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

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Example Questions

Example Question #81 : Applications Of Partial Derivatives

Write the equation of the plane tangent to the given function at the point (1, 3, 6) :

$f(x, y, z)=\ln(x)+y^2+z$

Possible Answers:

x+6y+z=-25

x+6y+z=25

x+3y+6z=25

x+3y+6z=0

Correct answer:

x+6y+z=25

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

$f_x=\frac{1}{x}$ , f_y=2y , f_z=1

The partial derivatives evaluated at the given point are

f_x=1 , f_y=6 , f_z=1

Plugging this into the above formula, we get

(x-1)+6(y-3)+(z-6)=0

which simplified becomes

x+6y+z=25

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Example Question #82 : Applications Of Partial Derivatives

Find the equation of the plane tangent to the following function at the point (5, 1, 1) :

f(x, y, z)=2x^2+6y+3z^2

Possible Answers:

20x+6y+6z=112

5x+1y+1z=56

20x+6y+6z=56

5x+1y+1z=112

Correct answer:

20x+6y+6z=112

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=4x , f_y=6 , f_z=6z

The partial derivatives evaluated at the given point are

f_x=20 , f_y=6 , f_y=6

Now, plug in the information into the formula above:

20(x-5)+6(y-1)+6(z-1)=0

which simplified becomes

20x+6y+6z=112

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Example Question #83 : Applications Of Partial Derivatives

Find $\bigtriangledown f$ of the following function:

$f(x, y, z)=2z+2x^2y^2+\cos(z)$

Possible Answers:

$\left \langle 4xy^2, 4x^2y, 2z-\sin(z)\right \rangle$

$\left \langle 4xy^2, 4x^2y, 2-\sin(z)\right \rangle$

$\left \langle 4x^2y, 4xy^2, 2-\sin(z)\right \rangle$

$\left \langle 4xy^2, 4x^2y, 2+\sin(z)\right \rangle$

Correct answer:

$\left \langle 4xy^2, 4x^2y, 2-\sin(z)\right \rangle$

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=4xy^2

f_y=4x^2y

$f_z=2-\sin(z)$

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Example Question #61 : Gradient Vector, Tangent Planes, And Normal Lines

Find the equation of the plane tangent to the following function at the point (4, 1, 1) :

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

Possible Answers:

x+8y+z=21

8x+2y+2z=21

x+16y+z=21

$x+16y+z=\frac{21}{2}$

Correct answer:

x+16y+z=21

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

$f_x=y^2(xz)^{-\frac{1}{2}}$ , $f_y=4y\sqrt{xz}$ , $f_z=y^2(xz)^{-\frac{1}{2}}$

The partial derivatives evaluated at the given point are

$f_x=\frac{1}{2}$ , f_y=8 , $f_z=\frac{1}{2}$

Plugging this into the formula above, we get

$\frac{1}{2}(x-4)+8(y-1)+\frac{1}{2}(z-1)=0$

which simplified becomes

x+16y+z=21

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Example Question #65 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\bigtriangledown f$ for the following function:

$f(x, y, z)=2xe^x\sin(y)+\tan(e^{yz})$

Possible Answers:

$\left \langle 2e^x\sin(y)(1+x), -2xe^x\cos(y)+ze^{yz}\sec^2(e^{yx}),ye^{yz}\sec^2(e^{yx}) \right \rangle$

$\left \langle 2e^x\sin(y)(1+x), 2xe^x\cos(y),ye^{yz}\sec^2(e^{yx}) \right \rangle$

$\left \langle 2e^x\sin(y)(1+x), 2xe^x\cos(y)+ye^{yz}\sec^2(e^{yx}),ze^{yz}\sec^2(e^{yx}) \right \rangle$

$\left \langle 2e^x\sin(y)(1+x), 2xe^x\cos(y)+ze^{yz}\sec^2(e^{yx}),ye^{yz}\sec^2(e^{yx}) \right \rangle$

Correct answer:

$\left \langle 2e^x\sin(y)(1+x), 2xe^x\cos(y)+ze^{yz}\sec^2(e^{yx}),ye^{yz}\sec^2(e^{yx}) \right \rangle$

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

$f_x= 2e^x\sin(y)(1+x)$

$f_y=2xe^x\cos(y)+ze^{yz}\sec^2(e^{yx})$

$f_z=ye^{yz}\sec^2(e^{yx})$

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Example Question #1621 : Calculus 3

Find $\bigtriangledown f$ for the following function:

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

Possible Answers:

$\left \langle yze^x(3x^2+x^3), x^3ye^x, x^3ze^x+2z\right \rangle$

$\left \langle yze^x(3x^2+x^3), x^3z, x^3y+2z\right \rangle$

$\left \langle yze^x(3x^2+x^3), x^3ze^x, x^3ye^x+2z\right \rangle$

$\left \langle yz(3x^2+x^3e^x), x^3ze^x, x^3ye^x+2z\right \rangle$

Correct answer:

$\left \langle yze^x(3x^2+x^3), x^3ze^x, x^3ye^x+2z\right \rangle$

Explanation:

The gradient vector 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= yze^x(3x^2+x^3)

f_y= x^3ze^x

f_z=x^3ye^x+2z

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Example Question #61 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\bigtriangledown f$ of the following function:

$f(x, y, z)=3x^3yz+y\cos(y)$

Possible Answers:

$\left \langle 9x^2yz, 3x^3z+\cos(y)+y\sin(y), 3x^3y\right \rangle$

$\left \langle 9x^2yz, 3x^3z+\cos(y)-y\sin(y), 3x^3y\right \rangle$

$\left \langle 27x^2yz, 3x^3z+\cos(y)-y\sin(y), 3x^3y\right \rangle$

$\left \langle 9x^2yz, \cos(y)-y\sin(y), 3x^3y\right \rangle$

Correct answer:

$\left \langle 9x^2yz, 3x^3z+\cos(y)-y\sin(y), 3x^3y\right \rangle$

Explanation:

The gradient vector 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=9x^2yz

$f_y=3x^3z+\cos(y)-y\sin(y)$

f_z=3x^3y

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Example Question #1622 : Calculus 3

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

$f(x, y, z)=\ln(x)+y^2z^2$

Possible Answers:

x+4y+8z=17

x+8y+4z=17

x+2y+z=17

x+4y+8z=0

Correct answer:

x+4y+8z=17

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

$f_x=\frac{1}{x}$ , f_y=2yz^2 , f_z=2y^2z

The partial derivatives evaluated at the given point are

f_x=1 , f_y=4 , f_z=8

Plugging these into the formula above, we get

1(x-1)+4(y-2)+8(z-1)=0

which simplifies to

x+4y+8z=17

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Example Question #62 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\bigtriangledown f$ of the following function:

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

Possible Answers:

$\left \langle -2xyz\sin(x^2z)e^{\cos(x^2z)}, e^{\cos(x^2z)}, -x^2\sin(x^2z)e^{\cos(x^2z)}\right \rangle$

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

$\left \langle -2xz\sin(x^2z)e^{\cos(x^2z)}, e^{\cos(x^2z)}, -x^2\sin(x^2z)e^{\cos(x^2z)}\right \rangle$

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

Correct answer:

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

Explanation:

The gradient vector 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

$f_x=-2xyz\sin(x^2z)e^{\cos(x^2z)}$

$f_y= e^{\cos(x^2z)}$

$f_z=-x^2y\sin(x^2z)e^{\cos(x^2z)}$

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Example Question #1623 : Calculus 3

Find the equation to the tangent plane to the following function at the point (-2, 1, -3) :

$f(x, y, z)=\ln(xy^2z^3)$

Possible Answers:

x+4y-2z=12

-2x+y-3z=12

-x+4y-2z=12

-x+4y-2z=6

Correct answer:

-x+4y-2z=12

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

$f_x=\frac{1}{x}$ , $f_y=\frac{2}{y}$ , $f_z=\frac{3}{z}$

The partial derivatives evaluated at the given point are

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

Plugging all of this into the formula above, we get

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

which simplified becomes

-x+4y-2z=12

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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 #81 : Applications Of Partial Derivatives

Example Question #82 : Applications Of Partial Derivatives

Example Question #83 : Applications Of Partial Derivatives

Example Question #61 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #65 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #1621 : Calculus 3

Example Question #61 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #1622 : Calculus 3

Example Question #62 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #1623 : Calculus 3

All Calculus 3 Resources

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