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17. x Section 7.4
sec4 y
/4 1, so the length is
/4 24. The area is 300 times the length of the arch. sec2 y dy
/4 1
/4 (sec4 y 1) dy y
25 /4 2 sin tan y
/4 2. to 1
25 x , so the length is 50 2 x sin2 dx, which evaluates, using NINT, 2 50 73.185. Multiply that by 300, then by $1.75 to obtain 18. y
1 3x 4 1
2 1, so the length is
1 the cost (rounded to the nearest dollar): $38,422. 3x 2 dx
1 3
1 2 (3x 4 1) dx
2 3 x3 19. (a) y So y 7 3 . 3 25. For track 1: y1 y1
10 10 5 5 0 at x 10 5 22.3607, and dy 2 dy 1 1 corresponds to here, so take as . Then dx dx 4x 2 x 0.2x 100 0.2x 2 . NINT fails to evaluate x C, and, since (1, 1) lies on the curve, C x. 0. 1 (y1 )2 dx because of the undefined slope at the limits, so use the track's symmetry, and "back away" from the upper limit a little, and find
22.36 (b) Only one. We know the derivative of the function and the value of the function at one value of x. 20. (a)
dx 2 dx 1 1 corresponds to 4 here, so take as 2 . dy dy y y 1 Then x C and, since (0, 1) lies on the curve, y 1 C 1. So y . 1 x 2
0 1 (y1 )2 dx 52.548. Then, pretending the last little stretch at each end is a straight line, add 2 100 track 1 as 0.2(22.36)2 0.156 to get the total length of (b) Only one. We know the derivative of the function and the value of the function at one value of x. 21. y
/4 0 52.704. Using a similar strategy, find the 32.274. Now length of the right half of track 2 to be enter Y1 52.704 and 1
150 0.2t cos 2x, so the length is
/4 1 cos 2x dx
0 2 cos x dx
/4 2 2
2 Y2 32.274 + NINT , t, x, 0 and 2 sin x
0 1. 0.2t graph in a [ 30, 0] by [0, 60] window to see the effect of the xcoordinate of the lane2 starting position on the length 22. y
1 (1 1
2/4 1 x 2/3)1/2x (1 x
2/3 1/3 , so the length is
2/3 of lane 2. (Be patient!) Solve graphically to find the intersection at x 19.909, which leads to starting point 8 8
24 1 x 2/3)x dx dx coordinates ( 19.909, 8.410).
1 2/3 x 3 2 1/3 x , but NINT fails on 3 26. f (x) x
2/4 8
3 2 3 2 1/3 dx 2 1
1 2/4
0 (f (x))2 dx because of the undefined slope at 8 x 2/3 8 x 6.
3 sin x for 0 20 0. So, instead solve for x in terms of y using the x1/3 y 0, and 3 1 2 2 quadratic formula. (x1/3)2 x1/3
1 2 1 ( 1 8 3 ( 1 8
2/3 2 1 4y . Using the positive values, 23. Find the length of the curve y y
20 0 x 20. x x
2 0 4y 4y 1 1)3. Then, 1)2
1 2 4y 3 3 cos x, so the length is 20 20 2 3 3 1 cos x dx, which evaluates, using NINT, 20 20 , and 1/3 (x )2 dy 3.6142. to 21.07 inches. ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.
 Spring '08
 GERMAN

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