Arc Length - Multivariable Calculus

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Question

Determine the length of the curve $\vec{r}(t)=\left \langle t,4\sin(2t),4\cos(2t)\right \rangle$ , on the interval $0\leq t\leq 2\pi$ .

Answer

First we need to find the tangent vector, and find its magnitude.

$\vec{v}(t)=\vec{r}\ '(t)=\left \langle 1, 8\cos(2t), -8\sin(2t)\right \rangle$

$\left | \vec{v}(t)\right |=\sqrt{1^2+(8\cos(2t))^2+(-8\sin(2t))^2}$

$\left | \vec{v}(t)\right |=\sqrt{1+64\cos^2(2t)+64\sin^2(2t)}$

$\left | \vec{v}(t)\right |=\sqrt{1+64(\cos^2(2t)+\sin^2(2t))}$

$\left | \vec{v}(t)\right |=\sqrt{1+64}=\sqrt{65}$

Now we can set up our arc length integral

$L=\int_{a}^{b}\left | \vec{v}(t)\right |dt$

$L=\int_{0}^{2\pi} \sqrt{65}\ dt=t\sqrt{65}\Big|_{0}^{2\pi}=2\pi\sqrt{65}$

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Question

Determine the length of the curve $\vec{r}(t)=\left \langle t,4\sin(2t),4\cos(2t)\right \rangle$ , on the interval $0\leq t\leq 2\pi$ .

Answer

First we need to find the tangent vector, and find its magnitude.

$\vec{v}(t)=\vec{r}\ '(t)=\left \langle 1, 8\cos(2t), -8\sin(2t)\right \rangle$

$\left | \vec{v}(t)\right |=\sqrt{1^2+(8\cos(2t))^2+(-8\sin(2t))^2}$

$\left | \vec{v}(t)\right |=\sqrt{1+64\cos^2(2t)+64\sin^2(2t)}$

$\left | \vec{v}(t)\right |=\sqrt{1+64(\cos^2(2t)+\sin^2(2t))}$

$\left | \vec{v}(t)\right |=\sqrt{1+64}=\sqrt{65}$

Now we can set up our arc length integral

$L=\int_{a}^{b}\left | \vec{v}(t)\right |dt$

$L=\int_{0}^{2\pi} \sqrt{65}\ dt=t\sqrt{65}\Big|_{0}^{2\pi}=2\pi\sqrt{65}$

Compare your answer with the correct one above

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