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Math Help - length of curve

  1. #1
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    length of curve

    from t=0 to t=pi/2
    x=e^-t cost
    y=e^-t sint

    dx/dt = -e^-t cost - e^-t sint
    dy/dt = -e^-t sint + e^-t cost

    (dx/dt)^2 = ?
    (dy/dt)^2 = ?

    L = (int 0 to pi/2) [sqrt(-2e^-2t)]dt

    how do you get that? when i square dx/dt and dy/dt using the (a+b)(a+b) = a^2 +2ab+b^2 method i get a wrong answer, but when i just slap on a ^2 on each term and simplify i get it right. does that squaring rule not apply here for some reason?
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  2. #2
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    Quote Originally Posted by jeph View Post
    from t=0 to t=pi/2
    x=e^-t cost
    y=e^-t sint

    dx/dt = -e^-t cost - e^-t sint
    dy/dt = -e^-t sint + e^-t cost

    (dx/dt)^2 = ?
    (dy/dt)^2 = ?

    L = (int 0 to pi/2) [sqrt(-2e^-2t)]dt

    how do you get that? when i square dx/dt and dy/dt using the (a+b)(a+b) = a^2 +2ab+b^2 method i get a wrong answer, but when i just slap on a ^2 on each term and simplify i get it right. does that squaring rule not apply here for some reason?
    You want
    sqrt{(dx/dt)^2 + (dy/dt)^2}

    So
    (dx/dt)^2 = [-e^{-t}cos(t) - e^{-t}sin(t)]^2

    = [e^{-t}cos(t) + e^{-t}sin(t)]^2

    = e^{-2t}cos^2(t) + 2e^{-2t}sin(t)cos(t) + e^{-2t}sin^2(t)

    = [e^{-2t}cos^2(t) + e^{-2t}sin^2(t)] + 2e^{-2t}sin(t)cos(t)

    = e^{-2t}[cos^2(t) + sin^2(t)] + 2e^{-2t}sin(t)cos(t)

    = e^{-2t} + 2e^{-2t}sin(t)cos(t)

    Similarly:
    (dy/dt)^2 = [-e^{-t}sin(t) + e^{-t}cos(t)]^2

    = e^{-2t} - 2e^{-2t}sin(t)cos(t)

    So
    sqrt{(dx/dt)^2 + (dy/dt)^2}

    = sqrt{e^{-2t} + 2e^{-2t}sin(t)cos(t) + e^{-2t} - 2e^{-2t}sin(t)cos(t)}

    = sqrt{2e^{-2t}}

    as you said you needed.

    -Dan
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  3. #3
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    wow...i wish i remembered what my thoughts were while working on this problem on the exam. i dunno if i just got lucky or i really knew what i was doing at the time. the problem works just as well if you just slap on the ^2 on each of the terms.
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