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Calculus 3 : Gradient Vector, Tangent Planes, and Normal Lines

Study concepts, example questions & explanations for Calculus 3

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Example Questions

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

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

Possible Answers:

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

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

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

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

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

Find $\bigtriangledown f$ of the following function:

$f(x, y, z)=xz^3\cos(y)$

Possible Answers:

$\left \langle 0, -\sin(y), z^2\right \rangle$

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

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

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

Correct answer:

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

Explanation:

The gradient of the function is

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

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

The partial derivatives are

$f_x=z^3\cos(y)$

$f_y=-xz^3\sin(y)$

$f_z=3xz^2\cos(y)$

The derivatives were found using the following rules:

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

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Example Question #44 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\bigtriangledown f$ for the following function:

$f(x, y, z)=4e^{xy}+\ln(z)$

Possible Answers:

$\left \langle 4xe^{xy}, 4ye^{xy}, \frac{1}{z}\right \rangle$

$\left \langle 4ye^{xy}, 4xe^{xy}, -\frac{1}{z}\right \rangle$

$\left \langle ye^{xy}, xe^{xy}, \frac{1}{z}\right \rangle$

$\left \langle 4ye^{xy}, 4xe^{xy}, \frac{1}{z}\right \rangle$

Correct answer:

$\left \langle 4ye^{xy}, 4xe^{xy}, \frac{1}{z}\right \rangle$

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

$f_x=4ye^{xy}$

$f_y=4xe^{xy}$

$f_z=\frac{1}{z}$

The rules used to find the derivatives are

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} \ln(x)=\frac{1}{x}$

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Example Question #45 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\bigtriangledown f$ of the following function:

$f(x, y, z)=z^2+\sec(xy)$

Possible Answers:

$(y+x)(\sec(xy)\tan(xy)) +2z$

$\left \langle -y\sec(xy)\tan(xy), -x\sec(xy)\tan(xy), 2z\right \rangle$

$\left \langle x\sec(xy)\tan(xy), y\sec(xy)\tan(xy), 2z\right \rangle$

$\left \langle y\sec(xy)\tan(xy), x\sec(xy)\tan(xy), 2z\right \rangle$

Correct answer:

$\left \langle y\sec(xy)\tan(xy), x\sec(xy)\tan(xy), 2z\right \rangle$

Explanation:

The gradient of the 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

$f_x=y\sec(xy)\tan(xy)$

$f_y=x\sec(xy)\tan(xy)$

f_z=2z

The derivatives were found using the following rules:

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

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Example Question #46 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\nabla f$ of the given function:

$f(x, y, z)=xz^2\cos(y^3)$

Possible Answers:

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

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

$\left \langle z^2\cos(y^3), 3xy^2z^2\sin(y^3), 2xz\cos(y^3)\right \rangle$

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

Correct answer:

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

Explanation:

The gradient of the 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

$f_x=z^2\cos(y^3)$

$f_y=-3xy^2z^2\sin(y^3)$

$f_z=2xz\cos(y^3)$

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

Find $\nabla f$ of the following function:

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

Possible Answers:

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

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

$e^x\tan(z) + 2yz^4+4y^2z^3$

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

Correct answer:

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

Explanation:

The gradient of the 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

$f_x=e^x\tan(z)$

f_y=2yz^4

f_z=4y^2z^3

The derivatives were found using the following rules:

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

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Calculus 3 : Gradient Vector, Tangent Planes, and Normal Lines

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

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

Example Question #1601 : Calculus 3

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

Example Question #44 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #45 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #46 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #47 : Gradient Vector, Tangent Planes, And Normal Lines

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