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Calculus 3 : Applications of Partial Derivatives

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Example Questions

Example Question #3 : Gradient Vector, Tangent Planes, And Normal Lines

Find the slope of the function f(x,y,z)=4x+5y+6z at the point (3,2,1)

Possible Answers:

(4,5,6)

(0,0,0)

(16,18,22)

(12,10,6)

(4,10,18)

Correct answer:

(4,5,6)

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.

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,z)=4x+5y+6z at the point (3,2,1)

$\frac{\delta f}{\delta x}=4$

$\frac{\delta f}{\delta y}=5$

$\frac{\delta f}{\delta z}=6$

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Example Question #3 : Gradient Vector, Tangent Planes, And Normal Lines

Find the slope of the function f(x,y,z)=2xz+3y^2z at the point (1,5,2)

Possible Answers:

(1,5,2)

(4,60,77)

(20,5,150)

(9,27,169)

(2,75,10)

Correct answer:

(4,60,77)

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.

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,z)=2xz+3y^2z at the point (1,5,2)

$\frac{\delta f}{\delta x}=2z=4$

$\frac{\delta f}{\delta y}=6yz=60$

$\frac{\delta f}{\delta z}=2x+3y^2=2+75=77$

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Example Question #1 : Gradient Vector, Tangent Planes, And Normal Lines

Find the slope of the function f(x,y)=x^3y^2 at the point (2,3)

Possible Answers:

(36,24)

(108,48)

(216,144)

(72,72)

(54,16)

Correct answer:

(108,48)

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.

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)=x^3y^2 at the point (2,3)

$\frac{\delta f}{\delta x}=3x^2y^2=108$

$\frac{\delta f}{\delta y}=2x^3y=48$

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Example Question #5 : Gradient Vector, Tangent Planes, And Normal Lines

Find the slope of the function f(x,y)=3xy+2y^3 at the point (2,2)

Possible Answers:

(6,30)

(6,24)

(28,12)

(30,6)

(12,28)

Correct answer:

(6,30)

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.

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)=3xy+2y^3 at the point (2,2)

$\frac{\delta f}{\delta x}=3y=6$

$\frac{\delta f}{\delta y}=3x+6y^2=30$

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Example Question #1 : Gradient Vector, Tangent Planes, And Normal Lines

Find the slope of the function f(x,y)=xsin(y) at the point $(3,\pi)$

Possible Answers:

(3,0)

(3,3)

(3,-3)

(0,3)

(0,-3)

Correct answer:

(0,-3)

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.

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)=xsin(y) at the point $(3,\pi)$

$\frac{\delta f}{\delta x}=sin(y)=0$

$\frac{\delta f}{\delta y}=xcos(y)=-3$

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Example Question #31 : Applications Of Partial Derivatives

Find the slope of the function f(x,y)=5xy^2-2x^2y at the point (2,4)

Possible Answers:

(-48,-72)

(72,-48)

(-72,48)

(72,48)

(48,72)

Correct answer:

(48,72)

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.

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)=5xy^2-2x^2y at the point (2,4)

$\frac{\delta f}{\delta x}=5y^2-4xy=80-32=48$

$\frac{\delta f}{\delta y}=10xy-2x^2=80-8=72$

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Example Question #11 : Gradient Vector, Tangent Planes, And Normal Lines

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

Possible Answers:

(1,2)

$(\pi,2\pi)$

$(2,\pi)$

$(2\pi,\pi)$

$(\pi,2)$

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 #1571 : Calculus 3

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

Possible Answers:

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

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

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

$(2,\pi)$

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

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 #13 : 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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Calculus 3 : Applications of Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #3 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #1 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #5 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #1 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #31 : Applications Of Partial Derivatives

Example Question #11 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #1571 : Calculus 3

Example Question #13 : Gradient Vector, Tangent Planes, And Normal Lines

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

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