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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 #42 : Applications Of Partial Derivatives

Find $\nabla f$ for the given function:

$f(x, y, z)=e^{\tan(xz)}+\sin(y)$

Possible Answers:

$\left \langle sec^2(xz)e^{\tan(xz)}, \cos(y), sec^2(xz)e^{\tan(xz)}\right \rangle$

$\left \langle x\sec^2(xz)e^{\tan(xz)}, \cos(y), z\sec^2(xz)e^{\tan(xz)}\right \rangle$

$\left \langle z\sec^2(xz)e^{\tan(xz)}, -\cos(y), x\sec^2(xz)e^{\tan(xz)}\right \rangle$

$\left \langle z\sec^2(xz)e^{\tan(xz)}, \cos(y), x\sec^2(xz)e^{\tan(xz)}\right \rangle$

Correct answer:

$\left \langle z\sec^2(xz)e^{\tan(xz)}, \cos(y), x\sec^2(xz)e^{\tan(xz)}\right \rangle$

Explanation:

The gradient of a function is given by

$\nabla f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=z\sec^2(xz)e^{\tan(xz)}$

$f_y=\cos(y)$

$f_z=x\sec^2(xz)e^{\tan(xz)}$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$ . $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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Example Question #43 : Applications Of Partial Derivatives

Find $\nabla f$ of the following function:

$f(x, y, z)=x^2+z^4+y^4\cos(z)$

Possible Answers:

$\left \langle 2x, 4y^3, 4z^3-\sin(z)\right \rangle$

$\left \langle 2x, 4y^3\cos(z), 4z^3+y^4\sin(z)\right \rangle$

$\left \langle 2x, 4y^3\cos(z), 4z^3-y^4\sin(z)\right \rangle$

$\left \langle 2x, 4y^3\cos(z), 4z^3\right \rangle$

Correct answer:

$\left \langle 2x, 4y^3\cos(z), 4z^3-y^4\sin(z)\right \rangle$

Explanation:

The gradient of a function is given by

$\nabla f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

f_x=2x

$f_y=4y^3\cos(z)$

$f_z=4z^3-y^4\sin(z)$

The derivatives were found using the following rules:

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

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Example Question #1581 : Calculus 3

Find $\nabla f$ for the following function:

$f(x,y,z)=e^{\sec(x)}+\tan(yz)$

Possible Answers:

$\left \langle \sec(x)\tan(x)e^{\sec(x)}, y\sec^2(yz), z\sec^2(yz)\right \rangle$

$\left \langle \sec(x)\tan(x)e^{\sec(x)}, z\sec(yz), y\sec(yz)\right \rangle$

$\left \langle \sec(x)\tan(x)e^{\sec(x)}, z\sec^2(yz), y\sec^2(yz)\right \rangle$

$\sec(x)\tan(x)e^{\sec(x)}+z\sec^2(yz)+y\sec^2(yz)$

Correct answer:

$\left \langle \sec(x)\tan(x)e^{\sec(x)}, z\sec^2(yz), y\sec^2(yz)\right \rangle$

Explanation:

The gradient of a function is given by

$\nabla f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=\sec(x)\tan(x)e^{\sec(x)}$

$f_y=z\sec^2(yz)$

$f_z=y\sec^2(yz)$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$

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Example Question #42 : Applications Of Partial Derivatives

Find $\bigtriangledown f$ for the following function:

$f(x, y, z)=xyz^4+y^2+e^{xz}$

Possible Answers:

$\left \langle z^4y+xe^{xz}, xz^4+2y, 4z^3xy+ze^{xz}\right \rangle$

$\left \langle xe^{xz}, 2y, ze^{xz}\right \rangle$

$\left \langle z^4y+ze^{xz}, xz^4+2y, 4z^3xy+xe^{xz}\right \rangle$

$z^4y+ze^{xz} + xz^4+2y+4z^3xy+xe^{xz}$

Correct answer:

$\left \langle z^4y+ze^{xz}, xz^4+2y, 4z^3xy+xe^{xz}\right \rangle$

Explanation:

The gradient of a function is given by

$\bigtriangledown f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=z^4y+ze^{xz}$

f_y=xz^4+2y

$f_z=4z^3xy+xe^{xz}$

The partial derivatives were found using the following rules:

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

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Example Question #21 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\bigtriangledown f$ of the function:

$f(x, y, z)=x\cos(e^y)+z^4y$

Possible Answers:

$\cos(e^y)-xe^y\sin(e^y) +4z^3y$

$\left \langle \cos(e^y), xe^y\sin(e^y), 4z^3y\right \rangle$

$\left \langle \cos(e^y), -xe^y\sin(e^y), 4z^3y\right \rangle$

$\left \langle \cos(e^y), -xe^y\sin(e^y), 4z^3\right \rangle$

Correct answer:

$\left \langle \cos(e^y), -xe^y\sin(e^y), 4z^3y\right \rangle$

Explanation:

The gradient of a function is given by

$\bigtriangledown f=\left \langle f_x, f_y, f_z\right \rangle$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=\cos(e^y)$

$f_y=-xe^y\sin(e^y)$

f_z=4z^3y

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #22 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\bigtriangledown F$ of the function x^3+2y^2+z

Possible Answers:

$\left \langle x^2,3y,z\right \rangle$

$\left \langle 2x,y,z\right \rangle$

$\left \langle 3x^2,4y,1\right \rangle$

$\left \langle 3x^2,y^2,z\right \rangle$

Correct answer:

$\left \langle 3x^2,4y,1\right \rangle$

Explanation:

The formula for the gradient of F is

$\left \langle \frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z}\right \rangle$ .

Using the rules for partial differentiation, we get

$\frac{\partial F}{\partial x}=3x^2, \frac{\partial F}{\partial y}=4y, \frac{\partial F}{\partial z}=1$ .

Putting into vector notation, we get

$\left \langle 3x^2,4y,1\right \rangle$ .

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Example Question #23 : Gradient Vector, Tangent Planes, And Normal Lines

Find the equation of the plane tangent to the point $\left ( 3,5,8\right )$ if the gradient vector $\bigtriangledown F=\left \langle 3,1,2\right \rangle$ .

Possible Answers:

3x+y+2z=30

x-4y-z=15

x+5y-2z=28

2x+3y+2z=30

Correct answer:

3x+y+2z=30

Explanation:

By definition, $\bigtriangledown F$ is the vector that orthogonal to the plane at the point we were given.

We then use the formula for a plane given a point $\left ( x_0,y_0,z_0\right )$ and normal vector $\bigtriangledown F$ .

We get

3(x-3)+1(y-5)+2(z-8)=0 .

Through algebraic manipulation, we get

3x+y+2z=30 .

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Example Question #1581 : Calculus 3

Find the gradient, $\bigtriangledown F$ , of the function F=x^2y+xyz .

Possible Answers:

$\left \langle x^3y,2x,xyz\right \rangle$

$\left \langle 2xy+yz,x^2+xz,xy\right \rangle$

$\left \langle 2x,2xy+yz,1\right \rangle$

$\left \langle 2xy,yz,z\right \rangle$

Correct answer:

$\left \langle 2xy+yz,x^2+xz,xy\right \rangle$

Explanation:

The gradient of a function F(x,y,z) is as follows:

$\bigtriangledown F=\left \langle \frac{\mathrm{dF} }{\mathrm{d} x},\frac{\mathrm{dF} }{\mathrm{d} y},\frac{\mathrm{dF} }{\mathrm{d} z} \right \rangle$ .

We compute the derivative of the function with respect to each of the variables and treat the others like constants.

Using the rule

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

we obtain

$\frac{\mathrm{dF} }{\mathrm{d} x}=2xy+yz$ ,

$\frac{\mathrm{dF} }{\mathrm{d} y}=x^2+xz$ ,

and

$\frac{\mathrm{dF} }{\mathrm{d} z}=xy$ .

Putting these expressions into the vector completes the problem, and you obtain

$\left \langle 2xy+yz,x^2+xz,xy\right \rangle$ .

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Example Question #25 : Gradient Vector, Tangent Planes, And Normal Lines

Find the gradient, $\bigtriangledown F$ , of the function F=x^3y^5z+yz+x^3 .

Possible Answers:

$\left \langle 3x^2y^5z+3x^2,5x^3y^4z+z,x^3y^5+y\right \rangle$

$\left \langle 5x^4y^5+5x^2,3xy+z,x^3y^5+y\right \rangle$

$\left \langle 5x^4yz,3xz,z\right \rangle$

$\left \langle 3y^5z+3x^2,5x^3y^3+y,x^3y^5+z\right \rangle$

Correct answer:

$\left \langle 3x^2y^5z+3x^2,5x^3y^4z+z,x^3y^5+y\right \rangle$

Explanation:

The gradient of a function F(x,y,z) is as follows:

$\bigtriangledown F=\left \langle \frac{\mathrm{dF} }{\mathrm{d} x},\frac{\mathrm{dF} }{\mathrm{d} y},\frac{\mathrm{dF} }{\mathrm{d} z} \right \rangle$ .

We compute the derivative of the function with respect to each of the variables and treat the others like constants.

Using the rule

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

we obtain

$\frac{\mathrm{dF} }{\mathrm{d} x}=3x^2y^5z+3x^2$ ,

$\frac{\mathrm{dF} }{\mathrm{d} y}=5x^3y^4z+z$ ,

and $\frac{\mathrm{dF} }{\mathrm{d} z}=x^3y^5+y$ .

Putting these expressions into the vector completes the problem, and you obtain

$\left \langle 3x^2y^5z+3x^2,5x^3y^4z+z,x^3y^5+y\right \rangle$ .

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Example Question #1582 : Calculus 3

Find the gradient, $\bigtriangledown F$ , of the function F=xy^3+y^4+xyz .

Possible Answers:

$\left \langle y^4+yz,3xy^2+4y^3,xy\right \rangle$

$\left \langle y^2+xyz,3xy^3+4y^2,xz\right \rangle$

$\left \langle y^4,3xy^3+4y^3+xy,yz\right \rangle$

$\left \langle y^3+yz,3xy^2+4y^3+xz,xy\right \rangle$

Correct answer:

$\left \langle y^3+yz,3xy^2+4y^3+xz,xy\right \rangle$

Explanation:

The gradient of a function F(x,y,z) is as follows:

$\bigtriangledown F=\left \langle \frac{\mathrm{dF} }{\mathrm{d} x},\frac{\mathrm{dF} }{\mathrm{d} y},\frac{\mathrm{dF} }{\mathrm{d} z} \right \rangle$ .

We compute the derivative of the function with respect to each of the variables and treat the others like constants.

Using the rule

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

we obtain

$\frac{\mathrm{dF} }{\mathrm{d} x}=y^3+yz$ ,

$\frac{\mathrm{dF} }{\mathrm{d} y}=3xy^2+4y^3+xz$ ,

and

$\frac{\mathrm{dF} }{\mathrm{d} z}=xy$ .

Putting these expressions into the vector completes the problem, and you obtain

$\left \langle y^3+yz,3xy^2+4y^3+xz,xy\right \rangle$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #42 : Applications Of Partial Derivatives

Example Question #43 : Applications Of Partial Derivatives

Example Question #1581 : Calculus 3

Example Question #42 : Applications Of Partial Derivatives

Example Question #21 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #22 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #23 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #1581 : Calculus 3

Example Question #25 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #1582 : Calculus 3

All Calculus 3 Resources

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