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

• Feb 24th 2010, 05:58 PM
Linnylou09
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.
• Feb 24th 2010, 06:17 PM
Scott H
You are correct so far. Now all we need to do is perform a simple substitution:

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

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

Hope this helps!