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Calculus 3 : Applications of Partial Derivatives

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Example Questions

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

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

Possible Answers:

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

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

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

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

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 #22 : 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:

2x+3y+2z=30

x-4y-z=15

3x+y+2z=30

x+5y-2z=28

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 #31 : Gradient Vector, Tangent Planes, And Normal Lines

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

Possible Answers:

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

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

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

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

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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Example Question #32 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\bigtriangledown F$ of the function x^2y+3xy^2z+z^4 .

Possible Answers:

$\left \langle 2x^2+4y^3z,x^3+6xy^2z,3xy^2+4z^2\right \rangle$

$\left \langle 2xy+3y^2z,x^2+6xyz,3xy^2+4z^3\right \rangle$

$\left \langle 2x+3y^2z,6yz,6y\right \rangle$

$\left \langle 3x+6y^3,xyz,4x^2z^3\right \rangle$

Correct answer:

$\left \langle 2xy+3y^2z,x^2+6xyz,3xy^2+4z^3\right \rangle$

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

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

$\frac{\mathrm{dF} }{\mathrm{d} y}=x^2+6xyz$ ,

and

$\frac{\mathrm{dF} }{\mathrm{d} z}=3xy^2+4z^3$ .

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

$\left \langle 2xy+3y^2z,x^2+6xyz,3xy^2+4z^3\right \rangle$ .

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Example Question #33 : Gradient Vector, Tangent Planes, And Normal Lines

Find the tangent plane to the surface given by

f(x, y, z)=x+y^2+z^3

at the point (1, 3, 1)

Possible Answers:

x+6y+3z=22

x+3y+z=22

x+6y+3z=0

x+6y+3z=-22

Correct answer:

x+6y+3z=22

Explanation:

The equation of the tangent plane is given by

f_x(x_0, y_0, z_0)(x-x_0)+f_y(x_0, y_0, z_0)(y-y_0)+f_z(x_0, y_0, z_0)(z-z_0)=0

So, we must find the partial derivatives for the function evaluated at the point given. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

f_x=1

f_y=2y

f_z=3z^2

The derivatives were found using the following rules:

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

Now, we evaluate them at the given point:

f_x=1

f_y=6

f_z=3

Finally, plug in all of our information into the formula and simplify:

x+6y+3z=22

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Example Question #31 : Gradient Vector, Tangent Planes, And Normal Lines

Find the tangent plane to the surface given by

f(x, y, z)=e^x+y^2z

at the point (0, 2, 5)

Possible Answers:

2y+5z=-60

x+20y+4z=-60

x+20y+4z=60

2y+5z=60

Correct answer:

x+20y+4z=60

Explanation:

The equation of the tangent plane is given by

f_x(x_0, y_0, z_0)(x-x_0)+f_y(x_0, y_0, z_0)(y-y_0)+f_z(x_0, y_0, z_0)(z-z_0)=0

So, we must find the partial derivatives for the function evaluated at the point given. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

f_x=e^x

f_y=2yz

f_z=y^2

The rules used to find the derivatives are

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

Evaluated at the given point, the partial derivatives are

f_x=1

f_y=20

f_z=4

Plugging all of this into the above formula, and simplifying, we get

x+20y+4z=60

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Example Question #51 : Applications Of Partial Derivatives

Find the tangent plane to the surface given by

f(x, y, z)=x^2+4y^2+5z^2

at the point (1, 3, 3)

Possible Answers:

2x+24y+30z=-164

2x+24y+30z=164

x+3y+3z=-164

x+3y+3z=164

Correct answer:

2x+24y+30z=164

Explanation:

The equation of the tangent plane is given by

f_x(x_0, y_0, z_0)(x-x_0)+f_y(x_0, y_0, z_0)(y-y_0)+f_z(x_0, y_0, z_0)(z-z_0)=0

So, we must find the partial derivatives for the function evaluated at the point given. 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=8y

f_z=10z

The derivatives were found using the following rules:

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

Next, we evaluate the partial derivatives at the given point:

f_x=2

f_y=24

f_z=30

Plugging in all our information into the formula above, we get

2x+24y+30z=164

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Calculus 3 : Applications of Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

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

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

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

Example Question #1581 : Calculus 3

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

Example Question #31 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #32 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #33 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #31 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #51 : Applications Of Partial Derivatives

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