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

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Example Questions

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

Find the equation of the tangent plane to $z=\ln(4x^3+10y^2)$ at z=(0,5) .

Possible Answers:

$z=-\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}-\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

$z=\frac{6}{127}x-\frac{50}{127}y+\frac{256}{127}+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}$

Correct answer:

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

Explanation:

First, we need to find the partial derivatives in respect to , and , and plug in z=(1,5) .

$f(x,y)=\ln(4x^3+10y^2)$ , $f(0,5)=\ln(4(1)^3+10(5)^2)=\ln(254)$

$f_x(x,y)=\frac{12x^2}{4x^3+10y^2}$ , $f_x(0,5)=\frac{12(1)^2}{4(1)^3+10(5)^2}=\frac{12}{254}=\frac{6}{127}$

$f_y(x,y)=\frac{20y}{4x^3+10y^2}$ , $f_y(0,5)=\frac{20(5)}{4(1)^3+10(5)^2}=\frac{100}{254}=\frac{50}{127}$

Remember that the general equation for a tangent plane is as follows:

z-f(x_0,y_0)=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)

Now lets apply this to our problem

$z-\ln(254)=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)$

$z=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)+\ln(254)$

$z=\frac{6}{127}x-\frac{6}{127}+\frac{50}{127}y-\frac{250}{127}+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

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Example Question #21 : Applications Of Partial Derivatives

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

Possible Answers:

$(\pi,\pi,0)$

$(\pi,-\pi,-\pi)$

$(0,\pi,0)$

$(0,-\pi,0)$

$(\pi,-\pi,0)$

Correct answer:

$(0,\pi,0)$

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

d[cos(u)]=-sin(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,z)=xsin(y)cos(z) at the point $(\pi,\pi,\pi)$

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

$\frac{\delta f}{\delta y}=xcos(y)cos(z)=\pi$

$\frac{\delta f}{\delta z}=-xsin(y)cos(y)=0$

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Example Question #2 : Gradient Vector, Tangent Planes, And Normal Lines

Find the slope of the function f(x,y,z)=xy^2sin(z) at the point $(3,2,2\pi)$

Possible Answers:

(0,0,0)

(0,0,12)

(4,4,4)

(4,12,0)

(4,12,12)

Correct answer:

(0,0,12)

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,z)=xy^2sin(z) at the point $(3,2,2\pi)$

$\frac{\delta f}{\delta x}=y^2sin(z)=0$

$\frac{\delta f}{\delta y}=2xysin(z)=0$

$\frac{\delta f}{\delta z}=xy^2cos(z)=12$

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

Find the slope of the function f(x,y,z)=6xyz+2xy+xz at the point (3,0,3)

Possible Answers:

(18,6,3)

(3,60,21)

(3,60,3)

(18,0,18)

(21,60,21)

Correct answer:

(3,60,3)

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,z)=6xyz+2xy+xz at the point (3,0,3)

$\frac{\delta f}{\delta x}=6yz+2y+z=3$

$\frac{\delta f}{\delta y}=6xz+2x=60$

$\frac{\delta f}{\delta z}=6xy+x=3$

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Example Question #2 : Gradient Vector, Tangent Planes, And Normal Lines

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

Possible Answers:

(60,40,30)

(60,30,40)

(40,60,30)

(30,40,60)

(40,30,60)

Correct answer:

(60,40,30)

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,z)=5xyz at the point (2,3,4)

$\frac{\delta f}{\delta x}=5yz=60$

$\frac{\delta f}{\delta y}=5xz=40$

$\frac{\delta f}{\delta z}=5xy=30$

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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:

(0,0,0)

(12,10,6)

(4,5,6)

(16,18,22)

(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 #6 : 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 #7 : 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)

(72,72)

(54,16)

(108,48)

(216,144)

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 #1561 : Calculus 3

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

Possible Answers:

(30,6)

(6,24)

(12,28)

(6,30)

(28,12)

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)

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

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

Report an Error

Example Question #4 : 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:

(0,-3)

(3,3)

(3,0)

(3,-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 $(3,\pi)$

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

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

Report an Error

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

Example Question #21 : Applications Of Partial Derivatives

Example Question #2 : Gradient Vector, Tangent Planes, And Normal Lines

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

Example Question #2 : Gradient Vector, Tangent Planes, And Normal Lines

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

Example Question #6 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #7 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #1561 : Calculus 3

Example Question #4 : Gradient Vector, Tangent Planes, And Normal Lines

All Calculus 3 Resources

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