We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

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

Study concepts, example questions & explanations for Calculus 3

Create An Account

All Calculus 3 Resources

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

Example Questions

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

Report an Error

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 , $f_y=2y\sin(z^2)$ , $f_z=2y^2z\cos(z^2)$

Report an Error

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

Report an Error

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

Report an Error

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

Report an Error

Example Question #1645 : Calculus 3

Find the gradient vector of the following function:

$f(x, y,z)=7z^2+z+x^2+2x+\ln(xy^2)$

Possible Answers:

$\left \langle 2x+2+\frac{1}{x}, \frac{1}{y}, 14z+1\right \rangle$

$\left \langle 2x+2+\frac{1}{x}, \frac{2}{y}, 15z\right \rangle$

$\left \langle 4x+\frac{1}{x}, \frac{2}{y}, 14z+1\right \rangle$

$\left \langle 2x+2+\frac{1}{x}, \frac{2}{y}, 14z+1\right \rangle$

Correct answer:

$\left \langle 2x+2+\frac{1}{x}, \frac{2}{y}, 14z+1\right \rangle$

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

$f_x=2x+2+\frac{1}{x}$ , $f_y= \frac{2}{y}$ , f_z=14z+1

Report an Error

Example Question #1646 : Calculus 3

Find the equation of the plane tangent to the following function at (7, 1, 1) :

$f(x,y,z)=z^3\ln(y)+\frac{x^2}{2}$

Possible Answers:

7x+y+z=51

7x+y=51

7x+y=0

7x+y+z=0

Correct answer:

7x+y+z=51

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=x , $f_y=\frac{z^3}{y}$ , $f_z=3z^2\ln(y)$

Evaluated at the given point, the partial derivatives are

f_x=7 , f_y=1 , f_z=0

Plugging this into our equation, we get

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

which simplified becomes

7x+y+z=51

Report an Error

Example Question #1647 : Calculus 3

Find the gradient vector of the following function:

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

Possible Answers:

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

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

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

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

Correct answer:

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

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

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

Report an Error

Example Question #111 : Applications Of Partial Derivatives

Find the equation of the tangent plane at the point (1,1,-2) to the surface

x^3+y^2+3z^2=5

Possible Answers:

3x+2y-12z-29=0

2x+3y+12z+19=0

3x-2y+12z+29=0

2x+3y+12z-18=0

Correct answer:

3x+2y-12z-29=0

Explanation:

The equation of the surface given in the problem is the level surface (with k=5 ) of the function F(x,y,z)=x^3+y^2+3z^2 . Hence, we can apply the definition of the tangent plane to a level surface to calculate the plane passing through the point (1,1,-2) at an orthogonal direction (that is, perpendicularly) to the point's normal vector.

Recall this definition of the tangent plane to a level surface F(x,y,z) :

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

We have F(x,y,z)=x^3+y^2+3z^2 and the point (1,1,-2) . Calculating the partial derivatives of as well as the values of each of these derivatives at yields the following information:

F_x(x,y,z)=3x^2 ,

F_y(x,y,z)=2y ,

F_z(x,y,z)=6z ,

F_x(1,1,-2)=3(1)^2=3 ,

F_y(1,1,-2)=2(1)=2 ,

F_z(1,1,-2)=6(-2)=-12 .

Therefore, the aforementioned definition of the tangent plane gives the equation for the tangent plane of F(x,y,z) at (1,1,2) as

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

$\Rightarrow 3x-3+2y-2-12z-24=0$

$\Rightarrow 3x+2y-12z-29=0$

Report an Error

Example Question #1648 : Calculus 3

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

$f(x,y,z)=xy\sin(x)+z$

Possible Answers:

$z=\frac{1}{2}$

z=2

z=0

x+y+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

$f_x=y(\sin(x)+x\cos(x))$

$f_y=x\sin(x)$

f_z=1

The partial derivatives evaluated at the given point are

f_x=0 , f_y=0 , f_z=1

Plugging this into the equation above, we get

1(z-2)=0

which simplifies to

z=2

Report an Error

View Calculus 3 Tutors

Dilip
Certified Tutor

University of Calcutta, Bachelor of Science, Physics. Indiana Institute of Technology, Master of Science, Physics.

View Calculus 3 Tutors

Jason
Certified Tutor

Brigham Young University, Bachelor, Neuroscience.

View Calculus 3 Tutors

Reza
Certified Tutor

University of Sherbrooke, Doctor of Philosophy, Mathematics. University of Manitoba, Master of Science, Mathematics.

All Calculus 3 Resources

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

Popular Subjects

Math Tutors in New York City, Chemistry Tutors in Washington DC, SSAT Tutors in New York City, SSAT Tutors in Seattle, ISEE Tutors in Phoenix, Reading Tutors in Houston, GMAT Tutors in Phoenix, Reading Tutors in Chicago, Math Tutors in Houston, French Tutors in Boston

Popular Courses & Classes

ACT Courses & Classes in Atlanta, Spanish Courses & Classes in Phoenix, SSAT Courses & Classes in Philadelphia, SAT Courses & Classes in Denver, ACT Courses & Classes in Miami, LSAT Courses & Classes in San Diego, SAT Courses & Classes in San Francisco-Bay Area, ACT Courses & Classes in San Diego, SSAT Courses & Classes in Los Angeles, Spanish Courses & Classes in Atlanta

Popular Test Prep

MCAT Test Prep in Denver, ISEE Test Prep in San Diego, ACT Test Prep in Atlanta, ISEE Test Prep in Phoenix, ACT Test Prep in San Diego, ISEE Test Prep in New York City, MCAT Test Prep in New York City, GRE Test Prep in Denver, ACT Test Prep in San Francisco-Bay Area, ISEE Test Prep in Seattle

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

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 : 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

Example Question #1645 : Calculus 3

Example Question #1646 : Calculus 3

Example Question #1647 : Calculus 3

Example Question #111 : Applications Of Partial Derivatives

Example Question #1648 : Calculus 3

All Calculus 3 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: