Pre-Calc Homework Solutions 412

Pre-Calc Homework Solutions 412 - 412 Section 10.3 1 0 e 4...

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Unformatted text preview: 412 Section 10.3 1 0 e 4 1 i 4 28. (a) r(0) 0 i j, 2 i (e0)j 31. a(t) r(t) 3i j, so v(t) 1 2 t j 2 (3t)i C1 t tj C1 and C2 i 2j, 3 2 t i 2 C2. r(0) r(2) Initial 1 8 e 4 (e4)j 1 8 e 4 and since v(0) must point directly from (1, 2) toward (4, 1) with magnitude 2, (4 1)i 1) 2 1 , 1 , terminal 4 2, e4 4t (1 (1 2)j 2)2 v(0) e 1, and C1 6 10 2 i (b) v(t) dy dt (e 4t 1)i dx (2e )j; dt 2t (4 2 10 j 2e2t. 2 Length 0 2 0 2 0 (e (e (e 4t 4t 1) 2 (2e ) dt 2t 2 4t 1) dt 1) dt 2 2 3 10 i 5 3 2 t So r(t) 2 10 j 5 3 10 t 5 2 t2 1 i 1 2 t 2 10 t 5 2 j. 32. (a) dx dt 1 0 when t 2 2 2. That corresponds to (2 2, 6). 1 equals 6. 1 4t e 4 e8 7 4 2 point (b) dy dx 2 y , 3( 2)2 t 0 746.989 dy/dt dx/dt 1 6t , which for t 2/t 2 (c) 0 2 1 4t e t (e4t 1)2 (2e2t)2 dt 4 2 1 4t e t (e4t 1) dt 2 0 4 2 1 8t 1 4t e e te4t t dt 2 4 0 4 1 2 2 1 8t 1 4t 1 e e (4t 1)e4t t 2 2 0 32 16 16 e16 12e8 69 1,737,746.456 16 (c) When y d 2y dx 2 dy dx 12, t dy /dt dx/dt 2. (1 2/t 2)6 (4/t 3)6t , (1 2/t 2)3 which for t 2 equals 24. 0 and t 160: 33. (a) The j-component is zero at t 160 seconds. (b) (c) d dt 3 (40)(40 64 160) 225 m 15 , which for t 2 29. (a) v(t) (b) v(t) at t (cos t)i (2 sin 2t)j 0 and sin 2t 0, 2 0 when both cos t 2 0. cos t 0 3 3 t(t 160) t 64 32 15 equals m per second. 4 3 t 32 15 equals 0 at t 2 40 and So v(t) 3 ; sin 2t 0 at t 2 3 0 at t , . 2 2 , , 3 , and 2 . 2 (d) v(t) 80 seconds (and is negative after that time). 34. (a) Solve t (t 3)2 3 3t 2 3t 2 (c) x sin t, y cos 2t. Relate the two using the identity cos 2u 1 2 sin2 u: y 1 2x 2, where as t ranges over all possible values, 1 x 1. When t increases from 0 to 2 , the particle starts at (0, 1), goes to (1, 1), then goes to ( 1, 1), and then goes to (0, 1), tracing the curve twice. dy 30. (a) dx dy/dt dx/dt 3t 2 6t 2 12 6t 2 4: t 2. Then check that 2: it does. 2(t 3)j, so v1(2) 1 2 for t i t2 2t 2 (b) First particle: v1(t) i . 2j 4 2t and the direction unit vector v1 is Second particle: v2(t) 3 i 2 , 2 (b) Horizontal tangents: t 4 0 for t 2. Vertical tangents: 2t 2 2t 0 for t 0, 1. Plugging the t-values into x 2t 3 3t 2 and y t 3 12t produces the x- and y-coordinates of the critical points. t 2: horizontal tangent at ( 28, 16) t 0: vertical tangent at (0, 0) t 1: vertical tangent at ( 1, 11) t 2: horizontal tangent at (4, 16) the direction unit vector is 5 5 3 j, which is constant, and 2 1 1 , . 2 2 ...
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