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$$x=\frac{t^2}{2} \text{ , } y=\frac{(2t+1)^{3/2}}{3} \text{ , } 0 \le t \le 20$$

The formula for the length of a parametric curve is $L=\int_{a}^{b}\sqrt{[f'(t)]^2+[g'(t)]^2}dt$. Taking the derivative of both functions I get $f'(t)=t$ and $g'(t)=(2t+1)$. Plugging these into the length formula gives $$\int_{0}^{20}\sqrt{[t]^2+[(2t+1)]^2}dt$$ $$\int_{0}^{20}\sqrt{5t^2+4t+1}dt$$

I am not sure how to proceed with this problem since I cannot see a perfect square for a term inside the square root. Could someone help me with this?

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Your derivative for $g$ is wrong. – John Hughes Dec 20 '13 at 18:56
I missed a power of 1/2. I fixed it and got the correct answer, thanks! – Kot Dec 20 '13 at 19:13
up vote 2 down vote accepted

The intended integrand is indeed: $$t^2+2t+1=(t+1)^2$$ And we note that if $t>-1$, $$\sqrt{(t+1)^2}=|t+1|=t+1,~~~0\le t\le 20$$

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