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Math Help - Finding Length

  1. #1
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    Finding Length

    What is the length of the plane curve defined by r(t) = <t^2 + 1, t^3 - 2> for t is an element of [0,2]?
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  2. #2
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    Arc length for a parametric curve is given by:

    \int_{a}^{b}\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{d  t})^{2}}dt
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  3. #3
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    So would it be <br />
\int_{a}^{b}\sqrt{(2t)^{2}+(3t^2)^{2}}dt<br />
    <br />
\int_{a}^{b}\sqrt{4t^{2}+9t^4}dt<br />
    <br />
\int_{a}^{b}t\sqrt{4+9t^2}dt<br />
    <br />
u = {4+9t^2}du<br />
    and <br />
du= 18t<br />
    So <br />
(1/18)\int_{a}^{b}\sqrt{u}du<br />
    <br />
(1/27)(u)^(3/2)<br />
    <br />
(1/27)(4+9b^2)^(3/2)-(1/27)(4+9a^2)^(3/2)<br />

    Is this the right answer?
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  4. #4
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    Not quite. Aren't your limits of integration 0 to 2?. Use those, not a and b. That was just for an illustration of the formula.
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  5. #5
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    Yes, so it would be
    <br />
(1/27)(4+9(2)^2)^{3/2}-(1/27)(4)^{3/2}<br />
    <br />
(1/27)(40)^{3/2}-(1/27)(4)^{3/2}<br />
    <br />
(1/27)(4+9(2)^2)^{3/2}-(1/27)4^{3/2}<br />

    <br />
\frac{40^{3/2}-8}{27}<br />

    Correct?
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  6. #6
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    Yep.
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