{sech}^{2} x}{\sqrt{1-\tanh ^{2} x}}\) and then make a change of variables.) c. Verify that \(\int \operatorname{sech} x d x=2 \tan ^{-1}\left(e^{x}\right)+C\) by proving \(\frac{d}{d x}\left(2 \tan ^{-1}\left(e^{x}\right)\right)=\operatorname{sech} x\).

Carry out the following steps to derive the formula \(\int \operatorname{csch} x d x=\ln |\tanh (x / 2)|+C\) (Theorem 9). a. Change variables with the substitution \(u=x / 2\) to show that $$\int \operatorname{csch} x d x=\int \frac{2 d u}{\sinh 2 u}$$. b. Use the identity for \(\sinh 2 u\) to show that \(\frac{2}{\sinh 2 u}=\frac{\operatorname{sech}^{2} u}{\tanh u}\). c. Change variables again to determine \(\int \frac{\operatorname{sech}^{2} u}{\tanh u} d u,\) and then express your answer in terms of \(x\).

Recall that the inverse hyperbolic tangent is defined as \(y=\tanh ^{-1} x \Leftrightarrow x=\tanh y,\) for \(-1

Use the substitution \(u=x^{r}\) to show that \(\int \frac{d x}{x \sqrt{1-x^{2 r}}}=-\frac{1}{r} \operatorname{sech}^{-1} x^{r}+C,\) for \(r>0,\) and \(0