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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 $z=\ln(4x^3+10y^2)$ $\displaystyle z=\ln(4x^3+10y^2)$ at z=(0,5) $\displaystyle z=(0,5)$.

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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

Now lets apply this to our problem

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

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

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

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

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

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

Possible Answers:

$(\pi,\pi,0)$ $\displaystyle (\pi,\pi,0)$

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

$(0,\pi,0)$ $\displaystyle (0,\pi,0)$

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

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

Correct answer:

$(0,\pi,0)$ $\displaystyle (0,\pi,0)$

Explanation:

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

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

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

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:

$\displaystyle d[sin(u)]=cos(u)du$

$\displaystyle d[cos(u)]=-sin(u)du$

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 $\displaystyle f(x,y,z)=xsin(y)cos(z)$ at the point $(\pi,\pi,\pi)$ $\displaystyle (\pi,\pi,\pi)$

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

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

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

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

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

Possible Answers:

(4,4,4) $\displaystyle (4,4,4)$

$\displaystyle (0,0,12)$

$\displaystyle (4,12,0)$

$\displaystyle (4,12,12)$

(0,0,0) $\displaystyle (0,0,0)$

Correct answer:

$\displaystyle (0,0,12)$

Explanation:

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

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

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:

$\displaystyle d[sin(u)]=cos(u)du$

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

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

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

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

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

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

Possible Answers:

$\displaystyle (3,60,21)$

$\displaystyle (18,6,3)$

$\displaystyle (18,0,18)$

$\displaystyle (21,60,21)$

$\displaystyle (3,60,3)$

Correct answer:

$\displaystyle (3,60,3)$

Explanation:

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

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

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 $\displaystyle f(x,y,z)=6xyz+2xy+xz$ at the point (3,0,3) $\displaystyle (3,0,3)$

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

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

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

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

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

Possible Answers:

$\displaystyle (60,40,30)$

$\displaystyle (30,40,60)$

$\displaystyle (40,60,30)$

$\displaystyle (60,30,40)$

$\displaystyle (40,30,60)$

Correct answer:

$\displaystyle (60,40,30)$

Explanation:

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

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

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 $\displaystyle f(x,y,z)=5xyz$ at the point (2,3,4) $\displaystyle (2,3,4)$

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

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

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

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Example Question #31 : Applications Of Partial Derivatives

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

Possible Answers:

$\displaystyle (16,18,22)$

(4,5,6) $\displaystyle (4,5,6)$

$\displaystyle (4,10,18)$

(0,0,0) $\displaystyle (0,0,0)$

$\displaystyle (12,10,6)$

Correct answer:

(4,5,6) $\displaystyle (4,5,6)$

Explanation:

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

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

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

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

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

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

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

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

Possible Answers:

$\displaystyle (4,60,77)$

(1,5,2) $\displaystyle (1,5,2)$

$\displaystyle (9,27,169)$

$\displaystyle (20,5,150)$

$\displaystyle (2,75,10)$

Correct answer:

$\displaystyle (4,60,77)$

Explanation:

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

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

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 $\displaystyle f(x,y,z)=2xz+3y^2z$ at the point (1,5,2) $\displaystyle (1,5,2)$

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

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

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

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

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

Possible Answers:

(36,24) $\displaystyle (36,24)$

(54,16) $\displaystyle (54,16)$

$\displaystyle (216,144)$

(72,72) $\displaystyle (72,72)$

$\displaystyle (108,48)$

Correct answer:

$\displaystyle (108,48)$

Explanation:

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

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

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 $\displaystyle f(x,y)=x^3y^2$ at the point (2,3) $\displaystyle (2,3)$

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

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

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

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

Possible Answers:

(12,28) $\displaystyle (12,28)$

(6,30) $\displaystyle (6,30)$

(30,6) $\displaystyle (30,6)$

(28,12) $\displaystyle (28,12)$

(6,24) $\displaystyle (6,24)$

Correct answer:

(6,30) $\displaystyle (6,30)$

Explanation:

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

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

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 $\displaystyle f(x,y)=3xy+2y^3$ at the point (2,2) $\displaystyle (2,2)$

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

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

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

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

Possible Answers:

(0,-3) $\displaystyle (0,-3)$

(3,3) $\displaystyle (3,3)$

(3,0) $\displaystyle (3,0)$

(3,-3) $\displaystyle (3,-3)$

(0,3) $\displaystyle (0,3)$

Correct answer:

(0,-3) $\displaystyle (0,-3)$

Explanation:

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

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

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

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

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

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

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

Example Question #1 : 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 #3 : Gradient Vector, Tangent Planes, And Normal Lines

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

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

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

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