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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 #1 : Line Integrals

Write the parametric equations of the line that passes through the points $\left ( 0,1,3\right )$ and $\left ( 2,5,8\right )$ .

Possible Answers:

x=5t+3,y=3t=1,z=-5t+1

x=2t+4,y+3t+1,z=t

x=-2t,y=4t+1,z=5t+3

x=-2t,y=-4t+1,z=-5t+3

Correct answer:

x=-2t,y=-4t+1,z=-5t+3

Explanation:

First, you must find the vector that is parallel to the line.

This vector is

$v=\left \langle x_1-x_2,y_1-y_2,z_1-z_2\right \rangle$ .

From the points we were given, this becomes

$\left \langle -2,-4,-5\right \rangle$ .

To form the parametric equations, we need to pick a point that lies on the line we want.

The point $\left ( 0,1,3\right )$ is used.

The vector form of the line is from the following equation

$r=\left ( x_1,y_1,z_1\right )+t\left \langle v\right \rangle=\left ( 0,1,3\right )+t\left \langle -2,-4,-5\right \rangle$ .

We then rewrite each expression in terms of the variables x, y, and z.

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Example Question #2 : Line Integrals

Evaluate the line integral $\int_C f(x,y)ds$ of the function

f(x,y)=y+cos(x) over the line segment from (0,0) to (4,5)

Possible Answers:

$\frac{11\sqrt{21}}{3}$

$\frac{9\sqrt{21}}{7}$

$\frac{\sqrt{31}}{3}$

$\frac{5\sqrt{41}}{2}$

$\frac{11\sqrt{41}}{4}$

Correct answer:

$\frac{5\sqrt{41}}{2}$

Explanation:

Evaluate the line integral $\int_C f(x,y)ds$ using the function

$f(x,y)=y+\cos(\pi x)$ over the line segment from (0,0) to (4,5)

Define the Parametric Equations to Represent

The points given lie on the line $y = \frac{5}{4}x$ . Define the parameter x = t , then can be written $y =\frac{5}{4}t$ . Therefore, the parametric equations for are:

x(t)=t

$y(t)=\frac{5}{4}t$

_________________________________________________________________

The line integral of a function f(x,y) along the curve with the parametric equation x(t) and y(t) with $a\leq t\leq b$ is defined by:

$\int_C f(x,y)ds = \int_a^bf(x(t),y(t))\left \| \vec{r'}(t)\right \|dt$ (1)

Where $\vec{r'(t)}$ is the vector derivative of the vector $\vec{r}(t)=\left<x(t),y(t)\right>$ , therefore $\left \| \vec{r'(t)}\right \|$ is simply the magnitude of the vector derivative.

______________________________________________________________

Write the vector $\vec{r(t)}$ :

$\vec{r(t)}=\left<t, \frac{5}{4}t\right>$

Differentiate,

$\vec{r'(t)}=\left<1,\frac{5}{4}\right>$

The absolute value (magnitude) of this vector is:

$\left \| \vec{r'}(t)\right \|=\left \| \left<1,\frac{5}{4 }\right>\right \|=\sqrt{1^2 +\left(\frac{5}{4} \right )^2}=\frac{\sqrt{41}}{4}$

Write the function in terms of the parameter :

$f(x(t),y(t))=f\left(t,\frac{5}{4}t\right) =\frac{5}{4}t+\cos(\pi t)$

Insert everything into Equation (1) noting that the limits of integration will be $0\leq t \leq 4$ due to the fact that the parameter x(t)=t varies from to over the line segment we are integrating over.

$\int_C(y+\cos(\pi x))ds=\int_0^4\left(\frac{5}{4}t+\cos(\pi t)\right)\frac{\sqrt{41}}{4}dt$

$=\frac{\sqrt{41}}{4}\left[\frac{5}{8}t^2+\frac{1}{\pi}\sin(\pi t) \right ]_{t=0}^{t=4}$

$=\frac{\sqrt{41}}{4}\left[\frac{5}{8}(16)+\frac{1}{\pi }\sin(4 \pi) -0-\frac{1}{\pi}\sin(0)\right ]$

$=\frac{5\sqrt{41}}{2}$

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Example Question #1 : Arc Length And Curvature

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

Possible Answers:

$\sqrt{65}$

$2\pi\sqrt{65}$

$\pi$

$2\pi$

Correct answer:

$2\pi\sqrt{65}$

Explanation:

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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Example Question #61 : 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:

$\sqrt{145}$

$2\pi\sqrt{145}$

$2\pi$

$2\sqrt{145}$

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 #71 : 3 Dimensional Space

Find the length of the curve $<e^{2t},e^{-2t},2\sqrt{2}t>$ , from t=0 , to t=5

Possible Answers:

$e^{10}$

None of the other answers

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

$e^{5}-e^{-5}$

Correct answer:

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

Explanation:

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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Example Question #3 : Arc Length And Curvature

Find the length of the arc drawn out by the vector function $F(t) = <a\cos t, a\sin t, at>$ with a>0 from t = 0 to $t =\pi$ .

Possible Answers:

$\pi$

$\sqrt2 a \pi$

None of the other answers

$2a^2\pi$

$\pi a$

Correct answer:

$\sqrt2 a \pi$

Explanation:

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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Example Question #72 : 3 Dimensional Space

Evaluate the curvature of the function y=10x^2 at the point (1,10) .

Possible Answers:

$\frac{20}{401^{3/2}}$

$\frac{20}{400^{2/3}}$

$\frac{20}{400^{3/2}}$

$\frac{20}{401^{2/3}}$

Correct answer:

$\frac{20}{401^{3/2}}$

Explanation:

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 f'(x) =20x, f''(x) = 20 , 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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Example Question #1 : Arc Length And Curvature

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 $0\leq t \leq 2$ .

Possible Answers:

$\frac{38}{3}$

$\frac{25}{3}$

$\frac{10}{3}$

Correct answer:

$\frac{38}{3}$

Explanation:

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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Example Question #7 : Arc Length And Curvature

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

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

Possible Answers:

Correct answer:

Explanation:

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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Example Question #8 : Arc Length And Curvature

Find the arc length of the curve

c(t)=<3t,2sin(2t),2cos(2t)>

on the interval

$0\leq t\leq 10$

Possible Answers:

Correct answer:

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

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

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

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

and following the arc length formula we solve for the integral

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

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

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

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

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

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

$=5t|_0^{10}$

=5(10)-5(0)

Hence the arc length is

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Line Integrals

Example Question #2 : Line Integrals

Example Question #1 : Arc Length And Curvature

Example Question #61 : 3 Dimensional Space

Example Question #71 : 3 Dimensional Space

Example Question #3 : Arc Length And Curvature

Example Question #72 : 3 Dimensional Space

Example Question #1 : Arc Length And Curvature

Example Question #7 : Arc Length And Curvature

Example Question #8 : Arc Length And Curvature

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