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Thread: Lengths of Curves

  1. #1
    Member VitaX's Avatar
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    Lengths of Curves

    Find the length of the curve

    $\displaystyle y= \frac{1}{3}(x^2 + 2)^{\frac{3}{2}}$ from $\displaystyle x=0$ to $\displaystyle x=3$

    $\displaystyle f'(x) = x(x^2 + 2)^{\frac{1}{2}}$

    $\displaystyle L = \int \sqrt{1 + \left(x(x^2+2)^\frac{1}{2}\right)}dx$

    $\displaystyle L = \int \sqrt{1 + x^4 + 2x^2}dx$

    Let $\displaystyle u = 1 + x^4 + 2x^2$

    After this I'm unsure of what to do. For example in the practice problems we had $\displaystyle 1 + 8x$ under the square root and what we did was let $\displaystyle u = 1 + 8x$ and $\displaystyle dx = \frac{1}{8}du$ and continue on from there. But in this problem I'm a little bit confused. I guess because the concept hasn't fully sank in yet.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by VitaX View Post
    Find the length of the curve

    $\displaystyle y= \frac{1}{3}(x^2 + 2)^{\frac{3}{2}}$ from $\displaystyle x=0$ to $\displaystyle x=3$

    $\displaystyle f'(x) = x(x^2 + 2)^{\frac{1}{2}}$

    $\displaystyle L = \int \sqrt{1 + \left(x(x^2+2)^\frac{1}{2}\right)}dx$

    $\displaystyle L = \int \sqrt{1 + x^4 + 2x^2}dx$

    Let $\displaystyle u = 1 + x^4 + 2x^2$

    After this I'm unsure of what to do. For example in the practice problems we had $\displaystyle 1 + 8x$ under the square root and what we did was let $\displaystyle u = 1 + 8x$ and $\displaystyle dx = \frac{1}{8}du$ and continue on from there. But in this problem I'm a little bit confused. I guess because the concept hasn't fully sank in yet.
    $\displaystyle x^4+2x^2+1=(x^2+1)^2$
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  3. #3
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    Quote Originally Posted by VitaX View Post
    Find the length of the curve

    $\displaystyle y= \frac{1}{3}(x^2 + 2)^{\frac{3}{2}}$ from $\displaystyle x=0$ to $\displaystyle x=3$

    $\displaystyle f'(x) = x(x^2 + 2)^{\frac{1}{2}}$

    $\displaystyle L = \int \sqrt{1 + \left(x(x^2+2)^\frac{1}{2}\right)}dx$

    $\displaystyle L = \int \sqrt{1 + x^4 + 2x^2}dx$

    Let $\displaystyle u = 1 + x^4 + 2x^2$

    After this I'm unsure of what to do. For example in the practice problems we had $\displaystyle 1 + 8x$ under the square root and what we did was let $\displaystyle u = 1 + 8x$ and $\displaystyle dx = \frac{1}{8}du$ and continue on from there. But in this problem I'm a little bit confused. I guess because the concept hasn't fully sank in yet.

    $\displaystyle x^4+2x^2+1=(x^2+1)^2$ ...and your integral is way easier than it looks!

    Tonio
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