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anonymous

  • 5 years ago

Find the length of the parametric curve curve x = et cos t, y = et sin t for 0 <= t <= 2.

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  1. anonymous
    • 5 years ago
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    you take the (dx/dt)^2+ (dy/dt)^2 amd then take the ontegral ove rthe inerval is this idea correct

  2. anonymous
    • 5 years ago
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    A differential element of a curve in the x-y plane, ds, is given in terms of the differentials of x and y by Pythagoras' Theorem. \[(ds)^2=(dx)^2+(dy)^2\]Dividing both sides of the equation by (dt)^2, you have\[\left( \frac{ds}{dt} \right)^2=\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2\]and from this,\[\frac{ds}{dt}=\sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2} \rightarrow ds =\sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2}dt\]

  3. anonymous
    • 5 years ago
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    Integrating,\[s=\int\limits_{t_1}^{t_2}\sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2}dt\]

  4. anonymous
    • 5 years ago
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    yup :)

  5. anonymous
    • 5 years ago
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    \[x=e^t \sin t \rightarrow \frac{dx}{dt}=e^t(\cos t - \sin t )\]\[y=e^t \sin t \rightarrow x=e^t \sin t \rightarrow \frac{dy}{dt}=e^t(\sin t + \cos t )\]

  6. anonymous
    • 5 years ago
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    Squaring and adding each, you have\[\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2=2e^2t\]

  7. anonymous
    • 5 years ago
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    So your arc length is\[s=\int\limits_{0}^{2}\sqrt{2}e^tdt=\sqrt{2}(e^2-1)\]

  8. anonymous
    • 5 years ago
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    *should be 2e^(2t) in second-last part of derivation.

  9. anonymous
    • 5 years ago
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    you've got some errors

  10. anonymous
    • 5 years ago
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    x= e^t cos t , not e^t sin t

  11. anonymous
    • 5 years ago
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    otherwise its great

  12. anonymous
    • 5 years ago
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    yeah i got that :D pretty neat calculus

  13. anonymous
    • 5 years ago
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    well, transcribing from notes to this site hardly ever goes to plan.

  14. anonymous
    • 5 years ago
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    :)

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