2.10E: Exercises for Section 2.10
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- Sep 9, 2021
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1) [T] Find expressions for coshx+sinhx and coshx−sinhx. Use a calculator to graph these functions and ensure your expression is correct.
- Answer
- ex 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 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)1−tanh(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) tanh−1(4x)
- Answer
- 41−16x2
30) sinh−1(x2)
31) sinh−1(cosh(x))
- Answer
- sinh(x)√cosh2(x)+1
32) cosh−1(x3)
33) tanh−1(cos(x))
- Answer
- −csc(x)
34) esinh−1(x)
35) ln(tanh−1(x))
- Answer
- −1(x2−1)tanh−1(x)
In exercises 36 - 42, find the antiderivatives for the functions.
36) ∫dx4−x2
37) ∫dxa2−x2
- Answer
- 1atanh−1(xa)+C
38) ∫dx√x2+1
39) ∫xdx√x2+1
- Answer
- √x2+1+C
40) ∫−dxx√1−x2
41) ∫ex√e2x−1
- Answer
- cosh−1(ex)+C
42) ∫−2xx4−1
In exercises 43 - 45, use the fact that a falling body with friction equal to velocity squared obeys the equation dvdt=g−v2.
43) Show that v(t)=√gtanh(√gt) satisfies this equation.
- Answer
- Answers may vary
44) Derive the previous expression for v(t) by integrating dvg−v2=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=c√1+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=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=sech−1(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 sinh−1(x). Multiply x=sinh(y)=ey−e−y2 by 2ey and solve for y. Does your expression match the textbook?
58) Prove the expression for cosh−1(x). Multiply x=cosh(y)=ey+e−y2 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.