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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #33 : Applications Of Partial Derivatives

Find the slope of the function f(x,y)=sin(xy) at the point $(2,\pi)$

Possible Answers:

$(2,\pi)$

$(\pi,2)$

$(2\pi,\pi)$

$(\pi,2\pi)$

(1,2)

Correct answer:

$(\pi,2)$

Explanation:

To consider finding the slope, let's discuss the topic of the gradient.

For a function f(x_1,x_2,...,x_n) , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point.

Knowledge of the following derivative rules will be necessary:

Trigonometric derivative:

d[sin(u)]=cos(u)du

Note that u may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at f(x,y)=sin(xy) at the point $(2,\pi)$

$\frac{\delta f}{\delta x}=ycos(xy)=\pi$

$\frac{\delta f}{\delta y}=xcos(xy)=2$

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Example Question #31 : Applications Of Partial Derivatives

Find the slope of the function f(x,y)=tan(xy) at the point $(\frac{1}{4},\pi)$

Possible Answers:

$(2,\pi)$

$(\frac{1}{4},2\pi)$

$(2\pi,\frac{1}{2})$

$(\frac{1}{4},\pi)$

$(2\pi,\frac{1}{4})$

Correct answer:

$(2\pi,\frac{1}{2})$

Explanation:

To consider finding the slope, let's discuss the topic of the gradient.

For a function f(x_1,x_2,...,x_n) , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point.

Knowledge of the following derivative rules will be necessary:

Trigonometric derivative:

$d[tan(u)]=\frac{du}{cos^2(u)}$

Note that u may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at f(x,y)=tan(xy) at the point $(\frac{1}{4},\pi)$

$\frac{\delta f}{\delta x}=\frac{y}{cos^2(xy)}=2\pi$

$\frac{\delta f}{\delta y}=\frac{x}{cos^2(xy)}=\frac{1}{2}$

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

Find the equation of the tangent plane to $z = \ln \left(x^2-3xy+y \right )$ at (5,1) .

Possible Answers:

$z_T=\ln (11)+\frac{7}{11}\left(x-1 \right )-\frac{14}{11}\left(y-5 \right )$

$z_T=\ln (11)-\frac{7}{11}\left(x-5 \right )+\frac{14}{11}\left(y-1 \right )$

$z_T=\ln (11)+\frac{7}{11}\left(x-5 \right )-\frac{14}{11}\left(y-1 \right)$

$z_T=-\ln (11)+\frac{7}{11}\left(x-5 \right )-\frac{14}{11}\left(y-1 \right )$

Correct answer:

$z_T=\ln (11)+\frac{7}{11}\left(x-5 \right )-\frac{14}{11}\left(y-1 \right)$

Explanation:

The definition of a tangent plane of a surface $z=f \left(x,y \right )$ is given by

$z_T = f \left(x_0, y_0 \right )+f_x\left(x_0, y_0 \right )\left(x-x_0\right )+f_y\left(x_0, y_0 \right )\left(y-y_0 \right )$ .

Therefore, we need to first find f_x,f_y , given below as

$f_x=\frac{2x-3y}{x^2-3xy+y}\Rightarrow f_x\left(5, 1 \right )=\frac{7}{11}$

$f_y=\frac{-3x+1}{x^2-3xy+y}\Rightarrow f_y\left(5, 1 \right )=-\frac{14}{11}$

Noting that

$f \left(5,1 \right )= \ln (11)$ ,

we can write our complete expression for the tangent line as

$z_T=\ln (11)+\frac{7}{11}\left(x-5 \right )-\frac{14}{11}\left(y-1 \right )$ .

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Example Question #15 : Gradient Vector, Tangent Planes, And Normal Lines

Calculate $\vec{R}$ if $\vec{\bigtriangledown} f = \vec{R}$ and f(x,y,z)= x^2yz^3-2xy+5yz

Possible Answers:

$\vec{R}= 2\left( xyz^3-y\right )\hat{x}+ \left( x^2z^3-2x+5z\right )\hat{j}+\left( 3x^2yz^2+5y\right )\hat{z}$

$\vec{R}= 2\left( xyz^3-y\right )\hat{x}- \left( x^2z^3-2x+5z\right )\hat{j}+\left( 3x^2yz^2+5y\right )\hat{z}$

$\vec{R}= 2\left( xyz^3\right )\hat{x}+ \left( x^2z^3-2x+5z\right )\hat{j}+\left( 3x^2yz^2+5y\right )\hat{z}$

$\vec{R}= \left( xyz^3-y\right )\hat{x}+ \left( x^2z^3-2x+5z\right )\hat{j}+\left( 3x^2yz^2+5y\right )\hat{z}$

Correct answer:

$\vec{R}= 2\left( xyz^3-y\right )\hat{x}+ \left( x^2z^3-2x+5z\right )\hat{j}+\left( 3x^2yz^2+5y\right )\hat{z}$

Explanation:

By definition $\vec{\bigtriangledown}=\hat{x}\frac{\partial}{\partial x}+\hat{y}\frac{\partial}{\partial y}+\hat{z}\frac{\partial}{\partial z}$ . Therefore,

$\vec{R}=\vec{\bigtriangledown}f=\hat{x}\frac{\partial f}{\partial x}+\hat{y}\frac{\partial f}{\partial y}+\hat{z}\frac{\partial f}{\partial z}$ , so we will need to find the partial derivatives of f(x,y,z) , shown below as

$\begin{align*} \frac{\partial f}{\partial x}&= 2(xyz^3-y) \\ \frac{\partial f}{\partial y}&= x^2z^3-2x+5z\\ \frac{\partial f}{\partial z}&= 3x^2yz^2+5y \end{align*}$

Therefore,

$\vec{R}= 2\left( xyz^3-y\right )\hat{x}+ \left( x^2z^3-2x+5z\right )\hat{j}+\left( 3x^2yz^2+5y\right )\hat{z}$

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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:

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

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

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

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

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 #19 : 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:

4yz

2xyz^2

2yz^2

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

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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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #33 : Applications Of Partial Derivatives

Example Question #31 : Applications Of Partial Derivatives

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

Example Question #15 : Gradient Vector, Tangent Planes, And Normal Lines

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

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

Example Question #41 : Applications Of Partial Derivatives

All Calculus 3 Resources

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