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

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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

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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.