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Calculus 3 : Applications of Partial Derivatives

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Example Questions

Example Question #41 : Applications Of Partial Derivatives

Calculate $R = \vec{\bigtriangledown } \cdot \vec{f}$ given $\vec{f}= (x+3y)\hat{x}+(x^2+\sqrt{z})\hat{y}+(xy\sqrt{z})\hat{z}$

Possible Answers:

$R = 1 \hat{x} +\frac{xy}{2\sqrt{z}} \hat{z}$

$R = 1 +\frac{xy}{2\sqrt{z}}$

$R = 1 +\frac{xy}{\sqrt{z}}$

$R = \frac{xy}{2\sqrt{z}}$

Correct answer:

$R = 1 +\frac{xy}{2\sqrt{z}}$

Explanation:

By definition,

$\vec{\bigtriangledown } \cdot \vec{f} = \frac{\partial f_x}{\partial x}+\frac{\partial f_y}{\partial y}+\frac{\partial f_z}{\partial z}$ , where f_x, f_y, f_z are the respective x,y,z components of $\vec{f}$ .

Therefore, we need to calculate the above terms, shown as

$\begin{align*} \frac{\partial f_x}{\partial x}&=1 \\ \frac{\partial f_y}{\partial y}&= 0\\ \frac{\partial f_z}{\partial z}&= \frac{xy}{2\sqrt{z}} \end{align*}$

Therefore,

$R = 1 +\frac{xy}{2\sqrt{z}}$ .

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Example Question #17 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\nabla f$ , where $f(x,y,z)=xzy+\tan(xz)$

Possible Answers:

$\left \langle yz+sec^2(xz), xz, xy+sec^2(xz)\right \rangle$

$\left \langle yz+z\sec(xz), xz, xy+x\sec(xz)\right \rangle$

$\left \langle yz+z\sec^2(xz), xz, xy+x\sec^2(xz)\right \rangle$

$\left \langle yz+z\sec^2(xz), 0, xy+x\sec^2(xz)\right \rangle$

Correct answer:

$\left \langle yz+z\sec^2(xz), xz, xy+x\sec^2(xz)\right \rangle$

Explanation:

The gradient vector of f, $\nabla f$ , is equal to $\left \langle f_x, f_y, f_z\right \rangle$ .

So, we must find the partial derivatives of the function with respect to x, y, and z, keeping the other variables constant for each partial derivative:

$f_x=yz+z\sec^2(xz)$

f_y=xz

$f_z=xy+x\sec^2(xz)$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$

Plugging this in to a vector, we get

$\left \langle yz+z\sec^2(xz), xz, xy+x\sec^2(xz)\right \rangle$

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Example Question #18 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\nabla f$ , where f is the following function:

$f(x, y,z)=z^2+e^z\cos(xy)+xyz$

Possible Answers:

$\left \langle ye^z\sin(xy)+yz, xe^z\sin(xy)+xz, 2z+e^z\cos(xy)+xy \right \rangle$

$(-x-y)(e^z\sin(xy)+yz e^z\sin(xy))+xz 2z+e^z\cos(xy)+xy$

$\left \langle -ye^z\sin(xy)+yz, -xe^z\sin(xy)+xz, 2z+e^z\cos(xy)+xy \right \rangle$

$\left \langle -ye^z\sin(xy), -xe^z\sin(xy), 2z+e^z\cos(xy) \right \rangle$

Correct answer:

$\left \langle -ye^z\sin(xy)+yz, -xe^z\sin(xy)+xz, 2z+e^z\cos(xy)+xy \right \rangle$

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.

Now, we find the partial derivatives:

$f_x=-ye^z\sin(xy)+yz$

$f_y=-xe^z\sin(xy)+xz$

$f_z=2z+e^z\cos(xy)+xy$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

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Example Question #14 : Gradient Vector, Tangent Planes, And Normal Lines

Find $f_{yxz}$ of the following function:

$f(x, y, z)=xy^2z^2-z\tan(x)$

Possible Answers:

2xyz^2

2yz^2

$4yz-\sec^2(x)$

4yz

Correct answer:

4yz

Explanation:

To find the given partial derivative of the function, we must treat the other variable(s) as constants. For higher order partial derivatives, we work from left to right for the given variables.

To start, we must find the partial derivative with respect to y:

f_y=2xyz^2

The following rules were used to find the derivative:

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

Next, we find the derivative of the above function above with respect to x:

$f_{yx}=2yz^2$

The rule used is stated above.

Finally, find the partial derivative of the above function with respect to z:

$f_{yxz}=4yz$

The rule used is already stated above.

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Example Question #20 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\nabla f$ of the given function:

$f(x, y, z)=z^2e^{x}+\cos(3y)$

Possible Answers:

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

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

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

$z^2e^x -3\sin(3y) +2ze^x$

Correct answer:

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

Explanation:

The gradient of a function is given by

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

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

f_x=z^2e^x

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

f_z=2ze^x

The derivatives were found using the following rules:

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

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Example Question #41 : Applications Of Partial Derivatives

Find $\nabla f$ of the function:

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

Possible Answers:

$\left \langle y\sec(xy), x\sec(xy), 2z+4\right \rangle$

$\left \langle xsec^2(xy), ysec^2(xy), 2z+4\right \rangle$

$\left \langle y\sec^2(xy), x\sec^2(xy), 2z+8\right \rangle$

$\left \langle y\sec^2(xy), x\sec^2(xy), 2z+4\right \rangle$

Correct answer:

$\left \langle y\sec^2(xy), x\sec^2(xy), 2z+4\right \rangle$

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

$f_x=y\sec^2(xy)$

$f_y=x\sec^2(xy)$

f_z=2z+4

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$

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Example Question #41 : Applications Of Partial Derivatives

Find $\nabla f$ for the given function:

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

Possible Answers:

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

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

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

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

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:

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

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

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

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

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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Calculus 3 : Applications of Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #41 : Applications Of Partial Derivatives

Example Question #17 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #18 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #14 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #20 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #41 : Applications Of Partial Derivatives

Example Question #41 : Applications Of Partial Derivatives

Example Question #43 : Applications Of Partial Derivatives

Example Question #1581 : Calculus 3

Example Question #42 : Applications Of Partial Derivatives

All Calculus 3 Resources

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