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Calculus 3 : Applications of Partial Derivatives

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Example Questions

Example Question #61 : Applications Of Partial Derivatives

Find $\nabla F$ of the function 4x^2yz+y^3+xz^4

Possible Answers:

$\left \langle 8xz+z^3,4x^2z+y^2,4x^2y\right \rangle$

$\left \langle 8xz+z^4,4x^2+3y^2,4x^2y+4xz^3\right \rangle$

$\left \langle 8xyz+z^4,4x^2z+3y^2,4x^2y+4xz^3\right \rangle$

$\left \langle 8xyz+z^4,4x^2z+3y^2,4x^2y+z^3\right \rangle$

Correct answer:

$\left \langle 8xyz+z^4,4x^2z+3y^2,4x^2y+4xz^3\right \rangle$

Explanation:

To find the gradient vector, you use the following definition: $\bigtriangledown F=\left \langle f_x,f_y,f_z\right \rangle$ . Taking the partial derivatives with respect to x, y, and z, we get $f_x=\frac{\partial }{\partial x}(4x^2yz+y^3+xz^4)=8xyz+z^4$ , $f_y=\frac{\partial }{\partial y}(4x^2yz+y^3+xz^4)=4x^2z+3y^2$ , and $f_z=\frac{\partial }{\partial z}(4x^2yz+y^3+xz^4)=4x^2y+4xz^3$ . Putting these into the vector produces the right answer.

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Example Question #62 : Applications Of Partial Derivatives

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

Possible Answers:

$\left \langle 6xy,3x^2+3y^2,1\right \rangle$

Correct answer:

$\left \langle 6xy,3x^2+3y^2,1\right \rangle$

Explanation:

To find the gradient vector for a function F(x,y,z) , we use the definition

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

$\frac{\partial }{\partial x}(3x^2y+y^3+z)=6xy$

$\frac{\partial }{\partial y}(3x^2y+y^3+z)=3x^2+3y^2$

$\frac{\partial }{\partial z}(3x^2y+y^3+z)=1$

Putting these answers into the vector gets us the correct answer

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Example Question #63 : Applications Of Partial Derivatives

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

Possible Answers:

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

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

$\left \langle x^4,y^4,z^4\right \rangle$

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

Correct answer:

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

Explanation:

To find the gradient vector for a function F(x,y,z) , we use the definition

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

$\frac{\partial }{\partial x}(x^3+y^3+z^3)=3x^2$

$\frac{\partial }{\partial y}(x^3+y^3+z^3)=3y^2$

$\frac{\partial }{\partial z}(x^3+y^3+z^3)=3z^2$

Putting these answers into the vector gets us the correct answer

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Example Question #1592 : Calculus 3

Find $\nabla F$ of the function x^5+2y^2+z

Possible Answers:

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

$\left \langle 5x^4,4y,1\right \rangle$

$\left \langle 1,4,1\right \rangle$

$\left \langle 5x,4y,1\right \rangle$

Correct answer:

$\left \langle 5x^4,4y,1\right \rangle$

Explanation:

To find the gradient vector for a function F(x,y,z) , we use the definition

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

$\frac{\partial }{\partial x}(x^5+2y^2+z)=5x^4$

$\frac{\partial }{\partial y}(x^5+2y^2+z)=4y$

$\frac{\partial }{\partial z}(x^5+2y^2+z)=1$

Putting these answers into the vector gets us the correct answer

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Example Question #1591 : Calculus 3

Compute the gradient of the following scalar function:

f(x,y,z)=x^4+y^3+z^2

Possible Answers:

$4x^3\mathbf{i}+3y^2\mathbf{j}+2z\mathbf{k}$

$\frac{x^5}{5}\mathbf{i}+\frac{y^4}{4}\mathbf{j}+\frac{z^3}{3}\mathbf{k}$

$4x^5\mathbf{i}+3y^4\mathbf{j}+2z^3\mathbf{k}$

$12x^2\mathbf{i}+6y\mathbf{j}+2\mathbf{k}$

$x^3\mathbf{i}+y^2\mathbf{j}+z\mathbf{k}$

Correct answer:

$4x^3\mathbf{i}+3y^2\mathbf{j}+2z\mathbf{k}$

Explanation:

The gradient of a function is defined as:

$\nabla f(x,y,z)=\frac{\partial f}{\partial x}\mathbf{i}+ \frac{\partial f}{\partial y}\mathbf{j} +\frac{\partial f}{\partial z}\mathbf{k}$

For our function:

$\frac{\partial}{\partial x}(x^4+y^3+z^2)=4x^3$

$\frac{\partial}{\partial y}(x^4+y^3+z^2)=3y^2$

$\frac{\partial}{\partial z}(x^4+y^3+z^2)=2z^2$

Thus, the gradient is:

$\nabla f(x,y,z)=4x^3\mathbf{i}+3y^2\mathbf{j}+2z\mathbf{k}$

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Example Question #41 : Gradient Vector, Tangent Planes, And Normal Lines

Compute the gradient of the following scalar function:

f(x,y,z)=x^2y^3z^4

Possible Answers:

$4x^2y^3z^3\mathbf{i}+3x^2y^2z^4\mathbf{j}+2xy^3z^4\mathbf{k}$

$8xy^3z^3\mathbf{i}+6x^2yz^4\mathbf{j}+6xy^3z^3\mathbf{k}$

$6xy^2z^4\mathbf{i}+6x^2yz^4\mathbf{j}+9x^2y^2z^3\mathbf{k}$

$12x^2y^3z^2\mathbf{i}+12x^2y^2z^3\mathbf{j}+8xy^3z^3\mathbf{k}$

$2xy^3z^4\mathbf{i}+3x^2y^2z^4\mathbf{j}+4x^2y^3z^3\mathbf{k}$

Correct answer:

$2xy^3z^4\mathbf{i}+3x^2y^2z^4\mathbf{j}+4x^2y^3z^3\mathbf{k}$

Explanation:

The gradient of a function is defined as:

$\nabla f(x,y,z)=\frac{\partial f}{\partial x}\mathbf{i}+ \frac{\partial f}{\partial y}\mathbf{j} +\frac{\partial f}{\partial z}\mathbf{k}$

For our function:

$\frac{\partial}{\partial x}(x^2y^3z^4)=2xy^3z^4$

$\frac{\partial}{\partial y}(x^2y^3z^4)=3x^2y^2z^4$

$\frac{\partial}{\partial z}(x^2y^3z^4)=4x^2y^3z^3$

Thus, the gradient is:

$\nabla f(x,y,z)=2xy^3z^4\mathbf{i}+3x^2y^2z^4\mathbf{j}+4x^2y^3z^3\mathbf{k}$

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Example Question #41 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\nabla f$ of the following function:

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

Possible Answers:

$\left \langle 0, 0, 0\right \rangle$

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

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

$z^4y+ xz^4+2y+4z^3xy+e^{xz}(x+z)$

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

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

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

The partial derivatives are

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

f_y=xz^4+2y

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

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Example Question #43 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\nabla f$ of the following function:

$f(x y,z)=x^3\sqrt{yz}$

Possible Answers:

$\left \langle 3x^2\sqrt{yz}, {x^3z}(yz)^{-\frac{1}{2}}, {x^3y}(yz)^{-\frac{1}{2}}\right \rangle$

$\left \langle 3x^2\sqrt{yz}, \frac{x^3z}{2}(yz)^{\frac{1}{2}}, \frac{x^3y}{2}(yz)^{\frac{1}{2}}\right \rangle$

$\left \langle 3x^2\sqrt{yz}, \frac{x^3}{2}(yz)^{-\frac{1}{2}}, \frac{x^3}{2}(yz)^{-\frac{1}{2}}\right \rangle$

$\left \langle 3x^2\sqrt{yz}, \frac{x^3z}{2}(yz)^{-\frac{1}{2}}, \frac{x^3y}{2}(yz)^{-\frac{1}{2}}\right \rangle$

Correct answer:

$\left \langle 3x^2\sqrt{yz}, \frac{x^3z}{2}(yz)^{-\frac{1}{2}}, \frac{x^3y}{2}(yz)^{-\frac{1}{2}}\right \rangle$

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

$f_x=3x^2\sqrt{yz}$

$f_y=\frac{x^3z}{2}(yz)^{-\frac{1}{2}}$

$f_z=\frac{x^3y}{2}(yz)^{-\frac{1}{2}}$

The rules used to find the derivatives are

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

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Example Question #41 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\nabla F$ of the function $2\sin(x)+3xy^2+z^5$

Possible Answers:

$\left \langle -2\cos(x),6xy,5z^4\right \rangle$

$\left \langle 2\cos(x),6xy,5z^4\right \rangle$

$\left \langle -2\sin(x),6xy,5z^4\right \rangle$

$2\sin(x)-6xy+5z^4$

Correct answer:

$\left \langle 2\cos(x),6xy,5z^4\right \rangle$

Explanation:

To find the gradient vector, you use the formula $\nabla F=\left \langle \frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z} \right \rangle$ , where F=f(x,y,z) . Using the function given and the rules for partial differentiation, we e $\frac{\partial F}{\partial x}=2\cos(x), \frac{\partial F}{\partial y}=6xy,\frac{\partial F}{\partial z}=5z^4$ . Plugging these values into vector notation gets you the correct answer.

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Example Question #1601 : Calculus 3

Find $\nabla F$ of the function 3xy+2z^5

Possible Answers:

3x+3y+10z^4

$\left \langle 3x,3y,10z^4\right \rangle$

$\left \langle 3y,3x,10z^4\right \rangle$

$\left \langle 3x,0,0\right \rangle$

Correct answer:

$\left \langle 3y,3x,10z^4\right \rangle$

Explanation:

To find the gradient vector, you use the formula $\nabla F=\left \langle \frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z} \right \rangle$ , where F=f(x,y,z) . Using the function given and the rules for partial differentiation, we e $\frac{\partial F}{\partial x}=3y, \frac{\partial F}{\partial y}=3x,\frac{\partial F}{\partial z}=10z^4$ . Plugging these values into vector notation gets you the correct answer.

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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 #61 : Applications Of Partial Derivatives

Example Question #62 : Applications Of Partial Derivatives

Example Question #63 : Applications Of Partial Derivatives

Example Question #1592 : Calculus 3

Example Question #1591 : Calculus 3

Example Question #41 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #41 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #43 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #41 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #1601 : Calculus 3

All Calculus 3 Resources

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