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Mathematics LibreTexts

2.10E: Exercises for Section 2.10

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1) [T] Find expressions for coshx+sinhx and coshxsinhx. Use a calculator to graph these functions and ensure your expression is correct.

Answer
ex and ex

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 ddx(tanh(x))=sech2(x).

5) Derive cosh2(x)+sinh2(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
3sinh(3x+1)

10) [T] sinh(x2)

11) [T] 1cosh(x)

Answer
tanh(x)sech(x)

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

13) [T] cosh2(x)+sinh2(x)

Answer
4cosh(x)sinh(x)

14) [T] cosh2(x)sinh2(x)

15) [T] tanh(x2+1)

Answer
xsech2(x2+1)x2+1

16) [T] 1+tanh(x)1tanh(x)

17) [T] sinh6(x)

Answer
6sinh5(x)cosh(x)

18) [T] ln(sech(x)+tanh(x))

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

19) cosh(2x+1)

Answer
12sinh(2x+1)+C

20) tanh(3x+2)

21) xcosh(x2)

Answer
12sinh2(x2)+C

22) 3x3tanh(x4)

23) cosh2(x)sinh(x)

Answer
13cosh3(x)+C

24) tanh2(x)sech2(x)

25) 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) tanh1(4x)

Answer
4116x2

30) sinh1(x2)

31) sinh1(cosh(x))

Answer
sinh(x)cosh2(x)+1

32) cosh1(x3)

33) tanh1(cos(x))

Answer
csc(x)

34) esinh1(x)

35) ln(tanh1(x))

Answer
1(x21)tanh1(x)

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

36) dx4x2

37) dxa2x2

Answer
1atanh1(xa)+C

38) dxx2+1

39) xdxx2+1

Answer
x2+1+C

40) dxx1x2

41) exe2x1

Answer
cosh1(ex)+C

42) 2xx41

In exercises 43 - 45, use the fact that a falling body with friction equal to velocity squared obeys the equation dvdt=gv2.

43) Show that v(t)=gtanh(gt) satisfies this equation.

Answer
Answers may vary

44) Derive the previous expression for v(t) by integrating dvgv2=dt.

45) [T] Estimate how far a body has fallen in 12seconds 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=dydx that satisfies dSdx=c1+S2. The constant c is the ratio of cable density to tension.

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

47) Integrate dydx=sinh(cx) to find the cable height y(x) if y(0)=1/c.

Answer
y=1ccosh(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 2m 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.

Answer
0.521095

50) [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).

51) [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?

Answer
10

52) 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?

53) Prove the formula for the derivative of y=sinh1(x) by differentiating x=sinh(y).

(Hint: Use hyperbolic trigonometric identities.)

54) Prove the formula for the derivative of y=cosh1(x) by differentiating x=cosh(y).

(Hint: Use hyperbolic trigonometric identities.)

55) Prove the formula for the derivative of y=sech1(x) by differentiating x=sech(y).

(Hint: Use hyperbolic trigonometric identities.)

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

57) Prove the expression for sinh1(x). Multiply x=sinh(y)=eyey2 by 2ey and solve for y. Does your expression match the textbook?

58) Prove the expression for cosh1(x). Multiply x=cosh(y)=ey+ey2 by 2ey 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.


This page titled 2.10E: Exercises for Section 2.10 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.

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