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 #12 : Multi Variable Chain Rule

Determine $\frac{\mathrm{dg} }{\mathrm{d} t}$ , where g=x^2+yz , x=t , y=e^t , z=4t

Possible Answers:

2t+4e^t+1

2t+4te^t(t+1)

2t+4e^t(t+1)

2x+ze^t+4y

Correct answer:

2t+4e^t(t+1)

Explanation:

To find the derivative of the function with respect to t, we must use the multivariable chain rule, which states that, for x, $\frac{\mathrm{d} g}{\mathrm{d} t}=\frac{\mathrm{dg} }{\mathrm{d} x}\cdot \frac{\mathrm{d} x}{\mathrm{d} t}$ . (The same rule applies for the other two variables, and we add the results of the three variables to get the total derivative.)

So, the derivatives are

$\frac{\mathrm{d} g}{\mathrm{d} x}=2x$ , $\frac{\mathrm{d} x}{\mathrm{d} t}=1$ , $\frac{\mathrm{d} g}{\mathrm{d} y}=z$ , $\frac{\mathrm{d} y}{\mathrm{d} t}=e^t$ , $\frac{\mathrm{d} g}{\mathrm{d} z}=y$ , $\frac{\mathrm{d} z}{\mathrm{d} t}=4$

Using the above rule, we get

2x(1)+z(e^t)+y(4)

which rewritten in terms of t, and simplified, becomes

2t+4e^t(t+1)

Report an Error

Example Question #13 : Multi Variable Chain Rule

Find $\frac{\mathrm{dk} }{\mathrm{d} t}$ , where k=x^2+xy , $x=e^{t^2}$ , $y=4t+\cos(t)$

Possible Answers:

$2te^{t^2+t}+8t^2e^t+2te^t\cos(t)+4e^{t^2}-e^{t^2}\sin(t)$

$4te^{t^2+t}+8t^2e^t+2t\cos(t)+4e^{t^2}-e^{t^2}\sin(t)$

$4te^{t^3}+8t^2e^t+2te^t\cos(t)+4e^{t^2}-e^{t^2}\sin(t)$

$4te^{t^2+t}+8t^2e^t+2te^t\cos(t)+4e^{t^2}-e^{t^2}\sin(t)$

Correct answer:

$4te^{t^2+t}+8t^2e^t+2te^t\cos(t)+4e^{t^2}-e^{t^2}\sin(t)$

Explanation:

To find the given derivative of the function, we must use the multivariable chain rule, which states that

$\frac{\mathrm{d}k }{\mathrm{d} t}=\frac{\mathrm{d} k}{\mathrm{d} x}\cdot \frac{\mathrm{d} x}{\mathrm{d} t}+\frac{\mathrm{d} k}{\mathrm{d} y}\cdot \frac{\mathrm{d} y}{\mathrm{d} t}$

So, we find all of these derivatives:

$\frac{\mathrm{d} k}{\mathrm{d} x}=2x+y$ , $\frac{\mathrm{d} x}{\mathrm{d} t}=2te^t$ , $\frac{\mathrm{d} k}{\mathrm{d} y}=x$ , $\frac{\mathrm{d} y}{\mathrm{d} t}=4-\sin(t)$

Plugging this into the formula above, and rewriting x and y in terms of t, we get

$(2e^{t^2}+4t+\cos(t))(2te^t)+e^{t^2}(4-\sin(t))$

which simplified becomes

$4te^{t^2+t}+8t^2e^t+2te^t\cos(t)+4e^{t^2}-e^{t^2}\sin(t)$

Report an Error

Example Question #16 : Multi Variable Chain Rule

Find $\frac{\mathrm{d} g}{\mathrm{d} t}$ of the following function:

g(x,y,z)=x+y^2+3z , where x=2t , y=t^2 , z=e^t

Possible Answers:

4t^2+4t^3+3e^t

2t+4yt+3e^t

2t+4t^3+3e^t

2t+4t^4+3e^t

Correct answer:

2t+4t^3+3e^t

Explanation:

To find the derivative of the function g(x,y,z) with respect to t we must use the multivariable chain rule, which states that

$\frac{\mathrm{d} g}{\mathrm{d} t}= \frac{\mathrm{d} g}{\mathrm{d} x}\cdot \frac{\mathrm{d} x}{\mathrm{d} t}+\frac{\mathrm{d} g}{\mathrm{d} y}\cdot \frac{\mathrm{d} y}{\mathrm{d} t}+\frac{\mathrm{d} g}{\mathrm{d} z}\cdot \frac{\mathrm{d} z}{\mathrm{d} t}$

Our partial derivatives are:

$\frac{\mathrm{d} g}{\mathrm{d} x}= 1$ , $\frac{\mathrm{d} g}{\mathrm{d} y}= 2y$ , $\frac{\mathrm{d} g}{\mathrm{d} z}= 3$ , $\frac{\mathrm{d} x}{\mathrm{d} t}= 2t$ , $\frac{\mathrm{d} y}{\mathrm{d} t}= 2t$ , $\frac{\mathrm{d} z}{\mathrm{d} t}= e^t$

Plugging all of this in - and rewriting any remaining variables in terms of t - we get

2t+4t^3+3e^t

Report an Error

Example Question #21 : Multi Variable Chain Rule

Find $\frac{dz}{dt}$ if z=x+y , x=t^2 and y=2t .

Possible Answers:

2t^2+2t

t^2

2t+2

Correct answer:

2t+2

Explanation:

Find $\frac{dz}{dt}$ if z=x+y , x=t^2 and y=2t .

We use the chain rule to find the total derivative of with respect to .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=1(2t)+1(2)=2t+2$

Report an Error

Example Question #373 : Partial Derivatives

Find $\frac{dz}{dt}$ if $z=e^{x+y}$ , x=lnt and y=1 .

Possible Answers:

$\frac{e^{lnt+1}}{t}$

2t^2+2t

$\frac{e^{lnt-1}}{7}$

$\frac{e^{lnt^2}}{4}$

Correct answer:

$\frac{e^{lnt+1}}{t}$

Explanation:

Find $\frac{dz}{dt}$ if $z=e^{x+y}$ , x=lnt and y=1 .

We use the chain rule to find the total derivative of with respect to .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=\left (e^{x+y}(1) \right )(1/t)+\left (e^{x+y}(1) \right )(0)=\frac{e^{x+y}}{t}$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=lnt and y=1 .

$\frac{dz}{dt}=\frac{e^{lnt+1}}{t}$

Report an Error

Example Question #1331 : Calculus 3

Find $\frac{dz}{dt}$ if z=x^3y , x=t^2+2t and y=5t-3 .

Possible Answers:

t^6-312t^5+224t^4-46t^3+117t^2

-5t^6+112t^5-424t^4-26t^3+72t^2

t^6+12t^5+24t^4-6t^3+7t^2

35t^6+162t^5+210t^4+16t^3-72t^2

Correct answer:

35t^6+162t^5+210t^4+16t^3-72t^2

Explanation:

Find $\frac{dz}{dt}$ if z=x^3y , x=t^2+2t and y=5t-3 .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(y3x^2)(2t+2)+(x^3)(5)=(3x^2y)(2t+2)+5x^3$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=t^2+2t and y=5t-3 .

$\frac{dz}{dt}=\left [ 3(t^2+2t)^2(5t-3) \right ](2t+2)+5(t^2+2t)^3$

$\frac{dz}{dt}=\left [ 3(t^4+4t^3+4t^2)(5t-3) \right ](2t+2)+5(t^6+6t^5+12t^4+8t^3)$

$=\left [ 3(5t^5+20t^4+20t^3-3t^4-12t^3-12t^2) \right ](2t+2)+(5t^6+30t^5+60t^4+40t^3)$

$=\left [ 3(5t^5+17t^4+8t^3-12t^2) \right ](2t+2) +(5t^6+30t^5+60t^4+40t^3)$

=(15t^5+51t^4+24t^3-36t^2)(2t+2) +(5t^6+30t^5+60t^4+40t^3)

=(30t^6+102t^5+48t^4-72t^3 +30t^5+102t^4+48t^3-72t^2) +(5t^6+30t^5+60t^4+40t^3)

=35t^6+162t^5+210t^4+16t^3-72t^2

Report an Error

Example Question #375 : Partial Derivatives

Find $\frac{dz}{dt}$ if z=x^2cosy , x=sin(t) and y=t^2 .

Possible Answers:

2sin(t^2)cos(t)cos(t^2)+tsin^2(t^2)sin(t^2)

2sin(t)cos(t)cos(t^2)-2tsin^2(t)sin(t^2)

2sin(t)cos(t)-2tsin^2(t)

cos(t)cos(t^2)-2sin(t^2)

Correct answer:

2sin(t)cos(t)cos(t^2)-2tsin^2(t)sin(t^2)

Explanation:

Find $\frac{dz}{dt}$ if z=x^2cosy , x=sin(t) and y=t^2 .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(2xcosy)(cost)+(x^2(-siny))(2t)$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=sin(t) and y=t^2 .

$\frac{dz}{dt}=(2sin(t)cos(t^2))(cost)+\left [ (sint)^2(-sin(t^2)) \right ](2t)$

=2sin(t)cos(t)cos(t^2)-2tsin^2(t)sin(t^2)

Report an Error

Example Question #1332 : Calculus 3

Find $\frac{dz}{dt}$ if z=ln(xy) , x=cos(t) and y=ln(t) .

Possible Answers:

$sin(t)+\frac{4}{tlnt}$

$-cos(t)+\frac{1}{t}$

$tan(2t)-\frac{1}{7lnt}$

$-tan(t)+\frac{1}{tlnt}$

Correct answer:

$-tan(t)+\frac{1}{tlnt}$

Explanation:

Find $\frac{dz}{dt}$ if z=ln(xy) , x=cos(t) and y=ln(t) .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=\frac{y}{xy}(-sin(t))+\frac{x}{xy}(\frac{1}{t})$

$\frac{dz}{dt}=\frac{1}{x}(-sin(t))+\frac{1}{y}(\frac{1}{t})$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=cos(t) and y=ln(t) .

$\frac{dz}{dt}=\frac{1}{cos(t)}(-sin(t))+\frac{1}{lnt}(\frac{1}{t})$

$\frac{dz}{dt}=\frac{-sin(t)}{cos(t)}+\frac{1}{tlnt}=-tan(t)+\frac{1}{tlnt}$

Report an Error

Example Question #24 : Multi Variable Chain Rule

Find $\frac{dz}{dt}$ if z=x^2y+cosy , x=-2t^2 and y=sin(t) .

Possible Answers:

$6t^3sin(t)+\left [ t^4-cos(sin(t)) \right ](sin(t))$

$t^3sin(t)-\left [ sin(cos(t)) \right ](cos(t))$

$16t^3sin(t)+\left [ 4t^4-sin(sin(t)) \right ](cos(t))$

$-56t^3sin(t)+\left [ 3t^4-sin(sin(t)) \right ](sin(t))$

Correct answer:

$16t^3sin(t)+\left [ 4t^4-sin(sin(t)) \right ](cos(t))$

Explanation:

Find $\frac{dz}{dt}$ if z=x^2y+cosy , x=-2t^2 and y=sin(t) .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(2xy)(-4t)+(x^2-siny)(cos(t))$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=-2t^2 and y=sin(t) .

$\frac{dz}{dt}=(2(-2t^2)sin(t))(-4t)+\left [ (-2t^2)^2-sin(sin(t)) \right ](cos(t))$

$\frac{dz}{dt}=16t^3sin(t)+\left [ 4t^4-sin(sin(t)) \right ](cos(t))$

Report an Error

Example Question #1340 : Calculus 3

Find $\frac{dz}{dt}$ if z=sin(xy) , x=-t^5+4t and y=2t-3 .

Possible Answers:

(-12t^5+13t^4+6t-1)sin(-2t^4+4t^2+3t^5-12t)

(-12t^5+15t^4+16t-12)cos(-2t^6+8t^2+3t^5-12t)

(4t^5+13t^4+3t+2)sin(-t^6+3t^2+9t^5-2t)

(-2t^5-5t^4+16t-42)cos(-t^6+8t^2+t^5-12t)

Correct answer:

(-12t^5+15t^4+16t-12)cos(-2t^6+8t^2+3t^5-12t)

Explanation:

Find $\frac{dz}{dt}$ if z=sin(xy) , x=-t^5+4t and y=2t-3 .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(ycos(xy))(-5t^4+4)+(xcos(xy))(2)$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=-t^5+4t and y=2t-3 .

$\frac{dz}{dt}=\left [ (2t-3)cos\left [ (-t^5+4t)(2t-3) \right ] \right ](-5t^4+4) +\left [ (-t^5+4t)cos\left [ (-t^5+4t)(2t-3) \right ] \right ](2)$

$\frac{dz}{dt}=\left [ (2t-3)(-5t^4+4)cos\left [ (-2t^6+8t^2+3t^5-12t) \right ] \right ] +\left [ 2(-t^5+4t)cos\left [ (-2t^6+8t^2+3t^5-12t) \right ] \right ]$

$\frac{dz}{dt}=(-10t^5+8t+15t^4-12)cos(-2t^6+8t^2+3t^5-12t) +(-2t^5+8t)cos(-2t^6+8t^2+3t^5-12t)$ $\frac{dz}{dt}=(-10t^5+8t+15t^4-12-2t^5+8t)cos(-2t^6+8t^2+3t^5-12t)$

=(-12t^5+15t^4+16t-12)cos(-2t^6+8t^2+3t^5-12t)

Report an Error

View Calculus 3 Tutors

Max
Certified Tutor

SHAHiDBEHESHTiUNiVERSiTY, Bachelor of Science, Mathematics. SHAHiDBEHESHTiUNiVERSiTY, Master of Science, Mathematics.

View Calculus 3 Tutors

Fatemeh
Certified Tutor

University of Isfahan, Doctor of Engineering, Chemical Engineering. University of Windsor, Doctor of Engineering, Environment...

View Calculus 3 Tutors

Muhammad
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Mechanical Engineering.

All Calculus 3 Resources

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

Popular Subjects

ACT Tutors in Miami, Algebra Tutors in Los Angeles, MCAT Tutors in Chicago, Chemistry Tutors in Philadelphia, Computer Science Tutors in San Francisco-Bay Area, Statistics Tutors in Boston, French Tutors in Los Angeles, Algebra Tutors in Miami, ISEE Tutors in Seattle, Computer Science Tutors in Atlanta

Popular Courses & Classes

ACT Courses & Classes in New York City, SSAT Courses & Classes in Washington DC, ISEE Courses & Classes in Miami, SSAT Courses & Classes in Boston, SAT Courses & Classes in Atlanta, ISEE Courses & Classes in New York City, LSAT Courses & Classes in Denver, Spanish Courses & Classes in San Francisco-Bay Area, ISEE Courses & Classes in San Diego, MCAT Courses & Classes in San Diego

Popular Test Prep

MCAT Test Prep in Washington DC, ACT Test Prep in New York City, MCAT Test Prep in Seattle, MCAT Test Prep in Phoenix, GRE Test Prep in Denver, MCAT Test Prep in Philadelphia, ISEE Test Prep in San Diego, GMAT Test Prep in Denver, SAT Test Prep in Denver, SAT Test Prep in Dallas Fort Worth

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 #12 : Multi Variable Chain Rule

Example Question #13 : Multi Variable Chain Rule

Example Question #16 : Multi Variable Chain Rule

Example Question #21 : Multi Variable Chain Rule

Example Question #373 : Partial Derivatives

Example Question #1331 : Calculus 3

Example Question #375 : Partial Derivatives

Example Question #1332 : Calculus 3

Example Question #24 : Multi Variable Chain Rule

Example Question #1340 : Calculus 3

All Calculus 3 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: