Pre-Calc Homework Solutions 412

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online