Pre-Calc Homework Solutions 403

# Pre-Calc Homework Solutions 403 - Section 10.1 20 x sec t 0...

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Unformatted text preview: Section 10.1 20. x sec t 0 0 403 cos t (see Ex. 15), y /3 sin t, so cos t)2 ( sin t)2 dx 25. In the first integral, replace t with x. Then becomes dx dx dx dt Length /3 2 (cos t) (sec t 2 cos t /3 1. g(y), y y, c y d. The tan t dt 26. Parameterize the curve as x . /3 2 2 0 sin t dt cos t 0 2 21. (a) x(t) (b) x Area 0 parameter is y itself, so replace t with y in the general t 1 formula. Then dy dy becomes dt dy 2t, y(t) 2, y 1 t 1, 0 1. 1, so 2 (t 1 27. x 1) 22 1) dt 1 t, y 2t 4 1, so t2 ( 2t 1)2 dt 12 dt Total length 0 4 2 2 3 (c) Slant height Area (1 5 (t 0 (t 0 1) dt 4 1 5 t2 2 t 0 5 22 2) 5 12 3 5 r x, h 1 2 t 2 t 0 m 12. 12 for m: 2 5, so Now solve 1 2 m 2 1 2 t 2 t 0 m 2 6, or m2 4 2 48 2m 1 12 0, and 22. (a) Because these values for x(t) and y(t) satisfy y m 13. Take the positive 13 1)3/2 1, which gives (3.394, 5.160). which is the equation of the line through the origin and (h, r), and this range of t-values gives the correct initial and terminal points. 28. solution. The midpoint is at t (x, y) ( 13 1)2 1 , (2 3 2 13 (b) x Area h, y 1 0 r, so 2 (rt) h2 r2 t 2 h2. r2 h2, so Area 2 cos 2t, so 2 sin 2t)2 . [0, 9] by [ 3, 3] r 2 dt 1 0 [ 4, 4] by [0, 10] r h2 r (c) Slant height 23. (a) x Length 0 0 /2 Use the right half of the curve, 0 x r r2 h2 Area 0 t . r2 3 cos t, y 6 cos 2t, so (3 cos t)2 (6 cos 2t)2 dt, 159.485. 2 (3 sin t) which using NINT evaluates to 2 sin 2t, y /2 ( 2 dt (2 cos 2t)2 dt 29. (b) x Length cos t, y 1/2 1/2 1/2 sin t, so ( cos t)2 dt . ( sin t)2 dt Use the top half of the curve, and make use of the shape's symmetry. x 3 cos t, y /2 1/2 6 cos 2t, so (6 cos 2t)2 dt 24. x Length 3 sin t, y 2 0 4 cos t, so ( 3 sin t)2 (4 cos t)2 dt 22.103. Area 2 0 2 (3 sin 2t) (3 cos t)2 which using NINT, evaluates to 144.513. which using NINT evaluates to ...
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