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Math Help - Length of a function using integrals

  1. #1
    Newbie alysha27's Avatar
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    Length of a function using integrals

    Find the length of the cardioid r=1+sinO (sorry I wish i knew how to use all the symbols)

    I know that

    L= integral of [r^2 +(dr/dO)^2]^1/2 dO (from 0 to 2pi)

    = integral of [1 + 2sinO + sin^2(O) +cos^2(O)]^1/2 dO...

    =integral of (2+2sinO)^1/2 dO from 0 to 2pi,

    ... then I get stuck on simplifying do I can something I can integrate. I know to multiplying and dividing by the integrand (2-2sinO)^1/2. Can't seem to remember how to do this from high school. Any help would be great. Thanks.
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  2. #2
    TD!
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    You could do it using substitution I assume (haven't checked), but here's an approach using some trigonometry tricks.

    I'm shifting t to pi/2-t. Of course, the boundaries shift too:

    \int\limits_0^{2\pi } {\sqrt {2 + 2\sin t} } dt = \int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\sqrt {2 + 2\sin \left( {\frac{\pi }{2} - t} \right)} } dt

    Now we can use \sin \left( \pi/2 - \alpha } \right) = \cos \alpha and then convert to the half angle using \cos \left( {2\alpha } \right) = 2\cos ^2 \alpha  - 1

    <br />
\int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\sqrt {2 + 2\cos t} } dt = \int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\sqrt {2 + 2\left( {2\cos ^2 \left( {\frac{t}{2}} \right) - 1} \right)} } dt<br />

    Now we solved two problems: we introduced a square and we can get rid off the constant 2:

    <br />
\int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\sqrt {2 + 2\left( {2\cos ^2 \left( {\frac{t}{2}} \right) - 1} \right)} } dt\int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\sqrt {4\cos ^2 \left( {\frac{t}{2}} \right)} } dt<br />

    Now we can lose the square root, but watch out that it's the absolute value which comes out of the square root. The problem now is that cos(t/2) becomes negative in our interval for t. But the area under cos(t/2) is the same on the interval [-pi/2,3pi/2] as on [-pi,pi] and on this last one, cos(t/2) is positive so we solved our problem of the absolute value without needing to integrate it.

    \int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\sqrt {4\cos ^2 \left( {\frac{t}{2}} \right)} } dt = 2\int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\left| {\cos \left( {\frac{t}{2}} \right)} \right|} dt = 2\int\limits_{ - \pi }^\pi  {\cos \left( {\frac{t}{2}} \right)} dt<br />

    And now it's easy:

    <br />
2\int\limits_{ - \pi }^\pi  {\cos \left( {\frac{t}{2}} \right)} dt = \left[ {4\sin \left( {\frac{t}{2}} \right)} \right]_{ - \pi }^\pi   = 4 + 4 = 8<br />
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TD!
    You could do it using substitution I assume (haven't checked), but here's an approach using some trigonometry tricks.

    I'm shifting t to pi/2-t. Of course, the boundaries shift too:

    \int\limits_0^{2\pi } {\sqrt {2 + 2\sin t} } dt = \int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\sqrt {2 + 2\sin \left( {\frac{\pi }{2} - t} \right)} } dt

    Now we can use \sin \left( \pi/2 - \alpha } \right) = \cos \alpha and then convert to the half angle using \cos \left( {2\alpha } \right) = 2\cos ^2 \alpha - 1

    <br />
\int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\sqrt {2 + 2\cos t} } dt = \int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\sqrt {2 + 2\left( {2\cos ^2 \left( {\frac{t}{2}} \right) - 1} \right)} } dt<br />

    Now we solved two problems: we introduced a square and we can get rid off the constant 2:

    <br />
\int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\sqrt {2 + 2\left( {2\cos ^2 \left( {\frac{t}{2}} \right) - 1} \right)} } dt\int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\sqrt {4\cos ^2 \left( {\frac{t}{2}} \right)} } dt<br />

    Now we can lose the square root, but watch out that it's the absolute value which comes out of the square root. The problem now is that cos(t/2) becomes negative in our interval for t. But the area under cos(t/2) is the same on the interval [-pi/2,3pi/2] as on [-pi,pi] and on this last one, cos(t/2) is positive so we solved our problem of the absolute value without needing to integrate it.

    \int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\sqrt {4\cos ^2 \left( {\frac{t}{2}} \right)} } dt = 2\int\limits_{ - \frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\left| {\cos \left( {\frac{t}{2}} \right)} \right|} dt = 2\int\limits_{ - \pi }^\pi {\cos \left( {\frac{t}{2}} \right)} dt<br />

    And now it's easy:

    <br />
2\int\limits_{ - \pi }^\pi {\cos \left( {\frac{t}{2}} \right)} dt = \left[ {4\sin \left( {\frac{t}{2}} \right)} \right]_{ - \pi }^\pi = 4 + 4 = 8<br />
    Hey that was cool! I've never seen that trick before.

    -Dan
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