Chapter 6 | Applications of Integration
755
414. ∫ dx
427. [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? 428. 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? 429. Prove the formula for the derivative of y = sinh −1 ( x ) by differentiating x = sinh( y ). ( Hint: Use hyperbolic trigonometric identities.) 430. Prove the formula for the derivative of y =cosh −1 ( x ) by differentiating x =cosh( y ). ( Hint: Use hyperbolic trigonometric identities.) 431. Prove the formula for the derivative of y = sech −1 ( x ) by differentiating x = sech( y ). ( Hint: Use hyperbolic trigonometric identities.) 432. Prove that ⎛ ⎝ cosh( x )+sinh( x ) ⎞ ⎠ n =cosh( nx )+sinh( nx ). 433. Prove the expression for sinh −1 ( x ). Multiply x = sinh( y ) = (1/2) ⎛ ⎝ e y − e − y ⎞ ⎠ by 2 e y and solve for y . Does your expression match the textbook? 434. Prove the expression for cosh −1 ( x ). Multiply x =cosh( y ) = (1/2) ⎛ ⎝ e y − e − y ⎞ ⎠ by 2 e y and solve for y . Does your expression match the textbook?
x 2 +1
415. ∫ xdx x 2 +1 416. ∫ − dx
x 1− x 2
417. ∫ e x
e 2 x −1
418. ∫ − 2 x
x 4 −1 For the following exercises, use the fact that a falling body with friction equal to velocity squared obeys the equation dv / dt = g − v 2 .
Show that v ( t ) = g tanh(( g ) t ) satisfies this
419.
equation. 420.
Derive the previous expression for v ( t ) by
dv g − v 2
= dt .
integrating
421. [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, use this scenario: A cable hanging under its own weight has a slope S = dy / dx that satisfies dS / dx = c 1+ S 2 . The constant c is the ratio of cable density to tension. 422. Show that S = sinh( cx ) satisfies this equation. 423. Integrate dy / dx = sinh( cx ) to find the cable height y ( x ) if y (0) =1/ c . 424. Sketch the cable and determine how far down it sags at x =0. For the following exercises, solve each problem. 425. [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. 426. [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).
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