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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 + et y(t) = 5  2t 0 t < 3 We have that dx dt = et  et and
3 dy dt = 2. Therefore, the exact length is
2 3 0 dx dt =
0 2 +
3 dy dt dt =
0 (et  et ) + 4dt =
3 2 2 3 0 3 e2t  2 + e2t + 4dt et + et dt
0 e2t + e2t + 2dt =
0 (et + et ) dt =
3 3 = e e t t t=3 t=0 = e e  (1  1) = e3  e3 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 + x1 )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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 Spring '08
 Keppelmann
 Math

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