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 $0\leq t\leq 2\pi$ .

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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 $0\leq t\leq 2\pi$ .

Tap to reveal answer

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}$$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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}$$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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}$$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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}$$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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}$$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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}$$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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}$$

← Didn't Know|Knew It →

Knew It

Arc Length - Multivariable Calculus

Determine the length of the curve , on the interval .

First we need to find the tangent vector, and find its magnitude. Now we can set up our arc length integral

Determine the length of the curve , on the interval .

First we need to find the tangent vector, and find its magnitude. Now we can set up our arc length integral

Determine the length of the curve , on the interval .

First we need to find the tangent vector, and find its magnitude. Now we can set up our arc length integral

Determine the length of the curve , on the interval .

First we need to find the tangent vector, and find its magnitude. Now we can set up our arc length integral

Determine the length of the curve , on the interval .

First we need to find the tangent vector, and find its magnitude. Now we can set up our arc length integral

Determine the length of the curve , on the interval .

First we need to find the tangent vector, and find its magnitude. Now we can set up our arc length integral

Determine the length of the curve , on the interval .

First we need to find the tangent vector, and find its magnitude. Now we can set up our arc length integral

Determine the length of the curve , on the interval .

First we need to find the tangent vector, and find its magnitude. Now we can set up our arc length integral

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

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

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

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

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

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

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

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