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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #601 : Calculus 3

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.

Possible Answers:

35.5

45.4

20.5

26.2

Correct answer:

26.2

Explanation:

To find the arc length of the curve function

c(t)

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

u=t+25 and du=dt

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 26.2

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Example Question #1 : Arc Length And Curvature

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

Possible Answers:

$4\sqrt{\pi}$

$\sqrt{2}\pi$

$4\sqrt{2}\pi$

$2\sqrt{2}\pi$

Correct answer:

$2\sqrt{2}\pi$

Explanation:

Untitled

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Example Question #603 : Calculus 3

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.

Possible Answers:

81.3

57.1

96.6

60.7

Correct answer:

60.7

Explanation:

To find the arc length of the curve function

c(t)

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 u=t+25 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

60.7 units

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Example Question #601 : Calculus 3

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

Possible Answers:

$k=\frac{30}{(36t^2+25)^{\frac{1}{2}}}$

$k=\frac{30}{(36t^2+25)}$

$k=\frac{30}{(36t^2+25)^{\frac{3}{2}}}$

$k=\frac{25}{(36t^2+25)^{\frac{3}{2}}}$

Correct answer:

$k=\frac{30}{(36t^2+25)^{\frac{3}{2}}}$

Explanation:

Using the formula for curvature $k=\frac{\left \| r'(t)\times r''(t) \right \|}{\left \| r'(t)\right \|^3}$ . r'(t)=6ti+5j, r''(t)=6i , $r'(t)\times r''(t) =-30k, \left \| r'(t)\times r''(t)\right \|=30$ , 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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Example Question #602 : Calculus 3

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

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

Possible Answers:

$15\sqrt{3}$

$\sqrt{29}$

$5\sqrt{26}$

$5\sqrt{29}$

Correct answer:

$5\sqrt{29}$

Explanation:

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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Example Question #603 : Calculus 3

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

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

Possible Answers:

$\sqrt{5}$

$2\sqrt{5}$

$4\sqrt{5}$

Correct answer:

$4\sqrt{5}$

Explanation:

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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Example Question #604 : Calculus 3

Given that

y=x^2

Find an expression for the curvature of the given conic

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

Possible Answers:

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

$\frac{2x}{(1+8x^2)^{\frac{3}{2}}}$

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

Correct answer:

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

Explanation:

Step 1: Find the first and the second derivative

y=x^2,y'=2x, y''=2

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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Example Question #605 : Calculus 3

Find an integral for the arc length of

$f(x)=(x+6)^\frac{3}{2}$ on the interval [-7,30] (Set up, DO NOT SOLVE)

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

Step 1:

Find the first derivative of the function f(x)

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

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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Example Question #606 : Calculus 3

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

Possible Answers:

$2\sqrt{145}$

$2\pi\sqrt{145}$

$2\pi$

$\sqrt{145}$

Correct answer:

$2\pi\sqrt{145}$

Explanation:

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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Example Question #81 : 3 Dimensional Space

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

Possible Answers:

$2\pi\sqrt{145}$

$2\pi$

$2\sqrt{145}$

$\sqrt{145}$

Correct answer:

$2\pi\sqrt{145}$

Explanation:

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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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #601 : Calculus 3

Example Question #1 : Arc Length And Curvature

Example Question #603 : Calculus 3

Example Question #601 : Calculus 3

Example Question #602 : Calculus 3

Example Question #603 : Calculus 3

Example Question #604 : Calculus 3

Example Question #605 : Calculus 3

Example Question #606 : Calculus 3

Example Question #81 : 3 Dimensional Space

All Calculus 3 Resources

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