Pre-Calc Homework Solutions 247

Pre-Calc Homework Solutions 247 - Section 6.2 dy dx dy dx...

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Section 6.2 dy dx dy dx dy ey dy ey 247 41. (cos x)e y sin x 44. (cos x)(e y esin x) cos x esin x dx dy dx dy y dy y 4 y ln x x ln x 4 dx x ln x dx x Integrate both sides. 4 Integrate both sides. cos x e sin x dx sin x On the right, let u du 2y1/2 2y1/2 2y 1/2 ln x On the right, let u du e e e e y y y y 1 dx x cos x dx e u du e e e u sin x sin x 4 u du 4 u2 2(ln x) (ln x) 2 2 1 2 C C C C]2 C]2 C) 2 C C C C.) sin x y y ) 1/2 [(ln x)2 (Note: technically C is now C y y dy 42. dx dy dx dy e y dy e y y(e) [(ln e)2 (1 1 1 0 ln (C ln (C e x y y e e sin x) C e xe e x dx y (ln x)4 Note: Absolute value signs are not needed because the original problem involved ln x, so we know that x 0. 45. (a) Let u x dx x 1 dx u1/2 du 1 Integrate both sides. e x dx e dx x x x du e dy e y y y e C C) ln (e 2 3/2 u C 3 2 (x 1)3/2 3 d 2 Alternatively, (x 1)3/2 dx 3 C C x 1. dy 43. 2xy 2 dx dy 2x dx y2 dy 2x dx y2 (b) By Part 1 of the Fundamental Theorem of Calculus, dy1 dx x 1 and dy2 dx x 1. 1, so both are y y 1 x2 1 x 2 antiderivatives of x C (c) Using NINT to find the values of y1 and y2, we have: C 1 y(1) 1 C y C 3 1 C 0.25 x y1 y2 y1 y2 4 2 3 0 0 4.667 4.667 1 1.219 3.448 4.667 2 2.797 1.869 4.667 3 4 4.667 6.787 0 2.120 4 1 x 2 4.667 4.667 3 C (d) C y1 x 0 x y2 x x x 0 3 1 dx 3 3 x x x 1 dx 1 dx 1 dx 1 dx x 0 ...
View Full Document

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.

Ask a homework question - tutors are online