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Calculus 3 : Applications of Partial Derivatives

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #72 : Gradient Vector, Tangent Planes, And Normal Lines

Find the gradient vector for the following function:

$f(x,y,z)=\sin^3(xz)+xy^2+z$

Possible Answers:

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

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

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

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

Correct answer:

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

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

$f_x= 3z\sin^2(xz)\cos(xz)+y^2$ , f_y=x , $f_z=3x\sin^2(xz)\cos(xz)+1$

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

Find the equation of the plane that passes through the points (2,2,4) , (3,5,2) and (1,2,1)

Possible Answers:

-5x+9y-2z+3=0

-9x-5y+3z+2=0

9x-5y+3z-2=0

-9x+5y-3z+2=0

Correct answer:

-9x+5y-3z+2=0

Explanation:

Step 1:

Let P=(2,2,4), Q=(3,5,2),R=(1,2,1)

Using these three points we will find two vectors and . [You can find PQ and QR too]

$PQ=(3,5,2)-(2,2,4)=(1,3,-2)\\ PR=(1,2,1)-(2,2,4)=(-1,0,-3)$

Step 2:

We are required to find a perpendicular (normal) vector to both and . So we need to take their cross product

$\begin{pmatrix}i &j &k \\ 1& 3& 2\\ 1& 0& -3\end{pmatrix}=(-9-0)i-(-3-2)j+(0-3)k$

=-9i+5j-3k

We have found our normal vector.

Step 3: We will use the following formula to find the final answer

$a(x-x_1)+b(y-y_1)+c(z-z_1)=0\\ -9(x-1)+5(y-2)-3(z-1)=0\\ -9x+9+5y-10-3z+3=0\\ -9x+5y-3z+2=0$

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

Find the gradient vector for

$\vec{F}=<e^xyz,ye^x,xyz>$

Possible Answers:

$\nabla\vec{F}=<e^xy^2z^2,x^2e^x,xyz^2>$

$\nabla\vec{F}=<e^xz,yze^x,xy>$

$\nabla\vec{F}=<e^xyz,e^x,xy>$

Correct answer:

$\nabla\vec{F}=<e^xyz,e^x,xy>$

Explanation:

Suppose that

$\vec{F}=<P,Q,R>$

then

$\nabla\vec{F}=\left<\frac{\partial{F}}{\partial{x}},\frac{\partial{F}}{\partial{y}},\frac{\partial{F}}{\partial{z}}\right>$

taking the respective partial derivatives and putting them into order as stated in the formula above yields

$\nabla\vec{F}=<e^xyz,e^x,xy>$

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Example Question #73 : Gradient Vector, Tangent Planes, And Normal Lines

Find the gradient vector for the following function:

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

Possible Answers:

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

$\left \langle \tan(y)-3z\sin(x), x\sec^2(y), 3z\cos(x)\right \rangle$

$\left \langle \tan(y)-3z\sin(x), x\sec^2(y), 3\cos(x)\right \rangle$

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

Correct answer:

$\left \langle \tan(y)-3z\sin(x), x\sec^2(y), 3\cos(x)\right \rangle$

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

$f_x= \tan(y)-3z\sin(x)$ , $f_y= x\sec^2(y)$ , $f_z=3\cos(x)$

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

Find the gradient vector of the following function:

$f(x,y,z)=7y^2\csc(x)+\tan(z)$

Possible Answers:

$\left \langle 7y^2\csc(x)\cot(x), 14y\csc(x), \sec^2(z)\right \rangle$

$\left \langle -7y^2\csc(x)\cot(x), 14y\csc(x), \sec(z)\right \rangle$

$\left \langle -7y^2\csc(x)\cot(x), 14y\csc(x), -\sec^2(z)\right \rangle$

$\left \langle -7y^2\csc(x)\cot(x), 14y\csc(x), \sec^2(z)\right \rangle$

Correct answer:

$\left \langle -7y^2\csc(x)\cot(x), 14y\csc(x), \sec^2(z)\right \rangle$

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

$f_x=-7y^2\csc(x)\cot(x)$ , $f_y=14y\csc(x)$ , $f_z=\sec^2(z)$

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Example Question #81 : Gradient Vector, Tangent Planes, And Normal Lines

Find the gradient vector of the following function:

$f(x,y,z)=x^2+3y^2+z^2+16z+\sqrt{yz}$

Possible Answers:

$\left \langle 2x, 6y+\frac{y}{2\sqrt{yz}}, 2z+16+\frac{z}{2\sqrt{yz}}\right \rangle$

$\left \langle 2x, 6y+\frac{z}{2\sqrt{yz}}, 2z+16+\frac{y}{2\sqrt{yz}}\right \rangle$

$\left \langle 2x, 6y+\frac{2z}{\sqrt{yz}}, 2z+16+\frac{2y}{\sqrt{yz}}\right \rangle$

$\left \langle 2x, 6y-\frac{z}{2\sqrt{yz}}, 2z+16-\frac{y}{2\sqrt{yz}}\right \rangle$

Correct answer:

$\left \langle 2x, 6y+\frac{z}{2\sqrt{yz}}, 2z+16+\frac{y}{2\sqrt{yz}}\right \rangle$

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=2x , $f_y=6y+\frac{z}{2\sqrt{yz}}$ , $f_z= 2z+16+\frac{y}{2\sqrt{yz}}$

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Example Question #1641 : Calculus 3

Find the gradient vector of the following function:

$f(x,y,z)=x^2+x+4+y^2\sin(z^2)$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

The gradient vector 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=2x+1 , $f_y=2y\sin(z^2)$ , $f_z=2y^2z\cos(z^2)$

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Example Question #1642 : Calculus 3

Determine the gradient vector for the following function:

f(x,y,z)=5x^3ye^x+z

Possible Answers:

$\left \langle 5x^2e^x(5+x), 5x^3e^x, 1\right \rangle$

$\left \langle 5x^2ye^x(5+x), 5x^3e^x, z\right \rangle$

$\left \langle 5x^2ye^x(5+x), 5x^3e^x, 1\right \rangle$

$\left \langle 5x^2e^x(5y+x), 5x^3e^x, 1\right \rangle$

Correct answer:

$\left \langle 5x^2ye^x(5+x), 5x^3e^x, 1\right \rangle$

Explanation:

The gradient vector 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= 5x^2ye^x(5+x) , f_y=5x^3e^x , f_z=1

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Example Question #1643 : Calculus 3

Write the equation of the tangent plane to the following function at (0, 4, 5) :

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

Possible Answers:

z=0

x=0

z=5

y=0

x+y+z=0

Correct answer:

z=5

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.

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

Evaluated at the given point, the partial derivatives are

f_x=0 , f_y=0 , f_z=10

Plugging this into our equation, we get

10(z-5)=0

which simplifies to

z=5

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Example Question #1644 : Calculus 3

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

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

Possible Answers:

x+2y+z=0

7x+4y+2z=-17

7x+4y+2z=0

x+2y+z=17

7x+4y+2z=17

Correct answer:

7x+4y+2z=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=3x^2z^2+y^2 , f_y=2xy , f_z=2x^3z

The partial derivatives evaluated at the given point are

f_x=7 , f_y=4 , f_z=2

Plugging these into the equation of the tangent plane, we get

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

which simplifies to

7x+4y+2z=17

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

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

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

Example Question #73 : Gradient Vector, Tangent Planes, And Normal Lines

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

Example Question #81 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #1641 : Calculus 3

Example Question #1642 : Calculus 3

Example Question #1643 : Calculus 3

Example Question #1644 : Calculus 3

All Calculus 3 Resources

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