Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 3 : Partial Derivatives

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 #351 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0+)}-\frac{(2xy^{3} - 5x^{4} + 3y^{4})}{(x^{4} + 2y^{4})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:-24\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{3}{2}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{51}{2}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{3}{10}\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }-\frac{3}{2}\\&y=x:0\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(2(0)y^{3} - 5(0)^{4} + 3y^{4})}{((0)^{4} + 2y^{4})}=\frac{-3y^{4}}{2y^{4}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{-3y^{4}}{2y^{4}}=-\frac{3}{2}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(2x(x)^{3} - 5x^{4} + 3(x)^{4})}{(x^{4} + 2(x)^{4})}=0\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

Report an Error

Example Question #351 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0+)}\frac{(26x^{3}y^{2} - 14x^{5})}{(18xy^{4} - 5x^{4}y + 3x^{5} + 3y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{6}{133}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{12}{19}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{36}{19}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{120}{19}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{12}{19}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(26(0)^{3}y^{2} - 14(0)^{5})}{(18(0)y^{4} - 5(0)^{4}y + 3(0)^{5} + 3y^{5})}=\frac{0}{3y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{3y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(26x^{3}(x)^{2} - 14x^{5})}{(18x(x)^{4} - 5x^{4}(x) + 3x^{5} + 3(x)^{5})}=\frac{12}{19}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

Report an Error

Example Question #351 : Partial Derivatives

Evaluate:

$\lim_{(x,y)\rightarrow (0,1)} \frac{x^2+xy}{y}$

Possible Answers:

$\infty$

$-\infty$

Does not exist

Correct answer:

Explanation:

Since we won't have any divided by zero problems, we can just evaluate at the point.

$\lim_{(x,y)\rightarrow (0,1)} \frac{x^2+xy}{y}=\frac{0^2+0\cdot1}{1}=0$

Report an Error

Example Question #352 : Partial Derivatives

Evaluate:

$\lim_{(x,y)\rightarrow (0,1)} \frac{x^2+xy}{y}$

Possible Answers:

Does not exist

$-\infty$

$\infty$

Correct answer:

Explanation:

Since we won't have any divided by zero problems, we can just evaluate at the point.

$\lim_{(x,y)\rightarrow (0,1)} \frac{x^2+xy}{y}=\frac{0^2+0\cdot1}{1}=0$

Report an Error

Example Question #1313 : Calculus 3

Compute $\frac{dz}{dt}$ for $z=2xe^{xy}$ , x=t^3 , $y=t^{-2}$ .

Possible Answers:

$\frac{dz}{dt}=6t^2e^{xy}-4t^{-3}x^2e^{xy}$

$\frac{dz}{dt}=(6t^2e^{xy}+2xye^{xy})-(4t^{-3}x^2e^{xy})$

$\frac{dz}{dt}=(6t^2e^{xy}-2xye^{xy})-(4t^{-3}x^2e^{xy})$

$\frac{dz}{dt}=(4t^{-3}x^2e^{xy})$

$\frac{dz}{dt}=(6t^2e^{xy}+2xye^{xy})$

Correct answer:

$\frac{dz}{dt}=(6t^2e^{xy}+2xye^{xy})-(4t^{-3}x^2e^{xy})$

Explanation:

All we need to do is use the formula for multivariable chain rule.

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

$\frac{\partial f}{\partial x}=2e^{xy}+2xye^{xy}$

$\frac{dx}{dt}=3t^2$

$\frac{\partial f}{\partial y}=2x^2e^{xy}$

$\frac{dy}{dt}=-2t^{-3}$

When we put this all together, we get.

$\frac{dz}{dt}=(6t^2e^{xy}+2xye^{xy})-(4t^{-3}x^2e^{xy})$

Report an Error

Example Question #2 : Multi Variable Chain Rule

Evaluate $\frac{\partial a}{\partial x}$ in terms of and/or if a= x-3y+z^2 , x=r^3 , y= s+2r , and z=6r .

Possible Answers:

3r^2-3s-6r+12r^2

3r^2+72r-6

3r^2+72rs-6

3r^6 +72r-6

Correct answer:

3r^2+72r-6

Explanation:

Expand the equation for chain rule.

$\frac{\partial a}{\partial x} = \frac{\partial a}{\partial x}\frac{\partial x}{\partial r}+ \frac{\partial a}{\partial y}\frac{\partial y}{\partial r} + \frac{\partial a}{\partial z}\frac{\partial z}{\partial r}$

Evaluate each partial derivative.

$\frac{\partial a}{\partial x} = 1$ , $\frac{\partial x}{\partial r} =3r^2$

$\frac{\partial a}{\partial y}= -3$ , $\frac{\partial y}{\partial r} = 2$

$\frac{\partial a}{\partial z} =2z$ , $\frac{\partial z}{\partial r} = 6$

Substitute the terms in the equation.

$\frac{\partial a}{\partial x} =(1)(3r^2)+(-3)(2)+(2z)(6) = 3r^2-6+12z$

Substitute z=6r .

3r^2-6+12(6r) = 3r^2+72r-6

The answer is: 3r^2+72r-6

Report an Error

Example Question #2 : Multi Variable Chain Rule

Use the chain rule to find $\frac{dr}{dt}$ when r=z^2-2y+1 , z=t^2 , y=t^3-2t .

Possible Answers:

$\frac{dr}{dt}=9-t$

$\frac{dr}{dt}=-t^3+5t^2+2t-3$

$\frac{dr}{dt}=12t^2+3t-7$

$\frac{dr}{dt}=4t^3-6t^2+4$

Correct answer:

$\frac{dr}{dt}=4t^3-6t^2+4$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem

$\frac{dr}{dt}=\frac{\partial r}{\partial z}\frac{dz}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$

$=\left ( 2z\right )\left ( 2t\right )+\left ( -2\right )\left ( 3t^2-2\right )$

$=\left ( 2t^2\right )\left ( 2t\right )+\left ( -2\right )\left ( 3t^2-2\right )$

=4t^3-6t^2+4

Report an Error

Example Question #4 : Multi Variable Chain Rule

Use the chain rule to find $\frac{dr}{dt}$ when r=3a^4+5b^2 , a=t^2-6t+12 , b=t-5 .

Possible Answers:

$\frac{dr}{dt}=t^7-4t^6-52t^5-25t^4+88t^3-186t^2+2281t-1266$

$\frac{dr}{dt}=24t^7-504t^6+4752t^5-25920t^4+88128t^3-186624t^2+228106t-124466$

$\frac{dr}{dt}=4t^3-6t^2+4$

$\frac{dr}{dt}=-54t^6+52t^5-250t^4+828t^3-1624t^2+2206t-1246$

Correct answer:

$\frac{dr}{dt}=24t^7-504t^6+4752t^5-25920t^4+88128t^3-186624t^2+228106t-124466$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem

$\frac{dr}{dt}=\frac{\partial r}{\partial a}\frac{da}{dt}+\frac{\partial r}{\partial b}\frac{db}{dt}$

$=\left ( 12a^3\right )\left ( 2t-6\right )+\left ( 10b\right )\left ( 1\right )$

$=\left ( 12\left ( t^2-6t+12\right )^3\right )\left ( 2t-6\right )+\left ( 10\left ( t-5\right )\right )\left ( 1\right )$

$=\left12 ( t^6-18t^5+144t^4-648t^3+1728t^2-2592t+1728\right )\left ( 2t-6\right )+10t-50$

=24t^7-504t^6+4752t^5-25920t^4+88128t^3-186624t^2+228096t-124416 +10t-50

=24t^7-504t^6+4752t^5-25920t^4+88128t^3-186624t^2+228106t-124466

Report an Error

Example Question #5 : Multi Variable Chain Rule

Use the chain rule to find $\frac{dr}{dt}$ when r=cos(ab) , a=t^3+7t , b=5t^2-12t+1 .

Possible Answers:

$\frac{dr}{dt}=4t^3-6t^2+4sin\left ( 4t^3-6t^2+4 \right )$

$\frac{dr}{dt}=(t^4-3t^3-128t^2-49t)sin(t^5-131t^4+66t^3-98t^2+17t-121)$

$\frac{dr}{dt}=(-25t^4+48t^3-108t^2+168t-7)sin(5t^5-12t^4+36t^3-84t^2+7t)$

$\frac{dr}{dt}=(5t^4+8t^3-18t^2-9t)cos(4t^5-11t^4+6t^3-8t^2+7t-11)$

Correct answer:

$\frac{dr}{dt}=(-25t^4+48t^3-108t^2+168t-7)sin(5t^5-12t^4+36t^3-84t^2+7t)$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem

$\frac{dr}{dt}=\frac{\partial r}{\partial a}\frac{da}{dt}+\frac{\partial r}{\partial b}\frac{db}{dt}$

=b*cos(ab)(3t^2+7)+a*cos(ab)(10t-12)

$=-(5t^2-12t+1)sin(5t^5-12t^4+36t^3-84t^2+7t)\left ( 3t^2+7\right )-(t^3+7t)sin(5t^5-12t^4+36t^3-84t^2+7t)\left ( 10t-12\right )$

=-(15^4-36t^3+38t^2-84t+7)sin(5t^5-12t^4+36t^3-84t^2+7t)-(10t^4-12t^3+70t^2-84t)sin(5t^5-12t^4+36t^3-84t^2+7t)

=(-25t^4+48t^3-108t^2+168t-7)sin(5t^5-12t^4+36t^3-84t^2+7t)

Report an Error

Example Question #6 : Multi Variable Chain Rule

Use the chain rule to find $\frac{dr}{dt}$ when r=sin(ab^2) , a=5t-1 , b=-9t^3 .

Possible Answers:

$\frac{dr}{dt}=(4t^3-6t^2+4)sin\left ( 4t^3-6t^2+4 \right )$

$\frac{dr}{dt}=(2745t^6-486t^5)cos(405t^7-81t^6)$

$\frac{dr}{dt}=(-90t^4+18t^3)cos(405t^7-81t^6)$

$\frac{dr}{dt}=(81t^6)sin(405t^7-81t^6)$

Correct answer:

$\frac{dr}{dt}=(2745t^6-486t^5)cos(405t^7-81t^6)$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem

$\frac{dr}{dt}=\frac{\partial r}{\partial a}\frac{da}{dt}+\frac{\partial r}{\partial b}\frac{db}{dt}$

$\frac{\partial r}{\partial a}=cos(ab^2)(b^2)$

=cos(405t^7-81t^6)(81t^6)

$\frac{da}{dt}=5$

$\frac{\partial r}{\partial b}=cos(ab^2)(2ab)$

=cos(405t^7-81t^6)(-90t^4+18t^3)

$\frac{db}{dt}=-27t^2$

$\frac{dr}{dt}=\left ( cos(405t^7-81t^6)(81t^6) \right )(5)+(cos(405t^7-81t^6)(-90t^4+18t^3))(-27t^2)$

=405t^6*cos(405t^7-81t^6)+(2430t^6-486t^5)cos(405t^7-81t^6)

=(2745t^6-486t^5)cos(405t^7-81t^6)

Report an Error

View Calculus 3 Tutors

Luis
Certified Tutor

State University of New York at New Paltz, Doctor of Science, Health Physics.

View Calculus 3 Tutors

Ryanto
Certified Tutor

Excelsior College, Bachelor of Science, Business, General. University of Houston, Master of Science, Accounting.

View Calculus 3 Tutors

Cole
Certified Tutor

Louisiana State University-Alexandria, Bachelor of Science, Computer Engineering, General. Louisiana State University-Alexand...

All Calculus 3 Resources

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

Popular Subjects

Physics Tutors in San Diego, Spanish Tutors in Houston, ISEE Tutors in Atlanta, ISEE Tutors in Houston, Calculus Tutors in Phoenix, Calculus Tutors in Miami, SSAT Tutors in Chicago, GMAT Tutors in Atlanta, French Tutors in Atlanta, French Tutors in Phoenix

Popular Courses & Classes

ACT Courses & Classes in Philadelphia, ISEE Courses & Classes in San Diego, Spanish Courses & Classes in Los Angeles, SSAT Courses & Classes in Denver, ACT Courses & Classes in Atlanta, Spanish Courses & Classes in Denver, MCAT Courses & Classes in Seattle, GRE Courses & Classes in Miami, GMAT Courses & Classes in Los Angeles, GRE Courses & Classes in Los Angeles

Popular Test Prep

SAT Test Prep in New York City, ACT Test Prep in Atlanta, GMAT Test Prep in Miami, GRE Test Prep in Philadelphia, ISEE Test Prep in Washington DC, SAT Test Prep in Washington DC, ACT Test Prep in Philadelphia, GRE Test Prep in Phoenix, LSAT Test Prep in San Diego, SSAT Test Prep in Denver

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 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #351 : Limits

Example Question #351 : Limits

Example Question #351 : Partial Derivatives

Example Question #352 : Partial Derivatives

Example Question #1313 : Calculus 3

Example Question #2 : Multi Variable Chain Rule

Example Question #2 : Multi Variable Chain Rule

Example Question #4 : Multi Variable Chain Rule

Example Question #5 : Multi Variable Chain Rule

Example Question #6 : Multi Variable Chain Rule

All Calculus 3 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: