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Calculus 3 : Applications of Partial Derivatives

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Example Questions

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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Example Question #91 : Applications Of Partial Derivatives

Find the equation to the tangent plane to the following function at the point (6, 0, 2) :

$f(x, y, z)=z^2+9y^2+\cos(x-6)$

Possible Answers:

z=-2

x+y+z=0

4z=2

z=2

Correct answer:

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

$f_x=-\sin(x-6)$ , f_y=18y , f_z=2z

Evaluated at the given point, the partial derivatives are

f_x=0 , f_y=0 , f_z=4

Plugging this into the formula above, we get

4(z-2)=0

which simplifies to

z=2

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Example Question #71 : Gradient Vector, Tangent Planes, And Normal Lines

Find $\bigtriangledown f$ for the following function:

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

Possible Answers:

$2xye^{x^2}\sec(e^{x^2})+\tan(e^{x^2})+\ln(x)$

$2xe^{x^2}\sec(e^{x^2})\hat{i}+\tan(e^{x^2})\hat{j}+\ln(x)\hat{k}$

$2xye^{x^2}\sec(e^{x^2})\hat{i}+\tan(e^{x^2})\hat{j}+\ln(x)\hat{k}$

$2xye^{x^2}\sec(e^{x^2})\hat{i}+y\tan(e^{x^2})\hat{j}+\ln(x)\hat{k}$

$2xye^{x^2}\sec(e^{x^2})\hat{i}+\tan(e^{x^2})\hat{j}+z\ln(x)\hat{k}$

Correct answer:

$2xye^{x^2}\sec(e^{x^2})\hat{i}+\tan(e^{x^2})\hat{j}+\ln(x)\hat{k}$

Explanation:

The gradient vector of a function is given by

$\bigtriangledown f=\left \langle f_x, f_y, f_z\right \rangle$

written in standard form as $f_x \hat{i}+f_y\hat{j}+f_z\hat{k}$

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=2xye^{x^2}\sec(e^{x^2})$

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

$f_z=\ln(x)$

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Example Question #92 : Applications Of Partial Derivatives

Find the gradient vector of the following function:

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

Possible Answers:

$\left \langle 1, \cos(z^2), -2yz\sin(z^2)+3z^2\right \rangle$

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

$\left \langle 1, \cos(z^2), -2z\sin(z^2)+3z^2\right \rangle$

$\left \langle 1, \cos(z^2), 2yz\sin(z^2)+3z^2\right \rangle$

Correct answer:

$\left \langle 1, \cos(z^2), -2yz\sin(z^2)+3z^2\right \rangle$

Explanation:

The gradient vector 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=1 , $f_y=-2yz\sin(z^2)$ , f_z=3z^2

Report an Error

Example Question #74 : Gradient Vector, Tangent Planes, And Normal Lines

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

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

Possible Answers:

x+18y+9z=84

3x+3y+3z=84

x+18y+9z=0

x+18y+9z=81

Correct answer:

x+18y+9z=84

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=1 , f_y=2yz , f_z=y^2

Evaluated at the given point, the partial derivatives are

f_x=1 , f_y=18 , f_z=9

Plugging these into the formula above, we get

1(x-3)+18(y-3)+9(z-3)=0

which simplifies to

x+18y+9z=84

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Example Question #75 : Gradient Vector, Tangent Planes, And Normal Lines

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

f(x,y,z)=x^2+7y^2+10z^2

Possible Answers:

-x-y-5z=516

-2x+14y+100z=0

2x+14y+100z=512

-2x+14y+100z=516

Correct answer:

-2x+14y+100z=516

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=14y , f_z=20z

Evaluated at the given point, the partial derivatives are

f_x=-2 , f_y=14 , f_z=100

Putting this into the equation above, we get

-2(x+1)+14(y-1)+100(z-5)=0

which simplified becomes

-2x+14y+100z=516

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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 #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

Example Question #91 : Applications Of Partial Derivatives

Example Question #71 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #92 : Applications Of Partial Derivatives

Example Question #74 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #75 : Gradient Vector, Tangent Planes, And Normal Lines

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