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Math Help - Arc Length and parametric equations

  1. #1
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    Arc Length and parametric equations

    I'm having a few problems with these questions... any help is appreciated! Thanks!

    find d^2y \over{dx^2} of x = \sqrt{t};  y = 2t + 4 at  t = 1



    Find the arc length of the curve  x = e^t cos(t),  y = e^tsin(t),  0 <= t <= {\pi\over{2}}
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  2. #2
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    Quote Originally Posted by akwingzero View Post
    I'm having a few problems with these questions... any help is appreciated! Thanks!

    find d^2y \over{dx^2} of x = \sqrt{t};  y = 2t + 4 at  t = 1
    [snip]
    The easy road: Eliminate the parameter to get the Cartesian equation:

    x = \sqrt{t} \Rightarrow x^2 = t .... (1)

    y = 2t + 4 .... (2)

    Substitute (1) into (2): y = 2x^2 + 4

    Getting the double derivative should be simple. For the record, you'd then note that t = 1 \Rightarrow x = 1 and do the substitution. But there's no need to substitute anything on this occassion.

    -----------------------------------------------------------------------------------------

    The hard road:

    \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = (2) (2 \sqrt{t}) = 4 \sqrt{t} since \frac{dx}{dt} = \frac{1}{2 \sqrt{t}}.

    Therefore:

    \frac{d^2 y}{dx^2} = \frac{d}{dx} \left[ \frac{dy}{dx} \right] = \frac{d}{dx} [4 \sqrt{t} ] = \frac{d}{dt}[4 \sqrt{t}] \times \frac{dt}{dx} = \left( \frac{2}{\sqrt{t}} \right) \, \left( 2 \sqrt{t}\right) = 4.


    Quote Originally Posted by akwingzero View Post
    [snip]
    Find the arc length of the curve  x = e^t cos(t),  y = e^tsin(t),  0 <= t <= {\pi\over{2}}
    I assume you're familiar with the arc length formula and I assume you can successfully differentiate both x and y with respect to t. That doesn't leave much to have trouble with.

    You might note that \left(e^t \cos t - e^t \sin t \right)^2 + \left(e^t \sin t + e^t \cos t \right)^2 =  e^{2t} \left( [\cos t - \sin t ]^2 + [\sin t + \cos t]^2 \right) = 2 e^{2t}

    after expanding and simplifying. Therefore the integrand in the arc length formula boils down to \sqrt{2} e^t. Now it should be blue sky all the way.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by akwingzero View Post
    I'm having a few problems with these questions... any help is appreciated! Thanks!

    find d^2y \over{dx^2} of x = \sqrt{t};  y = 2t + 4 at  t = 1



    Find the arc length of the curve  x = e^t cos(t),  y = e^tsin(t),  0 <= t <= {\pi\over{2}}
    For the second one, I would just note that this is the paramerization of the curve

    r=e^{\theta}

    Then we can uset that arc length in polar coordinates is

    \int_a^b\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^  2}

    So we would have

    \int_0^{\frac{\pi}{2}}\sqrt{(e^{\theta})^2+(e^{\th  eta})^2}d\theta=\int_0^{\frac{\pi}{2}}\sqrt{2}e^{\  theta}d\theta

    Same as Mr. F, just a little less messy.
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