Find the arclength of the curve defined by $r(t)=i+9t^2j+t^3k$ for $0 \leq t \leq \sqrt28$.
By Sophia Hammond
First I found $r'(t)=\langle 1,18t^2,3t^2\rangle$ and so the magnitude of $r'(t)= \sqrt{1+(18t)^2+(3t^2)^2}$ thus the integral from $0$ to $\sqrt{28}$ of $\sqrt{1+324t^2+9t^4} dt$. When I plugged $\sqrt{28}$ in, I get $\sqrt{1+324(28)+9(784)}$ to get $\sqrt{16,129} - \sqrt{1}$ which is $126$. This is not the correct answer.
$\endgroup$ 11 Answer
$\begingroup$Observe that\begin{align*} r'(t)&=18tj+3t^2k\\[4pt] \text{length }&=\int_{0}^{\sqrt{28}}\sqrt{(18t)^2+(3t^2)^2}dt=\int_{0}^{\sqrt{28}}3t\sqrt{36+t^2}dt=\left.\frac32\frac{(36+t^2)^{3/2}}{3/2}\right|_{0}^{\sqrt{28}} \end{align*}
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