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Question:

Consider the parametric curve defined by x(t) = t³ -3t+2 and y(t) = t³ -3t-10.
If a particle P(x(t); y(t)) travels along this curve, how many units of distance does the particle P cover between t = 0 and t = 5? Hint: When you get rid of the square root, you get an absolute value. Don't forget to split up the
integral to deal with it. (answer= 6√2 +3√2(124/3 -4) )
(P.S SHOW WORK B/C I NEED TO KNOW FOR FINAL .THANKS.)

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Answers:

zanthymx

Oracle

Karma Points: 32,833

Date Posted: 5/30/2008 10:14:02 PM Status: Live

Asker's Rating: Lifesaver

Response:

{Parametric Curve (given)} = r(t) =
     = < (t^3 - 3*t + 2), (t^3 - 3*t - 10) >

{Arc Length Between t = 0 & t = 5} =

     = ∫₀⁵ √{ (dx/dt)^2 + (dy/dt)^2 } dt

     = ∫₀⁵ √{ (3*t^2 - 3)^2 + (3^t^2 - 3)^2 } dt

     = ∫₀⁵ √{ 2*(3*t^2 - 3)^2 } dt

     = √2 ∫₀⁵ √{ (3*t^2 - 3)^2 } dt

     = √2 ∫₀⁵ | (3*t^2 - 3) | dt

     = √2 ∫₀¹ (3 - 3*t^2) dt +
               + √2 ∫₁⁵ (3*t^2 - 3) dt

     = √2 [ (3*t - t^3 ]₀¹ +
               + √2 [ (t^3 - 3*t) ]₁⁵

     = √2 { 3 - 1 } +
               + √2 { 125 - 15 - 1 + 3 }

     = (114)*√2

    = 161.22

.

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parameters-your help would be very much appreciated-will rate lifesaver

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