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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 #141 : Calculus 3

Find the equation of the plane that contains the point $\left ( 3,2,10\right )$ and has a normal vector $n=\left \langle 2,1,1\right \rangle$ .

Possible Answers:

2x+y+z=-18

2x-y+2z=9

x+y+3z=18

2x+y+z=18

Correct answer:

2x+y+z=18

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

2(x-3)+1(y-2)+1(z-10)=0 .

Manipulating this equation through algebra gives you the answer

2x+y+z=18

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Example Question #21 : 3 Dimensional Space

Determine the equation of the plane containing the three points

O=(0,0,0)

P=(5,4,7)

Q=(-2,5,6)

Possible Answers:

7x-y+z=0

-10x-20y+42z=0

-11x-44y+33z=0

3x-9y+13z=0

Correct answer:

-11x-44y+33z=0

Explanation:

The equation of a plane is defined as

$n\bullet <x-x_0,y-y_0,z-z_0>=0$

where

is the normal vector of the plane.

To find the normal vector, we first get two vectors on the plane

OP=<5,4,7>

and

OQ=<-2,5,6>

and find their cross product.

The cross product is defined as the determinant of the matrix

$\begin{vmatrix} i& j& k\\ 5& 4& 7\\ -2&5 &6 \end{vmatrix}$

Which is

$=det(\begin{vmatrix} 4& 7\\5 &6 \end{vmatrix})i-det(\begin{vmatrix} 5& 7\\-2 &6 \end{vmatrix})j+det(\begin{vmatrix} 5& 4\\ -2&5 \end{vmatrix})k$

=(4*6-5*7)i-(5*6-(-2)*7)j+(5*5-(-2)*4)k

=-11i-44j+33k

Which tells us the normal vector is

<-11,-44,33>

Using the point

O=(0,0,0)

and the normal vector to find the equation of the plane yields

$<-11,-44,33>\bullet <x-0,y-0,z-0> = 0$

$<-11,-44,33>\bullet <x,y,z> = 0$

-11x+(-44)y+33z=0

Simplified gives the equation of the plane

-11x-44y+33z=0

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Example Question #142 : Calculus 3

Find the unit tangent vector to the given curve

$\vec{r}(t)=\left \langle 10t, -\sin(t), e^t\right \rangle$

Possible Answers:

$\left \langle 10t, -\sin(t), e^t\right \rangle$

$\left \langle \frac{10t}{\sqrt{100t^2+\sin^2(t)+e^{2t}}}, \frac{-\sin(t)}{\sqrt{100t^2+\sin^2(t)+e^{2t}}}, \frac{e^{t}}{\sqrt{100t^2+\sin^2(t)+e^{2t}}}\right \rangle$

$\left \langle \frac{5t^2}{\sqrt{100t^2+\sin^2(t)+e^{2t}}}, \frac{\sin^2(t)}{\sqrt{100t^2+\sin^2(t)+e^{2t}}}, \frac{e^{2t}}{\sqrt{100t^2+\sin^2(t)+e^{2t}}}\right \rangle$

$\left \langle \frac{10t}{\sqrt{100t^2+\sin^2(t)+e^{2t}}}, \frac{-\sin(t)}{\sqrt{100t^2+\sin^2(t)+e^{2t}}}, \frac{e^{t}}{\sqrt{100t^2+\sin^2(t)+e^{2t}}}\right \rangle$

Correct answer:

$\left \langle \frac{10t}{\sqrt{100t^2+\sin^2(t)+e^{2t}}}, \frac{-\sin(t)}{\sqrt{100t^2+\sin^2(t)+e^{2t}}}, \frac{e^{t}}{\sqrt{100t^2+\sin^2(t)+e^{2t}}}\right \rangle$

Explanation:

To find the unit tangent vector, we must find the tangent vector and divide it by its magnitude.

To find the tangent vector, we take the derivative of each of the components:

$\vec{r}'(t)=\left \langle 10t, -\sin(t), e^t\right \rangle$

The derivatives were found using the following rules:

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

Now, we find the magnitude of the tangent vector by taking the square root of the sum of its components:

$\left | \vec{r}'(t)\right |=\sqrt{100t^2+\sin^2(t)+e^{2t}}$

Our final answer is

$\left \langle \frac{10t}{\sqrt{100t^2+\sin^2(t)+e^{2t}}}, \frac{-\sin(t)}{\sqrt{100t^2+\sin^2(t)+e^{2t}}}, \frac{e^{t}}{\sqrt{100t^2+\sin^2(t)+e^{2t}}}\right \rangle$

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Example Question #143 : Calculus 3

Find the equation of the plane that has a normal vector $n=\left \langle 3,2,8\right \rangle$ and a point on the plane $\left ( 5,-1,3\right )$ .

Possible Answers:

2x+3y+5z=20

x+4y+3z=15

3x+2y+8z=37

3x+4y-5z=19

Correct answer:

3x+2y+8z=37

Explanation:

If we are given both the normal vector to the plane and a point on the plane, we can use the formula A(x-x_0)+B(y-y_0)+C(z-z_0)=0 , where $n=\left \langle A,B,C\right \rangle$ and the point on the plane is (x_0,y_0,z_0) . Plugging in what we know, we get 3(x-5)+2(y+1)+8(z-3)=0 . Manipulating this equation through algebra and making the variables all on one side, we get 3x+2y+8z=37

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Example Question #144 : Calculus 3

Find the equation of the plane that contains the point (1,-3,5) and the normal vector $n=\left \langle 0,1,2\right \rangle$

Possible Answers:

y+z=10

y+2z=7

x+2y+z=7

y+2z=5

Correct answer:

y+2z=7

Explanation:

To find the equation of the plane, we use the formula 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 $n=\left \langle A,B,C\right \rangle$ . Plugging in what we were given in the problem statement, we get 0(x-1)+1(y+3)+2(z-5)=0 . Manipulating the equation through algebra to isolate the variables, we get y+2z=7

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Example Question #31 : Equations Of Lines And Planes

Find the equation of the plane given by a point on the plane $\left ( 2,-1,-4\right )$ and the normal vector to the plane $n=\left \langle 2,3,9\right \rangle$

Possible Answers:

2x+3y+8z=34

2x+3y-9z=36

2x+3y+9z=-35

2x+3y+9z=35

Correct answer:

2x+3y+9z=-35

Explanation:

To find the equation of the plane, we use the formula 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 $n=\left \langle A,B,C\right \rangle$ . Plugging in what we were given in the problem statement, we get 2(x-2)+3(y+1)+9(z+4)=0 . Manipulating the equation through algebra to isolate the variables, we get 2x+3y+9z=-35

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Example Question #141 : Calculus 3

Find the equation of the plane given by a point on the plane $\left ( -1,2,-4\right )$ and the normal vector to the plane $n=\left \langle 0,4,5\right \rangle$

Possible Answers:

x+4y+5z=12

x+5z=12

4y+5z=12

4y+5z=-12

Correct answer:

4y+5z=-12

Explanation:

To find the equation of the plane, we use the formula 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 $n=\left \langle A,B,C\right \rangle$ . Plugging in what we were given in the problem statement, we get 0(x+1)+4(y-2)+5(z+4)=0 . Manipulating the equation through algebra to isolate the variables, we get 4y+5z=-12

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Example Question #31 : 3 Dimensional Space

Find the equation of the line that passes through the points (0,2,1) and (5,4,3)

Note: The answer you obtain needs to be in vector form. Also, use the point (0,2,1) when forming the equation

Possible Answers:

$\left \langle 4t,3t+1,t\right \rangle$

$\left \langle 4t,t+1,-2t+2\right \rangle$

$\left \langle 3t,2t+1,3\right \rangle$

$\left \langle -5t,-2t+2,-2t+1\right \rangle$

Correct answer:

$\left \langle -5t,-2t+2,-2t+1\right \rangle$

Explanation:

To find the equation of the line, we need to find the vector that will be parallel to the line (its direction). This vector is formed by the points a=(x_1,y_1,z_1) and b=(x_2,y_2,z_2) (the ones from the problem statement.) We find the direction vector to be $\left \langle x_1-x_2,y_1-y_2,z_1-z_2\right \rangle=\left \langle 0-5,2-4,1-3\right \rangle=\left \langle -5,-2,-2\right \rangle$

We then pick a point on the line. We chose (0,2,1) . We then form the equation of the line by using the formula $\left \langle 0,1,2\right \rangle+t\left \langle-5,-2,-2 \right \rangle=\left \langle -5t,-2t+2,2t+1\right \rangle$

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Example Question #145 : Calculus 3

Find the equation of the plane given by a point on the plane $\left ( 5,-1,3\right )$ and the normal vector to the plane $n=\left \langle 2,3,1\right \rangle$

Possible Answers:

2x+3y+z=10

3x+2y-9z=20

x+2y+5z=-9

x+4y+2z=8

Correct answer:

2x+3y+z=10

Explanation:

To find the equation of the plane, we use the formula 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 $n=\left \langle A,B,C\right \rangle$ . Plugging in what we were given in the problem statement, we get 2(x-5)+3(y+1)+1(z-3)=0 . Manipulating the equation through algebra to isolate the variables, we get 2x+3y+z=10

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Example Question #144 : Calculus 3

Find the equation of the plane containing the point $\left ( 3,-2,1\right )$ , and is parallel to the plane with the equation 2x+5y+z=20

Possible Answers:

x+2y+5z=4

2x+5y+z=14

2x+5y+z=-3

2x+5y+z=3

Correct answer:

2x+5y+z=-3

Explanation:

We were given a point on the plane, and we need the normal vector to the plane. It is known that two planes that are parallel to each other have the same normal vector, so in this case $n=\left \langle 2,5,1\right \rangle$ (given by the equation of the other plane). To complete the problem, we use the equation A(x-x_0)+B(y-y_0)+C(z-z_0)=0 , where $n=\left \langle A,B,C\right \rangle$ and the point on the plane is (x_0,y_0,z_0) . Using the information we have, we get:

2(x-3)+5(y+2)+1(z-1)=0 . Through algebraic manipulation, we then get:

2x+5y+z=-3

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #141 : Calculus 3

Example Question #21 : 3 Dimensional Space

Example Question #142 : Calculus 3

Example Question #143 : Calculus 3

Example Question #144 : Calculus 3

Example Question #31 : Equations Of Lines And Planes

Example Question #141 : Calculus 3

Example Question #31 : 3 Dimensional Space

Example Question #145 : Calculus 3

Example Question #144 : Calculus 3

All Calculus 3 Resources

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