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Calculus 3 : Calculus 3

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

Example Questions

Example Question #11 : Equations Of Lines And Planes

Find the unit tangent vector to the curve given by

$\vec{r}(t)=3t\hat{i}+e^t\hat{j}+t^3\hat{k}$

Possible Answers:

$\frac{3\hat{i}+e^t\hat{j}+3t^2\hat{k}}{\sqrt{9+e^{2t}+9t^4}}$

$\frac{3\hat{i}+e^t\hat{j}+9t^4\hat{k}}{\sqrt{9+e^{2t}+9t^4}}$

${3\hat{i}+e^t\hat{j}+3t^2\hat{k}}$

$\frac{3\hat{i}+e^t\hat{j}+3t^2\hat{k}}{\sqrt{3+e^{t}+3t^2}}$

Correct answer:

$\frac{3\hat{i}+e^t\hat{j}+3t^2\hat{k}}{\sqrt{9+e^{2t}+9t^4}}$

Explanation:

To find the unit tangent vector for a given curve, we must find the derivative of each of the components of the curve, and then divide this vector - the tangent vector - by the magnitude of the tangent vector itself.

To start, let's find the derivatives:

$\vec{r}(t)=3\hat{i}+e^t\hat{j}+3t^2\hat{k}$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$

Next, we must find the magnitude of the tangent vector, by taking the square root of the sum of the squares the x, y, and z components. Note that the vector we were given was in unit vector notation, and i, j, and k are each equal to one unit, so they don't impact the magnitude of the vector (they simply indicate direction).

$\left |\vec{r}(t) \right | = \sqrt{3^2+(e^t)^2+(3t^2)^2}=\sqrt{9+e^{2t}+9t^4}$

Finally, divide the vector by its magnitude:

$\frac{3\hat{i}+e^t\hat{j}+3t^2\hat{k}}{\sqrt{9+e^{2t}+9t^4}}$

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Example Question #11 : Equations Of Lines And Planes

Find the unit tangent vector to the given curve:

$\vec{r}=19t\hat{i}+\cos(t)\hat{j}+\sin(t)\hat{k}$

Possible Answers:

$\frac{19t\hat{i}-\sin(t)\hat{j}+\cos(t)\hat{k}}{\sqrt{382}}$

$\frac{19\hat{i}-\sin(t)\hat{j}+\cos(t)\hat{k}}{19}$

$\frac{19\hat{i}+\sin(t)\hat{j}+\cos(t)\hat{k}}{\sqrt{382}}$

$\frac{19\hat{i}-\sin(t)\hat{j}+\cos(t)\hat{k}}{\sqrt{382}}$

Correct answer:

$\frac{19\hat{i}-\sin(t)\hat{j}+\cos(t)\hat{k}}{\sqrt{382}}$

Explanation:

To start, let's find the derivatives:

$\vec{r}'(t)=19\hat{i}-\sin(t)\hat{j}+\cos(t)\hat{k}$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$

$\left | \vec{r}'(t)\right |=\sqrt{(19)^2+(-\sin(t))^2+(\cos(t))^2}=\sqrt{19^2+1}=\sqrt{382}$

Finally, divide the tangent vector by its magnitude:

$\frac{19\hat{i}-\sin(t)\hat{j}+\cos(t)\hat{k}}{\sqrt{382}}$

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Example Question #13 : 3 Dimensional Space

Find the unit tangent vector given by the curve

$\vec{r}(t)=(te^t)\hat{i}+t^2\hat{j}+t^3\hat{k}$

Possible Answers:

$\frac{(2e^t)\hat{i}+2t\hat{j}+3t^2\hat{k}}{\sqrt{e^{2t}(1+2t+t^2)+4t^2+9t^4}}$

$\frac{(e^t+te^t)\hat{i}+2t\hat{j}+3t^2\hat{k}}{\sqrt{e^{2t}(1+2t+t^2)+4t^2+9t^4}}$

$\frac{(e^t+te^t)\hat{i}+2t\hat{j}+3t^2\hat{k}}{\sqrt{e^{2t}(2t+t^2)+4t^2+9t^4}}$

$(e^t+te^t)\hat{i}+2t\hat{j}+3t^2\hat{k}}$

Correct answer:

$\frac{(e^t+te^t)\hat{i}+2t\hat{j}+3t^2\hat{k}}{\sqrt{e^{2t}(1+2t+t^2)+4t^2+9t^4}}$

Explanation:

To start, let's find the derivatives:

$\vec{r}'(t)=(e^t+te^t)\hat{i}+2t\hat{j}+3t^2\hat{k}$

We used the following rules to find them:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$

$\left | \vec{r}'(t)\right |=\sqrt{(e^t+te^t)^2+(2t)^2+(3t^2)^2}=\sqrt{e^{2t}+2te^{2t}+t^2e^{2t}+4t^2+9t^4}$

Finally, divide the tangent vector by its magnitude:

$\frac{(e^t+te^t)\hat{i}+2t\hat{j}+3t^2\hat{k}}{\sqrt{e^{2t}(1+2t+t^2)+4t^2+9t^4}}$

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Example Question #131 : Calculus 3

Write the equation of the line - in vector form - passing through the two points

P_1=(1, 0, 0), P_2=(0, 3, 5)

Possible Answers:

$\left \langle t, 3, 5\right \rangle$

$\left \langle t, 3t, 5t\right \rangle$

$\left \langle t, 3-3t, 5-5t\right \rangle$

$\left \langle t, 3+3t, 5+5t\right \rangle$

Correct answer:

$\left \langle t, 3-3t, 5-5t\right \rangle$

Explanation:

To find the equation of the line, we first must find the vector passing between the two points:

$\vec{v}=(1, -3, -5)$

This was found by P_2-P_1 .

Next, we pick one point, and using the formula

$\vec{r}=\left \langle x_0, y_0, z_0\right \rangle+t\cdot \vec{v}$

we get our line:

$\vec{r}=\left \langle 0, 3, 5\right \rangle+t\left \langle 1, -3, -5\right \rangle=\left \langle t, 3-3t, 5-5t\right \rangle$

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Example Question #11 : Equations Of Lines And Planes

Write the equation for the plane containing the points

A=(2, 3, 4), B=(0, 0, 0), C=(3, -1, -4)

Possible Answers:

x=0

8x-20y+11z=0

3x-y-4z=0

-2x-3y-4z=0

Correct answer:

8x-20y+11z=0

Explanation:

To find the equation of the plane, we need a point on the plane, and the normal vector to the plane, which is given by the cross product of two vectors in the plane.

To start, let's find these vectors:

$\vec{AB}=\left \langle -2, -3, -4\right \rangle, \vec{BC}=\left \langle 3, -1, -4\right \rangle$

We found them by subtracting the terminal point (B, C) and the initial point (A, B, respectively).

Now, we can write the determinant in order to take the cross product of the two vectors:

$\vec{AB}X\vec{BC}=\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ -2& -3&-4 \\ 3& -1&-4 \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

Next, we take the cross product. One can do this by multiplying across from the top left to the lower right, and continuing downward, and then subtracting the terms multiplied from top right to the bottom left:

$12\hat{i}+2\hat{k}-12\hat{j}-(-9\hat{k}+4\hat{i}+8\hat{j})=8\hat{i}-20\hat{j}+11\hat{k}$

The equation of a plane is given by

a(x-x_0)+b(y-y_0)+c(z-z_0)=0 , where $\left \langle a, b, c\right \rangle$ is the normal vector and (x_0, y_0, z_0) is any point on the plane.

Using our normal vector $\left \langle 8, -20, 11\right \rangle$ and the point (0, 0, 0) , we get the equation

8x-20y+11z=0

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Example Question #131 : Calculus 3

Find the equation of the plane given by the three points

(2, 2, 0), (1, 5, 3), (4, 2, 0)

Possible Answers:

6y-6z=-12

-18x+12y+12z=24

6y-6z=0

6y-6z=12

Correct answer:

6y-6z=12

Explanation:

To find the equation of the plane given by three points, we need the normal vector to the plane and a point on the plane.

We first must find two vectors on the plane:

$\left \langle 1, -3, -3 \right \rangle, \left \langle -3, 3, 3 \right \rangle$

These were found by taking the difference between the first and second, and second and third points respectively.

Now, we can write the determinant in order to take the cross product of the two vectors:

$\begin{vmatrix} \hat{i} &\hat{j} & \hat{k}\\ 1 & -3 &-3 \\ -3 &3 &3 \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

$6\hat{j}-6\hat{k}$ $6\hat{j}-6\hat{k} = \left \langle 0, 6, 6 \right \rangle$

Now, we plug this, and a point of our choosing, into the equation for a plane:

a(x-x_0)+b(y-y_0)+c(z-z_0)=0 , where $\left \langle a, b, c \right \rangle$ is the normal vector and (x_0, y_0, z_0) is any given point.

Plugging this in, we get

0(x-2)+6(y-2)-6(z-0)=0

6y-6z=12

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Example Question #131 : Calculus 3

Find the equation of the plane that contains the point (1,-2,3) , and has a normal vector $n=\left \langle 3,2,5 \right \rangle$ .

Possible Answers:

3x+3y-2z=-8

3x+2y+4z=8

3x-2y+3z=-8

3x+2y+3z=8

Correct answer:

3x+2y+3z=8

Explanation:

Using the formula for a plane A(x-x_0)+B(y-y_0)+C(z-z_0)=0 , where the point given is (x_0, y_0, z_0) and the normal vector is $\left \langle A,B,C\right \rangle$ . Plugging in the known values, you get 3(x-1)+2(y+2)+3(z-3)=0 . Manipulating this equation through algebra gives you the answer 3x+2y+3z=8

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Example Question #132 : Calculus 3

Write the equation of the plane given by the three points

A=(0, -2, -2), B=(0, 1, 5), C=(3, 2, 3)

Possible Answers:

6x-6y+6z=24

6x-6y+6z=60

6x-6y+6z=0

2x-6y-6z=44

Correct answer:

6x-6y+6z=24

Explanation:

To find the equation of a plane, we need a point on the plane and the normal vector to the plane.

First, we must find two vectors on the plane using the given points (taking the difference between the two points in each vector):

$\vec{AB}=\left \langle 0, -2, -2\right \rangle, \vec{BC}=\left \langle 3, 1, -2\right \rangle$

Now, we can write the determinant in order to take the cross product of the two vectors:

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 0&-2 &-2 \\ 3& 1& -2 \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

$4\hat{i}-6\hat{j}-(-6\hat{k}-2\hat{i})=6\hat{i}-6\hat{j}+6\hat{k}=\left \langle 6, -6, 6\right \rangle$

Now, we choose a point (for example, (2, 3, 5) ) and use the normal vector given by $\left \langle a, b, c\right \rangle$ from the cross product:

a(x-x_0)+b(y-y_0)+c(z-z_0)=0

Plugging all of this in, and simplifying, we get

6x-6y+6z=24

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Example Question #133 : Calculus 3

Find the equation of the plane that contains the point $\left ( 0,3,4\right )$ and has a normal vector $\left \langle 1,1,5\right \rangle$ .

Possible Answers:

x-y-5z=20

x+y+5z=20

x+y+5z=23

x+y-5z=23

Correct answer:

x+y+5z=23

Explanation:

The formula of a plane containing a point $\left ( x_0,y_0,z_0\right )$ and with a normal vector $\left \langle A,B,C\right \rangle$ is

A(x-x_0)+B(y-y_0)+C(z-z_0)=0 ,

Using what we were given, we get

$1\left ( x-0\right )+1(y-3)+5(z-4)=0$ .

Through algebra, we get

x+y+5z=23 .

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Example Question #134 : Calculus 3

Find the equation of the plane that contains the point $\left ( 3,2,8\right )$ and has a normal vector $n=\left \langle 3,0,5\right \rangle$ .

Possible Answers:

3x+5z=-49

2x-2y+5z=49

3x+5z=49

x+6y=31

Correct answer:

3x+5z=49

Explanation:

Using the formula for a plane, we have

A(x-x_0)+B(y-y_0)+C(z-z_0)=0 ,

where the point given is (x_0, y_0, z_0) and the normal vector is $\left \langle A,B,C\right \rangle$ .

Plugging in the known values, you get

3(x-3)+0(y-2)+5(z-8)=0 .

Manipulating this equation through algebra gives you the answer

3x+5z=49

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #11 : Equations Of Lines And Planes

Example Question #11 : Equations Of Lines And Planes

Example Question #13 : 3 Dimensional Space

Example Question #131 : Calculus 3

Example Question #11 : Equations Of Lines And Planes

Example Question #131 : Calculus 3

Example Question #131 : Calculus 3

Example Question #132 : Calculus 3

Example Question #133 : Calculus 3

Example Question #134 : Calculus 3

All Calculus 3 Resources

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