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Math Help - Help on a Calc Problem? Find the length of the curve?

  1. #1
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    Help on a Calc Problem? Find the length of the curve?

    Find the length of the curve, L.
    y^2= 16(x+1)^3 where 0<=x<=2 and y>0
    ______________________________________

    Ok, so first I solved the equation for y
    y= 4(x+1)^3/2

    I know that the equation for the length of a curve is:
    s= int (a to b) sqrt(1+ (f'(x))^2)

    I then took the derivative of y= 4(x+1)^3/2 with respect to x and came up with:
    6(x+1)^(1/2)

    Then, I plugged what I knew into the formula...
    s= int (0 to 2) sqrt( 1+ (6(x+1)^1/2)^2) dx and simplified...
    s= int (0 to 2( sqrt( 1+36(x+1)) dx

    Now I am stuck! If I tried distributing the 36 and combining it with the 1 but I get 36x+37 under the square root which I don't know how to simply to find its AntiD.

    (The only problems like this I know how to work are the ones where you get the + or - 1/2 in the derivative and replace it with the difference/sum of the derivative squared. This one is different, unfortunately!) Oh, and Thanks very much in advance.
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  2. #2
    Senior Member
    Joined
    Dec 2008
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    You are correct so far. Now all we need to do is perform a simple substitution:

    \int_0^2\sqrt{36x+37}\,dx=\frac{1}{36}\int_0^2 36\sqrt{36x+37}\,dx.

    Now we let u=36x+37 and the antiderivative of 36\sqrt{36x+37} follows at once.

    Hope this helps!
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