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

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #51 : Gradient Vector, Tangent Planes, And Normal Lines

Find the equation of the tangent plane to the following function at (2, 5, 4) :

f(x, y, z)=xy+4z

Possible Answers:

5x+2y+4z=36

2x+5y+4z=36

5x+2y+4z=0

2x+5y+4z=0

Correct answer:

5x+2y+4z=36

Explanation:

The equation of the tangent plane is given by

f_x(x_0, y_0, z_0)(x-x_0)+f_y(x_0, y_0, z_0)(y-y_0)+f_z(x_0, y_0, z_0)(z-z_0)=0

So, we must find the partial derivatives for the function evaluated at the point given. 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

f_y=x

f_z=4

The derivatives were found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Evaluated at the given point, the partial derivatives are

f_x=5

f_y=2

f_z=4

Note that the partial derivative with respect to z was 4 to begin with; the fact that the point has a z coordinate of 4 is a coincidence.

Now, plug all of this into our given formula:

5(x-2)+2(y-5)+4(z-4)=0

which simplified becomes

5x+2y+4z=36

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Example Question #52 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\nabla f$ of the following function:

f(x, y, z)=ze^z+xy

Possible Answers:

$\left \langle y, x, 2e^z\right \rangle$

$\left \langle x, y, e^z+ze^z\right \rangle$

$\left \langle y, x, e^z\right \rangle$

$\left \langle y, x, e^z+ze^z\right \rangle$

Correct answer:

$\left \langle y, x, e^z+ze^z\right \rangle$

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=y

f_y=x

f_z=e^z+ze^z

The derivatives were found using the following rules:

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

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Example Question #53 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\nabla f$ of the following function:

f(x, y, z)=xyz^4+ze^y

Possible Answers:

$\left \langle yz^4, xz^4+ze^y, 4xyz^3+e^y\right \rangle$

$\left \langle yz^4, xz^4+ze^y, 4xyz+ze^y\right \rangle$

yz^4+xz^4+4xyz^3+2ze^y

$\left \langle yz^4, xz^4+ze^y, 4xyz^3+ze^y\right \rangle$

Correct answer:

$\left \langle yz^4, xz^4+ze^y, 4xyz^3+e^y\right \rangle$

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=yz^4

f_y=xz^4+ze^y

f_z=4xyz^3+ze^y

The derivatives were found using the following rules:

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

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Example Question #54 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\nabla f$ of the following function:

$f(x,y,z)=\sec(xyz^2)$

Possible Answers:

$\left \langle yz^2\sec(xyz^2)\tan(xyz^2), xz^2\sec(xyz^2)\tan(xyz^2), xyz\sec(xyz^2)\tan(xyz^2)\right \rangle$

$\left \langle yz^2\sec(xyz^2)\tan(xyz^2), xz^2\sec(xyz^2)\tan(xyz^2), 2xyz^2\sec(xyz^2)\tan(xyz^2)\right \rangle$

$\left \langle x\sec(xyz^2)\tan(xyz^2), y\sec(xyz^2)\tan(xyz^2), 2z\sec(xyz^2)\tan(xyz^2)\right \rangle$

$\left \langle yz^2\sec(xyz^2)\tan(xyz^2), xz^2\sec(xyz^2)\tan(xyz^2), 2xyz\sec(xyz^2)\tan(xyz^2)\right \rangle$

Correct answer:

$\left \langle yz^2\sec(xyz^2)\tan(xyz^2), xz^2\sec(xyz^2)\tan(xyz^2), 2xyz\sec(xyz^2)\tan(xyz^2)\right \rangle$

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

$f_x=yz^2\sec(xyz^2)\tan(xyz^2)$

$f_y=xz^2\sec(xyz^2)\tan(xyz^2)$

$f_z=2xyz\sec(xyz^2)\tan(xyz^2)$

The derivatives were found using the following rules:

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

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Example Question #51 : Gradient Vector, Tangent Planes, And Normal Lines

Find the equation of the tangent plane to the given function at (2, 3, 0) :

f(x,y,z)=xyz+y^2z

Possible Answers:

x+y+z=0

z=0

$z=\frac{1}{6}$

z=6

Correct answer:

z=0

Explanation:

The equation of the tangent plane is given by

f_x(x_0, y_0, z_0)(x-x_0)+f_y(x_0, y_0, z_0)(y-y_0)+f_z(x_0, y_0, z_0)(z-z_0)=0

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

The partial derivatives are

f_x=yz

f_y=xz+2yz^2

f_z=xy+2y^2z

The partial derivatives evaluated at the given point are

f_x=0

f_y=0

f_z=6

Plugging all of this into the above formula, we get

6(z-0)=0

z=0

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Example Question #81 : Applications Of Partial Derivatives

Find the equation of the tangent plane to the given function at (1, 1, 3) :

$f(x, y, z)=\frac{2xy}{z}$

Possible Answers:

3x+3y+z=2

6x+6y-2z=2

3x+3y-z=2

3x+3y-z=-2

Correct answer:

3x+3y-z=2

Explanation:

The equation of the tangent plane is given by

f_x(x_0, y_0, z_0)(x-x_0)+f_y(x_0, y_0, z_0)(y-y_0)+f_z(x_0, y_0, z_0)(z-z_0)=0

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

The partial derivatives are

$f_x=\frac{2y}{z}$

$f_y=\frac{2x}{z}$

$f_z=\frac{-2xy}{z^2}$

The derivatives evaluated at the given point are

$f_x=\frac{2}{3}$

$f_y=\frac{2}{3}$

$f_z=\frac{-2}{9}$

Plugging all of this into the above formula, we get

$\frac{2}{3}(x-1)+\frac{2}{3}(y-1)-\frac{2}{9}(z-3)=0$

which simplifies to

3x+3y-z=2

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Example Question #81 : Applications Of Partial Derivatives

Find $\bigtriangledown f$ of the following function:

$f(x, y, z)=x\sqrt{y+z^2}$

Possible Answers:

$\left \langle \sqrt{y+z^2}, \frac{x}{2\sqrt{y+z^2}}, -\frac{2z}{\sqrt{y+z^2}} \right \rangle$

$\left \langle \sqrt{y+z^2}, \frac{x}{\sqrt{y+z^2}}, -\frac{z}{\sqrt{y+z^2}} \right \rangle$

$\left \langle \sqrt{y+z^2}, \frac{x}{2\sqrt{y+z^2}}, \frac{z}{\sqrt{y+z^2}} \right \rangle$

$\left \langle \sqrt{y+z^2}, -\frac{x}{2\sqrt{y+z^2}}, \frac{2z}{\sqrt{y+z^2}} \right \rangle$

Correct answer:

$\left \langle \sqrt{y+z^2}, \frac{x}{2\sqrt{y+z^2}}, \frac{z}{\sqrt{y+z^2}} \right \rangle$

Explanation:

The gradient of the 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

$f_x= \sqrt{y+z^2}$

$f_y=\frac{x}{2\sqrt{y+z^2}}$

$f_z=\frac{z}{\sqrt{y+z^2}}$

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Example Question #57 : Gradient Vector, Tangent Planes, And Normal Lines

Find the equation of the plane tangent to the following function at (10, 1, 2) :

f(x, y, z)=x^2+y^2+2z^3

Possible Answers:

10x+y+6z=226

20x+2y+12z=113

10x+y+6z=113

10x+y+12z=119

Correct answer:

10x+y+6z=113

Explanation:

The equation of the tangent plane is given by

f_x(x_0, y_0, z_0)(x-x_0)+f_y(x_0, y_0, z_0)(y-y_0)+f_z(x_0, y_0, z_0)(z-z_0)=0

So, we must find the partial derivatives for the function evaluated at the point given. 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=2y

f_z=3z^2

The partial derivatives evaluated at the given point are

f_x=20

f_y=2

f_z=12

Plugging this into the formula above, we get

20(x-10)+2(y-1)+12(z-2)=0

which simplifies to

10x+y+6z=113

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Example Question #58 : Gradient Vector, Tangent Planes, And Normal Lines

Write the equation of the plane tangent to the given function at the point (2, 2, 1) :

$f(x, y, z)=\frac{xz^2}{y^2}$

Possible Answers:

4x-2y+4z=2

x+2y+4z=2

x-2y+4z=4

x-2y+4z=2

Correct answer:

x-2y+4z=2

Explanation:

The equation of the tangent plane is given by

f_x(x_0, y_0, z_0)(x-x_0)+f_y(x_0, y_0, z_0)(y-y_0)+f_z(x_0, y_0, z_0)(z-z_0)=0

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

The partial derivatives are

$f_x=\frac{z^2}{y^2}$ , $f_y=-\frac{2xz^2}{y^2}$ , $f_z=\frac{2xz}{y^2}$

The partial derivatives evaluated at the given point are

$f_x=\frac{1}{4}$ , $f_y=-\frac{1}{2}$ , f_z=1

Plugging this into the above formula, we get

$\frac{1}{4}(x-2)-\frac{1}{2}(y-2)+(z-1)=0$

which simplified becomes

x-2y+4z=2

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Example Question #85 : Applications Of Partial Derivatives

Find $\bigtriangledown f$ of the following function:

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

Possible Answers:

$\left \langle y\cos(xy)e^{\cos(xy)+z^2}, x\cos(xy)e^{\cos(xy)+z^2}, 2ze^{\cos(xy)+z^2}\right \rangle$

$\left \langle -y\sin(xy)e^{\cos(xy)+z^2}, -x\sin(xy)e^{\cos(xy)+z^2}, 2ze^{\cos(xy)+z^2}\right \rangle$

$\left \langle y\sin(xy)e^{\cos(xy)+z^2}, x\sin(xy)e^{\cos(xy)+z^2}, 2ze^{\cos(xy)+z^2}\right \rangle$

$\left \langle -x\sin(xy)e^{\cos(xy)+z^2}, -y\sin(xy)e^{\cos(xy)+z^2}, 2ze^{\cos(xy)+z^2}\right \rangle$

Correct answer:

$\left \langle -y\sin(xy)e^{\cos(xy)+z^2}, -x\sin(xy)e^{\cos(xy)+z^2}, 2ze^{\cos(xy)+z^2}\right \rangle$

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

$f_x= -y\sin(xy)e^{\cos(xy)+z^2}$

$f_y= -x\sin(xy)e^{\cos(xy)+z^2}$

$f_z= 2ze^{\cos(xy)+z^2}$

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

Example Question #52 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #53 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #54 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #51 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #81 : Applications Of Partial Derivatives

Example Question #81 : Applications Of Partial Derivatives

Example Question #57 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #58 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #85 : Applications Of Partial Derivatives

All Calculus 3 Resources

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