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Calculus 3 : 3-Dimensional Space

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 #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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Example Question #82 : 3 Dimensional Space

Convert the following into Cylindrical coordinates.

$x=\sin(t)$

$y=\cos(t)$

$z=\tan(t)$

Possible Answers:

r=t

$\theta=\tan^{-1}(\tan(t))$

z=1

r=t

$\theta=\tan(\cot(t))$

$z=\tan(t)$

r=1

$\theta=\tan^{-1}(\tan(t))$

$z=\tan(t)$

r=1

$\theta=\tan^{-1}(\cot(t))$

$z=\tan(t)$

r=1

$\theta=\tan^{-1}(\cot(t))$

z=t

Correct answer:

r=1

$\theta=\tan^{-1}(\cot(t))$

$z=\tan(t)$

Explanation:

In order to convert to cylindrical coordinates, we need to recall the conversion equations.

$r=\sqrt{x^2+y^2}$

$\theta=\tan^{-1}(\frac{y}{x})$

z=z

Now lets apply this to our problem.

$r=\sqrt{(\sin(t))^2+(\cos(t))^2}=\sqrt{1}=1$

$\theta=\tan^{-1}\Big(\frac{\cos(t)}{\sin(t)}\Big)=\tan^{-1}(\cot(t))$

$z=\tan(t)$

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Example Question #2 : Cylindrical Coordinates

When converting rectangular coordinates to cylindrical coordinates, which variable remains fixed?

Possible Answers:

x,y

None of them are fixed.

Correct answer:

Explanation:

To convert a point (x,y,z) into cylindrical corrdinates, the transformation equations are

$x = r\cos\theta$

$y = r\sin\theta$

z = z .

Choices for $r, \theta$ may vary depending on the situation, but the coordinate remains the same.

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Example Question #2 : Cylindrical Coordinates

A point in space is located, in Cartesian coordinates, at (5,12,7) . What is the position of this point in cylindrical coordinates?

Possible Answers:

$(17,112.62^{\circ},7)$

$(17,67.38^{\circ},7)$

$(17,67.38^{\circ},13)$

$(13,112.62^{\circ},7)$

$(13,67.38^{\circ},7)$

Correct answer:

$(13,67.38^{\circ},7)$

Explanation:

When given Cartesian coordinates of the form (x,y,z) to cylindrical coordinates of the form $(r,\theta,z)$ , the first and third terms are the most straightforward.

$r=\sqrt{x^2+y^2}$

z=z

Care should be taken, however, when calculating $\theta$ . The formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

For our coordinates (5,12,7)

$r=\sqrt{x^2+y^2}\rightarrow13$

$\theta=arctan(\frac{y}{x})\rightarrow67.38^{\circ}$ (Bearing in mind sign convention)

z=7

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Example Question #2 : Cylindrical Coordinates

A point in space is located, in Cartesian coordinates, at (4,-4,8) . What is the position of this point in cylindrical coordinates?

Possible Answers:

$(5.66,-45^{\circ},8)$

$(5.66,45^{\circ},8)$

$(5.66,-135^{\circ},8)$

$(5.66,135^{\circ},8)$

Correct answer:

$(5.66,-45^{\circ},8)$

Explanation:

When given Cartesian coordinates of the form (x,y,z) to cylindrical coordinates of the form $(r,\theta,z)$ , the first and third terms are the most straightforward.

$r=\sqrt{x^2+y^2}$

z=z

Care should be taken, however, when calculating $\theta$ . The formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

For our coordinates (4,-4,8)

$r=\sqrt{x^2+y^2}\rightarrow5.66$

$\theta=arctan(\frac{y}{x})\rightarrow -45^{\circ}$ (Bearing in mind sign convention)

z=8

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Calculus 3 : 3-Dimensional Space

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

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

Example Question #82 : 3 Dimensional Space

Example Question #2 : Cylindrical Coordinates

Example Question #2 : Cylindrical Coordinates

Example Question #2 : Cylindrical Coordinates

All Calculus 3 Resources

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