1. [T] Find expressions for coshx+sinhxcoshx+sinhx and coshx−sinhx.coshx−sinhx. Use a calculator to graph these functions and ensure your expression is correct.
2. From the definitions of cosh(x)cosh(x) and sinh(x),sinh(x), find their antiderivatives.
3. Show that cosh(x)cosh(x) and sinh(x)sinh(x) satisfy y″=y.
4. Use the quotient rule to verify that tanh(x)′=sech2(x).
5. Derive cosh2(x)+sinh2(x)=cosh(2x) from the definition.
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.
8. Take the derivative of the previous expression to find an expression for cosh(x+y).
For the following 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)
10. [T] sinh(x2)
11. [T] 1cosh(x)
12. [T] sinh(ln(x))
13. [T] cosh2(x)+sinh2(x)
14. [T] cosh2(x)−sinh2(x)
15. [T] tanh(√x2+1)
16. [T] 1+tanh(x)1−tanh(x)
17. [T] sinh6(x)
18. [T] ln(sech(x)+tanh(x))
For the following exercises (19-29), find the antiderivatives for the given functions.
19. cosh(2x+1)
20. tanh(3x+2)
21. xcosh(x2)
23. 3x3tanh(x4)
24. cosh2(x)sinh(x)
25. tanh2(x)sech2(x)
26. sinh(x)1+cosh(x)
27. coth(x)
28. cosh(x)+sinh(x)
29. (cosh(x)+sinh(x))n
For the following exercises (30-36), find the derivatives for the functions.
30. tanh−1(4x)
31. sinh−1(x2)
32. sinh−1(cosh(x))
33. cosh−1(x3)
34. tanh−1(cos(x))
35. esinh−1(x)
36. ln(tanh−1(x))
For the following exercises (37-43), find the antiderivatives for the functions.
37. ∫dx4−x2
38. ∫dxa2−x2
39. ∫dx√x2+1
40. ∫xdx√x2+1
41. ∫−dxx√1−x2
42. ∫ex√e2x−1
43. ∫−2xx4−1
For the following exercises (44-46), use the fact that a falling body with friction equal to velocity squared obeys the equation dv/dt=g−v2.
44. Show that v(t)=√gtanh(√gt) satisfies this equation.
45. Derive the previous expression for v(t) by integrating dvg−v2=dt.
46. [T] Estimate how far a body has fallen in 12 seconds by finding the area underneath the curve of v(t).
For the following exercises (47-49), use this scenario: A cable hanging under its own weight has a slope S=dy/dx that satisfies dS/dx=c√1+S2. The constant c is the ratio of cable density to tension.
47. Show that S=sinh(cx) satisfies this equation.
48. Integrate dy/dx=sinh(cx) to find the cable height y(x) if y(0)=1/c.
49. Sketch the cable and determine how far down it sags at x=0.
For the following exercises (50-53), solve each problem.
50. [T] A chain hangs from two posts 2 m apart to form a catenary described by the equation y=2cosh(x/2)−1. Find the slope of the catenary at the left fence post.
51. [T] A chain hangs from two posts four meters apart to form a catenary described by the equation y=4cosh(x/4)−3. Find the total length of the catenary (arc length).
52. [T] A high-voltage power line is a catenary described by y=10cosh(x/10). Find the ratio of the area under the catenary to its arc length. What do you notice?
53. A telephone line is a catenary described by y=acosh(x/a). Find the ratio of the area under the catenary to its arc length. Does this confirm your answer for the previous question?
54. Prove the formula for the derivative of y=sinh−1(x) by differentiating x=sinh(y). (Hint: Use hyperbolic trigonometric identities.)
55. Prove the formula for the derivative of y=cosh−1(x) by differentiating x=cosh(y).
(Hint: Use hyperbolic trigonometric identities.)
56. Prove the formula for the derivative of y=sech−1(x) by differentiating x=sech(y). (Hint: Use hyperbolic trigonometric identities.)
57. Prove that (cosh(x)+sinh(x))n=cosh(nx)+sinh(nx).
58. Prove the expression for sinh−1(x). Multiply x=sinh(y)=(1/2)(ey−e−y) by 2ey and solve for y. Does your expression match the textbook?
59. Prove the expression for cosh−1(x). Multiply x=cosh(y)=(1/2)(ey−e−y) by 2ey and solve for y. Does your expression match the textbook?
Candela Citations
- Calculus Volume 1. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: https://openstax.org/details/books/calculus-volume-1. License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at https://openstax.org/books/calculus-volume-1/pages/1-introduction