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Math Help - Lengths of Curves

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

    Find the length of the curve

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

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

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

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

    Let u = 1 + x^4 + 2x^2

    After this I'm unsure of what to do. For example in the practice problems we had 1 + 8x under the square root and what we did was let u = 1 + 8x and 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

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

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

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

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

    Let u = 1 + x^4 + 2x^2

    After this I'm unsure of what to do. For example in the practice problems we had 1 + 8x under the square root and what we did was let u = 1 + 8x and 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.
    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

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

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

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

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

    Let u = 1 + x^4 + 2x^2

    After this I'm unsure of what to do. For example in the practice problems we had 1 + 8x under the square root and what we did was let u = 1 + 8x and 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.

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

    Tonio
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