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

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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 ...
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