Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 3 : Calculus 3

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 #3 : Tangent Planes And Linear Approximations

Find the equation of the plane tangent to z=2x^2+4y^3 at the point (1,5) .

Possible Answers:

z=4x+300y-1002

z=x+30y-10

z=2x+300y-100

z=4x^2+3y^3-1002

Correct answer:

z=4x+300y-1002

Explanation:

To find the equation of the tangent plane, we find: f_x,f_y,z_0 and evaluate f_x, f_y at the point given. $f_x=\frac{\partial }{\partial x}(2x^2+4y^3)=4x$ , $f_y=\frac{\partial }{\partial y}(2x^2+4y^3)=12y^2$ , and z_0=2(1^2)+4(5^3)=502 . Evaluating at the point (1,5) gets us f_x=4, f_y=300 . We then plug these values into the formula for the tangent plane: z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0) . We then get z-502=4(x-1)+300(y-5) . The equation of the plane then becomes, through algebra, z=4x+300y-1002

Report an Error

Example Question #5 : Tangent Planes And Linear Approximations

Find the equation of the plane tangent to $z=3\sin(x)+2\cos(y)$ at the point $(\frac{\pi}{2},\frac{\pi}{6})$

Possible Answers:

$z=y-\frac{\pi}{6}$

$z=-y+\frac{\pi}{6}+3+\sqrt3$

$z=x-\frac{\pi}{3}+3+\sqrt3$

$z=y-\frac{\pi}{3}+2+\sqrt3$

Correct answer:

$z=-y+\frac{\pi}{6}+3+\sqrt3$

Explanation:

To find the equation of the tangent plane, we find: f_x,f_y,z_0 and evaluate f_x, f_y at the point given. $f_x=\frac{\partial }{\partial x}(3\sin(x)+2(\cos(y))=3\cos(x)$ , $f_y=\frac{\partial }{\partial y}(3\sin(x)+2\cos(y))=-2\sin(y)$ , and $z_0=3\sin(\frac{\pi}{2})+2\cos(\frac{\pi}{6})=3+\sqrt3$ . Evaluating at the point $(\frac{\pi}{2},\frac{\pi}{6})$ gets us f_x=0, f_y=-1 . We then plug these values into the formula for the tangent plane: z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0) . We then get $z-(3+\sqrt3)=0(x-\frac{\pi}{2})+-1(y-\frac{\pi}{6})$ . The equation of the plane then becomes, through algebra, $z=-y+\frac{\pi}{6}+3+\sqrt3$

Report an Error

Example Question #6 : Tangent Planes And Linear Approximations

Find the equation of the tangent plane to z=3x^2y at the point (2,4)

Possible Answers:

z=x+36y

z=48x+12y-96

z=48x-12

z=5x+50y

Correct answer:

z=48x+12y-96

Explanation:

To find the equation of the tangent plane, we need 5 things: f_x,f_y,z(x,y),f_x(x,y),f_y(x,y)

$f_x=\frac{\partial }{\partial x}(3x^2y)=6xy$

f_x(2,4)=6*2*4=48

$f_y=\frac{\partial }{\partial y}(3x^2y)=3x^2$

f_y(2,4)=3(2)^2=12

z(2,4)=3(2)^2(4)=48

Using the equation of the tangent plane

z-z(x,y)=f_x(x,y)(x-x_0)+f_y(x,y)(y-y_0) , we get

z-48=48(x-2)+12(y-4)

Through algebraic manipulation to get z by itself, we get

z=48x+12y-96

Report an Error

Example Question #1 : Applications Of Partial Derivatives

Find the absolute minimums and maximums of f(x,y)=10x^2-3y^2+10y on the disk of radius , $x^2+y^2\leq16$ .

Possible Answers:

Absolute Minimum: (0,-4)

Absolute Maximum: $\Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$ , $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$

Absolute Minimum: (0,-4)

Absolute Maximum: $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$

Absolute Minimum: (0,-4)

Absolute Maximum: $\Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$

Absolute Minimum: (0,4)

Absolute Maximum: $\Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$ , $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$

Absolute Minimum: (0,4)

Absolute Maximum: $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$

Correct answer:

Absolute Minimum: (0,-4)

Absolute Maximum: $\Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$ , $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$

Explanation:

The first thing we need to do is find the partial derivative in respect to , and .

$\frac{\partial}{\partial x}=20x$ , $\frac{\partial}{\partial y}=-6y+10$

We need to find the critical points, so we set each of the partials equal to .

$0=20x\rightarrow x=0$

$0=-6y+10 \rightarrow y=\frac{5}{3}$

We only have one critical point at $f\Big(0, \frac{5}{3}\Big)$ , now we need to find the function value in order to see if it is inside or outside the disk.

$f\Big(0, \frac{5}{3}\Big)=10(0)^2-3\Big(\frac{5}{3}\Big)^2+10\Big(\frac{5}{3}\Big)=\frac{25}{3}\approx8.33333$

This is within our disk.

We now need to take a look at the boundary, x^2+y^2=16 . We can solve for x^2 , and plug it into .

x^2+y^2=16

x^2=16-y^2

g(y)=10(16-y^2)-3y^2+10y=-13y^2+10y+160

We will need to find the absolute extrema of this function on the range $-4\leq y \leq 4$ . We need to find the critical points of this function.

g'(y)=-26y+10

$0=-26y+10 \rightarrow y=\frac{5}{13}$

The function value at the critical points and end points are:

g(-4)=-13(-4)^2+10(-4)+160=-88

g(4)=-13(4)^2+10(4)+160=-8

$g\Big(\frac{5}{13}\Big)=-13\Big(\frac{5}{13}\Big)^2+10\Big(\frac{5}{13}\Big)+160=\frac{2105}{13}\approx161.92$

Now we need to figure out the values of these correspond to.

$y=-4: x^2=(-4)^2-16 \rightarrow x^2=0\rightarrow x=0$

$y=4: x^2=4^2-16 \rightarrow x^2=0\rightarrow x=0$

$y=\frac{5}{13}: x^2=\Big(\frac{5}{13}\Big)^2-16 \rightarrow x^2=\frac{2679}{169}\rightarrow x\approx\pm\ 3.98$

Now lets summarize our results as follows:

$g(-4)=-88 \rightarrow f(0,-4)=-88$

$g(4)=-8 \rightarrow f(0,4)=-8$

$g\Big(\frac{5}{13}\Big)=\frac{2105}{13} \rightarrow f\Big(-\sqrt{\frac{2679}{169}},\frac{5}{13}\Big)=\frac{2105}{13}$

$g\Big(\frac{5}{13}\Big)=\frac{2105}{13} \rightarrow f\Big(\sqrt{\frac{2679}{169}},\frac{5}{13}\Big)=\frac{2105}{13}$

From this we can conclude that there is an absolute minimum at (0,-4) , and two absolute maximums at $\Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$ and $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$ .

Report an Error

Example Question #1 : Applications Of Partial Derivatives

Find the minimum and maximum of f(x,y)=2x-5y , subject to the constraint x^2+y^2=144 .

Possible Answers:

f(4.46,11.14) is a maximum

f(-4.46,-11.14) is a minimum

f(-4.46,11.14) is a maximum

f(4.46,-11.14) is a minimum

f(4.46,11.14) is a maximum

f(4.46,-11.14) is a minimum

There are no maximums or minimums

f(4.46,-11.14) is a maximum

f(-4.46,11.14) is a minimum

Correct answer:

f(4.46,-11.14) is a maximum

f(-4.46,11.14) is a minimum

Explanation:

First we need to set up our system of equations.

$2=2\lambda x \rightarrow x=\frac{1}{\lambda}$

$-5=2\lambda y \rightarrow y=-\frac{5}{2 \lambda}$

x^2+y^2=144

Now lets plug in these constraints.

$(\frac{1}{\lambda})^2+(-\frac{5}{2\lambda})^2=144$

$\frac{1}{\lambda ^2}+\frac{25}{4\lambda^2}=144$

$\frac{29}{4\lambda^2}=144$

Now we solve for $\lambda$

$\frac{29}{4\lambda^2}=144\rightarrow \lambda^2=\frac{29}{576}\rightarrow\lambda=\pm\sqrt{\frac{29}{576}}$

$\lambda=\sqrt{\frac{29}{576}}$

$x=\frac{1}{\sqrt{\frac{29}{576}}}\approx 4.46$ , $y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx -11.14$

$\lambda=-\sqrt{\frac{29}{576}}$

$x=\frac{1}{\sqrt{\frac{29}{576}}}\approx -4.46$ , $y=-\frac{5}{2\sqrt{\frac{29}{576}}}\approx 11.14$

Now lets plug in these values of , and into the original equation.

f(4.46,-11.14)=2(4.46)-5(-11.14)=64.62

f(-4.46,11.14)=2(-4.46)-5(11.14)=-64.62

We can conclude from this that f(4.46,-11.14) is a maximum, and f(-4.46,11.14) is a minimum.

Report an Error

Example Question #1 : Lagrange Multipliers

Find the absolute minimum value of the function f(x,y) = x^2+y^2 subject to the constraint x^2 +2y^2 = 1 .

Possible Answers:

$\sqrt2$

$2\sqrt2$

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

Let g = x^2 +2y^2 To find the absolute minimum value, we must solve the system of equations given by

$\triangledown f = \lambda\triangledown g, g =1$ .

So this system of equations is

$f_x = \lambda g_x$ , $f_y = \lambda g_y$ , g =1 .

Taking partial derivatives and substituting as indicated, this becomes

$2x = \lambda(2x), 2y = \lambda(4y), x^2+2y^2 = 1$ .

From the left equation, we see either x=0 or $\lambda = 1$ . If x=0 , then substituting this into the other equations, we can solve for $y, \lambda$ , and get $y = \pm \sqrt{2}/2$ , $\lambda = 1/2$ , giving two extreme candidate points at $(0, \frac{\sqrt{2}}{2}), (0, -\frac{\sqrt{2}}{2})$ .

On the other hand, if instead $\lambda =1$ , this forces y = 0 from the 2nd equation, and $x = \pm 1$ from the 3rd equation. This gives us two more extreme candidate points; (-1,0),(1,0) .

Taking all four of our found points, and plugging them back into , we have

$f(-1,0) = 1, f(1,0) = 1, f(0,\frac{\sqrt{2}}{2}) = \frac{1}{2}, f(0,-\frac{\sqrt{2}}{2}) = \frac{1}{2}$ .

Hence the absolute minimum value is $\frac{1}{2}$ .

Report an Error

Example Question #1 : Lagrange Multipliers

Find the dimensions of a box with maximum volume such that the sum of its edges is cm.

Possible Answers:

$6 \times 5 \times 4$

$5 \times 5 \times 5$

$5 \times 4 \times 3$

$6 \times 6 \times 6$

Correct answer:

$5 \times 5 \times 5$

Explanation:

Untitled

Report an Error

Example Question #1 : Lagrange Multipliers

Optimize $f(x)=\sqrt{6-x^2-y^2}$ using the constraint x+y-2=0

Possible Answers:

$\pm\frac{1}{2}$

$\pm2$

Correct answer:

$\pm2$

Explanation:

Untitled

Report an Error

Example Question #1 : Lagrange Multipliers

Maximize f(x,y,z)=xyz with constraint x^2+2y^2+3z^2=6

Possible Answers:

$\frac{3}{2\sqrt{2}}$

$\frac{3}{2}$

$3\sqrt{2}$

$\frac{\sqrt{3}}{2}$

Correct answer:

$\frac{3}{2\sqrt{2}}$

Explanation:

Untitled

Report an Error

Example Question #2 : Lagrange Multipliers

A company has the production function $P=100L^{\frac{3}{4}}K^{\frac{1}{4}}$ , where represents the number of hours of labor, and represents the capital. Each labor hour costs $150 and each unit capital costs $250. If the total cost of labor and capital is is $50,000, then find the maximum production.

Possible Answers:

$\$12300$

$\$12500$

$\$12000$

none of the above

Correct answer:

$\$12300$

Explanation:

Untitled

Report an Error

View Calculus 3 Tutors

Twaha
Certified Tutor

Lakehead University, Bachelor of Engineering, Chemical Engineering. Lakehead University, Master of Science, Chemical Engineer...

View Calculus 3 Tutors

Pankaj
Certified Tutor

University of Delhi, Electrical Engineer, Electrical Engineering. Columbia University in the City of New York, Master of Arts...

View Calculus 3 Tutors

Glen
Certified Tutor

Gonzaga University, Bachelor of Science, Computer Science. Bowling Green State University-Main Campus, Master of Science, Com...

All Calculus 3 Resources

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

Popular Subjects

Spanish Tutors in Denver, LSAT Tutors in Washington DC, Algebra Tutors in Chicago, LSAT Tutors in Atlanta, LSAT Tutors in San Diego, GMAT Tutors in Phoenix, LSAT Tutors in Philadelphia, Chemistry Tutors in New York City, Computer Science Tutors in San Diego, GRE Tutors in San Francisco-Bay Area

Popular Courses & Classes

SAT Courses & Classes in Dallas Fort Worth, ISEE Courses & Classes in Seattle, MCAT Courses & Classes in Dallas Fort Worth, ACT Courses & Classes in San Diego, MCAT Courses & Classes in New York City, Spanish Courses & Classes in Chicago, Spanish Courses & Classes in Los Angeles, SAT Courses & Classes in Washington DC, ISEE Courses & Classes in Washington DC, Spanish Courses & Classes in Atlanta

Popular Test Prep

ISEE Test Prep in San Diego, GMAT Test Prep in Dallas Fort Worth, MCAT Test Prep in Phoenix, SSAT Test Prep in Los Angeles, LSAT Test Prep in San Diego, GMAT Test Prep in Boston, ACT Test Prep in Philadelphia, MCAT Test Prep in Miami, GRE Test Prep in Los Angeles, SAT 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 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3 : Tangent Planes And Linear Approximations

Example Question #5 : Tangent Planes And Linear Approximations

Example Question #6 : Tangent Planes And Linear Approximations

Example Question #1 : Applications Of Partial Derivatives

Example Question #1 : Applications Of Partial Derivatives

Example Question #1 : Lagrange Multipliers

Example Question #1 : Lagrange Multipliers

Example Question #1 : Lagrange Multipliers

Example Question #1 : Lagrange Multipliers

Example Question #2 : Lagrange Multipliers

All Calculus 3 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: