# 6.3 - Math 182 Homework 6.3 Problem 6 Graph the curve and...

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: Math 182 Homework 6.3 Problem 6 Graph the curve and find its exact length: x(t) = et + e-t y(t) = 5 - 2t 0 t < 3 We have that dx dt = et - e-t and 3 dy dt = -2. Therefore, the exact length is 2 3 0 dx dt = 0 2 + 3 dy dt dt = 0 (et - e-t ) + 4dt = 3 2 2 3 0 3 e2t - 2 + e-2t + 4dt et + e-t dt 0 e2t + e-2t + 2dt = 0 (et + e-t ) dt = 3 -3 = e -e t -t t=3 t=0 = e -e - (1 - 1) = e3 - e-3 Problem 8 Graph the curve and find its exact length: y= Here we have that dy dx x2 ln x - 2x4 2 4 =x- 1 4x so that dy dx 2 = x2 - 2 1 1 1 1 x+ = x2 - + 2 4x 16x 2 16x2 and therefore the exact length is 4 2 dy dx 2 4 + 1dx = 2 x2 + ln x x2 + 2 4 1 1 + dx = 2 16x2 x=4 4 x+ 2 1 4x 2 4 dx = 2 1 (x + x-1 )dx 4 = = x=2 16 ln 4 + 2 4 1 - 4 ln 2 + 2 4 Since ln 4 = ln 22 = 2 ln 2 this all simplifies to 6 + ln 2 4 . Problem 12 Use Simpson's Rule with n = 10 to estimate the arc length of the curve x=y+ y 1y2 Compare your answer with with the value of the integral produced by your calculator. Now dx dy = 1 + 1 y- 2 so the length as an integral is 2 2 1 = 1 dy dx 2 2 + 1dy = 1 1 1+ 2 y 2 2 + 1dy = 1 1 1 2 + + dy y 4y 2 In comparison, the Maple Package gives to the same accuracy 1.7322. Extra Credit Problem 14 Find the length of the loop of the curve x(t) = 3t - t3 and y(t) = 3t2 . We must first find the loop (i.e. the values t0 , t1 of t for which x(t0 ) = x( t1 ) and y(t0 ) = y(t1 ). We note first that since y is an even function of ty it means that for all s we will have that y(s) = y(-s). Now if we can find a specific such s for which x(s) = x(-s) we will have succeeded. 3(-s) - (-s)3 = 3s - s3 -3s + s3 = 3s - s3 2s3 - 6s = 2s(s2 - 3) = 2s(s - So the loop occurs for - 3 t 3. Now dx = 3 - 3t2 and dy = 6t so dt dt dx dt 2 3)(s + 3) = 0 + dy dt 2 = (3 - 3t2 )2 + (6t)2 = 9 - 18t2 + 9t4 + 36t2 = 9t4 + 18t2 + 9 = 9(t2 + 1)2 Therefore, the length is 3 3 = 3 9(t2 + 1)2 dt 3 =3 3 3 (t2 + 1)dt = t3 + 3t 3 3 t= 3 t=- 3 = 32 + 32 - -3 2 - 3 2 = 12 3 3 4 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online