2.10E: Exercises for Section 2.10

1) [T] Find expressions for $\cosh x+\sinh x$ and $\cosh x−\sinh x.$ Use a calculator to graph these functions and ensure your expression is correct.

Answer

$e^x$ and $e^{−x}$

2) From the definitions of $\cosh(x)$ and $\sinh(x)$, find their antiderivatives.

3) Show that $\cosh(x)$ and $\sinh(x)$ satisfy $y''=y$.

Answer

Answers may vary

4) Use the quotient rule to verify that $\dfrac{d}{dx}\big(\tanh(x)\big)=\text{sech}^2(x).$

5) Derive $\cosh^2(x)+\sinh^2(x)=\cosh(2x)$ from the definition.

Answer

Answers may vary

6) Take the derivative of the previous expression to find an expression for $\sinh(2x)$.

7) Prove $\sinh(x+y)=\sinh(x)\cosh(y)+\cosh(x)\sinh(y)$ by changing the expression to exponentials.

Answer

Answers may vary

8) Take the derivative of the previous expression to find an expression for $\cosh(x+y).$

In exercises 9 - 18, find the derivatives of the given functions and graph along with the function to ensure your answer is correct.

9) [T] $\cosh(3x+1)$

Answer

$3\sinh(3x+1)$

10) [T] $\sinh(x^2)$

11) [T] $\dfrac{1}{\cosh(x)}$

Answer

$−\tanh(x)\text{sech}(x)$

12) [T] $\sinh(\ln(x))$

13) [T] $\cosh^2(x)+\sinh^2(x)$

Answer

$4\cosh(x)\sinh(x)$

14) [T] $\cosh^2(x)−\sinh^2(x)$

15) [T] $\tanh(\sqrt{x^2+1})$

Answer

$\dfrac{x\text{sech}^2(\sqrt{x^2+1})}{\sqrt{x^2+1}}$

16) [T] $\dfrac{1+\tanh(x)}{1−\tanh(x)}$

17) [T] $\sinh^6(x)$

Answer

$6\sinh^5(x)\cosh(x)$

18) [T] $\ln(\text{sech}(x)+\tanh(x))$

In exercises 19 - 28, find the antiderivatives for the given functions.

19) $\cosh(2x+1)$

Answer

$\frac{1}{2}\sinh(2x+1)+C$

20) $\tanh(3x+2)$

21) $x\cosh(x^2)$

Answer

$\frac{1}{2}\sinh^2(x^2)+C$

22) $3x^3\tanh(x^4)$

23) $\cosh^2(x)\sinh(x)$

Answer

$\frac{1}{3}\cosh^3(x)+C$

24) $\tanh^2(x)\text{sech}^2(x)$

25) $\dfrac{\sinh(x)}{1+\cosh(x)}$

Answer

$\ln(1+\cosh(x))+C$

26) $\coth(x)$

27) $\cosh(x)+\sinh(x)$

Answer

$\cosh(x)+\sinh(x)+C$

28) $(\cosh(x)+\sinh(x))^n$

In exercises 29 - 35, find the derivatives for the functions.

29) $\tanh^{−1}(4x)$

Answer

$\dfrac{4}{1−16x^2}$

30) $\sinh^{−1}(x^2)$

31) $\sinh^{−1}(\cosh(x))$

Answer

$\dfrac{\sinh(x)}{\sqrt{\cosh^2(x)+1}}$

32) $\cosh^{−1}(x^3)$

33) $\tanh^{−1}(\cos(x))$

Answer

$−\csc(x)$

34) $e^{\sinh^{−1}(x)}$

35) $\ln(\tanh^{−1}(x))$

Answer

$−\dfrac{1}{(x^2−1)\tanh^{−1}(x)}$

In exercises 36 - 42, find the antiderivatives for the functions.

36) $\displaystyle ∫\frac{dx}{4−x^2}$

37) $\displaystyle ∫\frac{dx}{a^2−x^2}$

Answer

$\dfrac{1}{a}\tanh^{−1}\left(\dfrac{x}{a}\right)+C$

38) $\displaystyle ∫\frac{dx}{\sqrt{x^2+1}}$

39) $\displaystyle ∫\frac{xdx}{\sqrt{x^2+1}}$

Answer

$\sqrt{x^2+1}+C$

40) $\displaystyle ∫−\frac{dx}{x\sqrt{1−x^2}}$

41) $\displaystyle ∫\frac{e^x}{\sqrt{e^{2x}−1}}$

Answer

$\cosh^{−1}(e^x)+C$

42) $\displaystyle ∫−\frac{2x}{x^4−1}$

In exercises 43 - 45, use the fact that a falling body with friction equal to velocity squared obeys the equation $\dfrac{dv}{dt}=g−v^2$.

43) Show that $v(t)=\sqrt{g}\tanh(\sqrt{g}t)$ satisfies this equation.

Answer

Answers may vary

44) Derive the previous expression for $v(t)$ by integrating $\dfrac{dv}{g−v^2}=dt$.

45) [T] Estimate how far a body has fallen in $12$seconds by finding the area underneath the curve of $v(t)$.

Answer

$37.30$

In exercises 46 - 48, use this scenario: A cable hanging under its own weight has a slope $S=\dfrac{dy}{dx}$ that satisfies $\dfrac{dS}{dx}=c\sqrt{1+S^2}$. The constant $c$ is the ratio of cable density to tension.

46) Show that $S=\sinh(cx)$ satisfies this equation.

47) Integrate $\dfrac{dy}{dx}=\sinh(cx)$ to find the cable height $y(x)$ if $y(0)=1/c$.

Answer

$y=\frac{1}{c}\cosh(cx)$

48) Sketch the cable and determine how far down it sags at $x=0$.

In exercises 49 - 52, solve each problem.

49) [T] A chain hangs from two posts $2$m apart to form a catenary described by the equation $y=2\cosh(x/2)−1$. Find the slope of the catenary at the left fence post.

Answer

$−0.521095$

50) [T] A chain hangs from two posts four meters apart to form a catenary described by the equation $y=4\cosh(x/4)−3.$ Find the total length of the catenary (arc length).

51) [T] A high-voltage power line is a catenary described by $y=10\cosh(x/10)$. Find the ratio of the area under the catenary to its arc length. What do you notice?

Answer

$10$

52) A telephone line is a catenary described by $y=a\cosh(x/a).$ Find the ratio of the area under the catenary to its arc length. Does this confirm your answer for the previous question?

53) Prove the formula for the derivative of $y=\sinh^{−1}(x)$ by differentiating $x=\sinh(y).$

(Hint: Use hyperbolic trigonometric identities.)

54) Prove the formula for the derivative of $y=\cosh^{−1}(x)$ by differentiating $x=\cosh(y).$

(Hint: Use hyperbolic trigonometric identities.)

55) Prove the formula for the derivative of $y=\text{sech}^{−1}(x)$ by differentiating $x=\text{sech}(y).$

(Hint: Use hyperbolic trigonometric identities.)

56) Prove that $\cosh(x)+\sinh(x))^n=\cosh(nx)+\sinh(nx).$

57) Prove the expression for $\sinh^{−1}(x).$ Multiply $x=\sinh(y)=\dfrac{e^y−e^{−y}}{2}$ by $2e^y$ and solve for $y$. Does your expression match the textbook?

58) Prove the expression for $\cosh^{−1}(x).$ Multiply $x=\cosh(y)=\dfrac{e^y+e^{−y}}{2}$ by $2e^y$ and solve for $y$. Does your expression match the textbook?

Contributors

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.