Arc Length and Curvature - AP Calculus BC

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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 $\vec{r}(t)=\left \langle t,3\sin(4t),3\cos(4t)\right \rangle$ , on the interval $0\leq t\leq 2\pi$

Answer

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

$\vec{v}(t)=\vec{r}\ '(t)=\left \langle 1, 12\cos(4t), -12\sin(4t)\right \rangle$

$\left | \vec{v}(t)\right |=\sqrt{1^2+(12\cos(4t))^2+(-12\sin(4t))^2}$

$\left | \vec{v}(t)\right |=\sqrt{1+144\cos^2(4t)+144\sin^2(4t)}$

$\left | \vec{v}(t)\right |=\sqrt{1+144(\cos^2(2t)+\sin^2(2t))}$

$\left | \vec{v}(t)\right |=\sqrt{1+144}=\sqrt{145}$

Now we can set up our arc length integral

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

$L=\int_{0}^{2\pi} \sqrt{145}\ dt=t\sqrt{145}\Big|_{0}^{2\pi}=2\pi\sqrt{145}$

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Question

Find the length of the curve $<e^{2t},e^{-2t},2\sqrt{2}t>$ , from , to

Answer

The formula for the length of a parametric curve in 3-dimensional space is $l = \int_{t_0}^{t_1}\sqrt{(\frac{dx_1}{dt})^2+(\frac{dx_1}{dt})^2+(\frac{dx_3}{dt})^2} dt$

Taking dervatives and substituting, we have

$l = \int_{0}^{5}\sqrt{(2e^{2t})^2+(-2e^{-2t})^2+(2\sqrt{2})^2} dt$

$l = \int_{0}^{5}\sqrt{4e^{4t}+4e^{-4t}+8} dt$

$l = 2\int_{0}^{5}\sqrt{e^{4t}+e^{-4t}+2} dt$ . Factor a out of the square root.

$l = 2\int_{0}^{5}\sqrt{e^{4t}+e^{-4t}+2e^{2t}e^{-2t}} dt$ . "Uncancel" an $e^{2t}, e^{-2t}$ next to the . Now there is a perfect square inside the square root.

$l = 2\int_{0}^{5}\sqrt{(e^{2t}+e^{-2t})^2} dt$ . Factor

$l = 2[\frac{1}{2}e^{2t}-\frac{1}{2}e^{-2t}]_{0}^{5}$ . Take the square root, and integrate.

$=2(\frac{e^{10}}{2}-\frac{e^{-10}}{2})-2({0})$

$=e^{10}-e^{-10}$

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Question

Find the length of the arc drawn out by the vector function $F(t) = <a\cos t, a\sin t, at>$ with from to $t =\pi$ .

Answer

To find the arc length of a function, we use the formula

$\int_{t_{0}}^{t_{1}} \sqrt{(\frac{dx}{dt})^2 +(\frac{dy}{dt})^2 +(\frac{dz}{dt})^2} dt$ .

Using $t_0 =0, t_1 =\pi$ we have

$=\int_0^\pi \sqrt{(-a\sin t)^2+(a\cos t)^2 +(a)^2} dt$

$=\int_0^\pi \sqrt{a^2\sin^2 t+a^2\cos^2 t +a^2} dt$

$=\int_0^\pi \sqrt{a^2(\sin^2 t+\cos^2 t) +a^2} dt$

$=\int_0^\pi \sqrt{a^2 +a^2} dt$

$=[\sqrt2at]_0^\pi$

$=\sqrt{2} a \pi$

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Question

Evaluate the curvature of the function at the point .

Answer

The formula for curvature of a Cartesian equation is $\kappa(x)= \frac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}$ . (It's not the easiest to remember, but it's the most convenient form for Cartesian equations.)

We have , hence

$\kappa(x)= \frac{|20|}{[1+(20x)^2]^{3/2}}$

and $\kappa(1)= \frac{|20|}{[1+(20(1))^2]^{3/2}} = \frac{20}{401^{3/2}}$ .

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Question

Find the length of the parametric curve

$\vec{r}(t)= \left< \frac{\sqrt{10}}{2}t^2, \frac{1}{3}t^3,5t+14\right>$

for $0\leq t \leq 2$ .

Answer

To find the solution, we need to evaluate

$L = \int_0^2 \left| \vec{r}(t)' \right| dt$ .

First, we find

$\vec{r}(t)' = \left< \sqrt{10}t, t^2, 5\right>$ , which leads to

$\left| \vec{r}(t)' \right| = \sqrt{(\sqrt{10}t)^2+(t^2)^2+(5)^2}$

$\left| \vec{r}(t)' \right| = \sqrt{t^4+10t^2+25}=\sqrt{(t^2+5)^2}=t^2+5$ .

So we have a final expression to integrate for our answer

$L= \int_0^2 \left| \vec{r}(t)' \right| dt = \int_0^2 \left(t^2+5 \right )dt= \frac{(2)^3}{3}+5(2)=\frac{38}{3}$

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Question

Determine the length of the curve given below on the interval 0<t<2

$\mathbf{r}=[2\sin t,\sqrt{5} t, 2\cos t]$

Answer

The length of a curve r is given by:

$L=\int_{t_1}^{t_2}|\mathbf{r'}(t)|dt$

To solve:

$\mathbf{r}'(t)=(2\cos t, \sqrt{5}, 2 \sin t)$

$|\mathbf{r'}(t)|=\sqrt{4\cos^2t+5+4\sin^2t}=\sqrt{4+5}=\sqrt{9}=3$

$L=\int^2_0 3 dt=3t|^2_0=6$

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Question

Find the arc length of the curve

on the interval

$0\leq t\leq 10$

Answer

To find the arc length of the curve function

on the interval $a\leq t\leq b$

we follow the formula

$\int_a^b \sqrt{(\frac{\mathrm{dx} }{\mathrm{d} t})^2+(\frac{\mathrm{dy} }{\mathrm{d} t})^2+(\frac{\mathrm{dz} }{\mathrm{d} t})^2},dt$

For the curve function in this problem we have

$\frac{\mathrm{dx} }{\mathrm{d} t}=3$

$\frac{\mathrm{dy} }{\mathrm{d} t}=4cos(2t)$

$\frac{\mathrm{dz} }{\mathrm{d} t}=-4sin(2t)$

and following the arc length formula we solve for the integral

$\int_0^{10} \sqrt{(3)^2+(4cos(2t))^2+(-4sin(2t))^2},dt$

$=\int_0^{10} \sqrt{9+16sin^2(2t)+16cos^2(2t)},dt$

$=\int_0^{10} \sqrt{9+16(sin^2(2t)+cos^2(2t))},dt$

$=\int_0^{10} \sqrt{9+16},dt$

$=\int_0^{10} \sqrt{25},dt$

$=\int_0^{10} 5,dt$

$=5t|_0^{10}$

Hence the arc length is

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Question

Find the arc length of the curve function

$c(t)=<3t,4t,\frac{2}{3}t^{\frac{3}{2}}>$

On the interval $0\leq t\leq5$

Round to the nearest tenth.

Answer

To find the arc length of the curve function

on the interval $a\leq t\leq b$

we follow the formula

$\int_a^b \sqrt{(\frac{\mathrm{dx} }{\mathrm{d} t})^2+(\frac{\mathrm{dy} }{\mathrm{d} t})^2+(\frac{\mathrm{dz} }{\mathrm{d} t})^2},dt$

For the curve function in this problem we have

$\frac{\mathrm{dx} }{\mathrm{d} t}=3$

$\frac{\mathrm{dy} }{\mathrm{d} t}=4$

$\frac{\mathrm{dz} }{\mathrm{d} t}=t^{\frac{1}{2}}$

and following the arc length formula we solve for the integral

$\int_0^{5} \sqrt{(3)^2+(4)^2+(t^{\frac{1}{2}})^2},dt$

$=\int_0^{10} \sqrt{t+25},dt$

Using u-substitution, we have

and

The integral then becomes

$=\int_{25}^{30} \sqrt{u},du$

$=\frac{2}{3}u^{\frac{3}{2}}|_{25}^{30}$

$=\frac{2}{3}([30^{\frac{3}{2}}]-[25^{\frac{3}{2}}])$

Hence the arc length is

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Question

Given that a curve is defined by $\vec{r}=<2\sin 2t,2\cos 2t,4t>$ , find the arc length in the interval $0\leq t \leq \frac{\pi}{2}$

Answer

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Question

Given that

Find an expression for the curvature of the given conic

$\frac{2}{(1+4x^2)^{\frac{3}{2}}}$

Answer

Step 1: Find the first and the second derivative

Step 2:

Radius of curvature is given by

$\kappa=\frac{y''}{(1+(y')^2)^\frac{3}{2}}$

Now substitute the calculated expressions into the equation to find the final answer

$\kappa=\frac{2}{(1+(2x)^2)^{\frac{3}{2}}}=\frac{2}{(1+4x^2)^{\frac{3}{2}}}$

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Question

Find the arc length of the parametric curve

$c(t)=<\frac{2}{3}t^{3/2}, 5t, 3>$

on the interval $0\leq t \leq 11$ .

Round to the nearest tenth.

Answer

To find the arc length of the curve function

on the interval

$a\leq t \leq b$

we follow the formula

$\int_{a}^{b}\sqrt{(\frac{\mathrm{dx} }{\mathrm{d} t})^2+(\frac{\mathrm{dy} }{\mathrm{d} t})^2+(\frac{\mathrm{dz} }{\mathrm{d} t})^2},dt$

For the curve function in this problem we have

$\frac{\mathrm{d} x}{\mathrm{d} t}=t^{1/2}$

$\frac{\mathrm{d} y}{\mathrm{d} t}=5$

$\frac{\mathrm{d} z}{\mathrm{d} t}=0$

and following the arc length formula we solve for the integral

$\int_{0}^{11}\sqrt{(t^{1/2})^2+(5)^2+(0)^2},dt$

$\int_{0}^{11}\sqrt{t+25},dt$

And using u-substitution, we set and then solve the integral

$\int_{25}^{36}\sqrt{u},du$

$=\frac{2}{3}u^{3/2}|_{25}^{36}$

$=\frac{2}{3}([36]^{3/2}-[25]^{3/2})$

$=\frac{2}{3}(216-125)$

$=\frac{2}{3}(91)$

Which is approximately

units

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Question

Determine the curvature of the vector $\left \langle 3t^2,5t,0\right \rangle$ .

Answer

Using the formula for curvature $k=\frac{\left | r'(t)\times r''(t) \right |}{\left | r'(t)\right |^3}$ . , $r'(t)\times r''(t) =-30k, \left | r'(t)\times r''(t)\right |=30$ , and $\left | r'(t)\right |\sqrt{36t^2+25}$ . Plugging into the formula, we get $k=\frac{30}{(36t^2+25)^{\frac{3}{2}}}$

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Question

Find the arc length of the given curve on the interval $0\leq t \leq 5$ :

$\vec{r}(t)=\left \langle 2\cos(t), 2\sin(t), 5t\right \rangle$

Answer

The arc length on the interval $b\leq t \leq a$ is given by

$\int_{b}^{a} \left | \vec{r'(t)}\right |dt$ , where $\left | \vec{r}'(t)\right |$ is the magnitude of the tangent vector.

The tangent vector is given by

$\vec{r}'(t)=\left \langle -2\sin(t), 2\cos(t), 5\right \rangle$

The magnitude of the vector is

$\left | \vec{r}'(t)\right |= \sqrt{(-2\sin(t))^2+(2\cos(t))^2+5^2}=\sqrt{29}$

This is the integrand.

Finally, integrate:

$\int_{0}^{5}\sqrt{29}dt=5\sqrt{29}$

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Question

Determine the arc length of the following vector on the interval $0\leq t \leq 4$ :

$\vec{r}(t)=\left \langle \cos(t), 2t, \sin(t)\right \rangle$

Answer

The arc length of a curve on some interval $b\leq t \leq a$ is given by

$\int_{a}^{b}\left | \vec{r}'(t)\right |dt$

where $\vec{r}'(t)$ is the tangent vector to the curve.

The tangent vector to the curve is found by taking the derivative of each component:

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

The magnitude of the vector is found by taking the square root of the sum of the squares of each component:

$\left | \vec{r}'(t)\right |=\sqrt{(-\sin(t))^2+4+(\cos(t))^2}=\sqrt{5}$

Now, plug this into the integral and integrate:

$\int_{0}^{4}\sqrt{5}dt=t\sqrt{5}\left.\begin{matrix} 4\0 \end{matrix}\right| = 4\sqrt{5}$

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Question

Find an integral for the arc length of

$f(x)=(x+6)^\frac{3}{2}$ on the interval (Set up, DO NOT SOLVE)

Answer

Step 1:

Find the first derivative of the function

$f'(x)=\frac{3}{2}(x+6)^{\frac{1}{2}}$

Step 2:

Use the formula to calculate arc length

$L=\int_{a}^{b}\sqrt{1+[f'(x)]^2} $dx$

$\int_{-7}^{30}\sqrt{1+\left(\frac{3}{2}(x+6)^\frac{1}{2}\right)^2}$dx$

$\int_{-7}^{30}\sqrt{1+\frac{9}{4}(x+6)}$dx$

$\int_{-7}^{30}\frac{1}{2}\sqrt{9x+58}$dx$

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Question

Determine the length of the curve $\vec{r}(t)=\left \langle t,3\sin(4t),3\cos(4t)\right \rangle$ , on the interval $0\leq t\leq 2\pi$

Answer

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

$\vec{v}(t)=\vec{r}\ '(t)=\left \langle 1, 12\cos(4t), -12\sin(4t)\right \rangle$

$\left | \vec{v}(t)\right |=\sqrt{1^2+(12\cos(4t))^2+(-12\sin(4t))^2}$

$\left | \vec{v}(t)\right |=\sqrt{1+144\cos^2(4t)+144\sin^2(4t)}$

$\left | \vec{v}(t)\right |=\sqrt{1+144(\cos^2(2t)+\sin^2(2t))}$

$\left | \vec{v}(t)\right |=\sqrt{1+144}=\sqrt{145}$

Now we can set up our arc length integral

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

$L=\int_{0}^{2\pi} \sqrt{145}\ dt=t\sqrt{145}\Big|_{0}^{2\pi}=2\pi\sqrt{145}$

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Question

Determine the length of the curve $\vec{r}(t)=\left \langle t,3\sin(4t),3\cos(4t)\right \rangle$ , on the interval $0\leq t\leq 2\pi$

Answer

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

$\vec{v}(t)=\vec{r}\ '(t)=\left \langle 1, 12\cos(4t), -12\sin(4t)\right \rangle$

$\left | \vec{v}(t)\right |=\sqrt{1^2+(12\cos(4t))^2+(-12\sin(4t))^2}$

$\left | \vec{v}(t)\right |=\sqrt{1+144\cos^2(4t)+144\sin^2(4t)}$

$\left | \vec{v}(t)\right |=\sqrt{1+144(\cos^2(2t)+\sin^2(2t))}$

$\left | \vec{v}(t)\right |=\sqrt{1+144}=\sqrt{145}$

Now we can set up our arc length integral

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

$L=\int_{0}^{2\pi} \sqrt{145}\ dt=t\sqrt{145}\Big|_{0}^{2\pi}=2\pi\sqrt{145}$

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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 $\vec{r}(t)=\left \langle t,3\sin(4t),3\cos(4t)\right \rangle$ , on the interval $0\leq t\leq 2\pi$

Answer

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

$\vec{v}(t)=\vec{r}\ '(t)=\left \langle 1, 12\cos(4t), -12\sin(4t)\right \rangle$

$\left | \vec{v}(t)\right |=\sqrt{1^2+(12\cos(4t))^2+(-12\sin(4t))^2}$

$\left | \vec{v}(t)\right |=\sqrt{1+144\cos^2(4t)+144\sin^2(4t)}$

$\left | \vec{v}(t)\right |=\sqrt{1+144(\cos^2(2t)+\sin^2(2t))}$

$\left | \vec{v}(t)\right |=\sqrt{1+144}=\sqrt{145}$

Now we can set up our arc length integral

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

$L=\int_{0}^{2\pi} \sqrt{145}\ dt=t\sqrt{145}\Big|_{0}^{2\pi}=2\pi\sqrt{145}$

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