Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : How to find rate of flow

Study concepts, example questions & explanations for Calculus 1

Create An Account

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #21 : Rate

During a period of rainfall, the volume of fluid in a lake is given by the formula: $V(t)=15+\sqrt[3]{t^2}$

What is the rate of downpour into a lake if it is the only contributing factor to the volume's change?

Possible Answers:

$\frac{3}{2t^{\frac{2}{3}}}$

$\frac{2}{3t^{\frac{1}{2}}}$

$\frac{3}{2t^{\frac{1}{3}}}$

$\frac{2}{3t^{\frac{1}{3}}}$

Correct answer:

$\frac{2}{3t^{\frac{1}{3}}}$

Explanation:

Since the change in volume is due solely to the downpour, the rate change of the volume is equal to the rate of downpour.

For this probelm we will use the power rule to solve which states,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

Applying this rule we are able to find the rate of downpour.

$V'(t)=\frac{d}{dt}(15+\sqrt[3]{t^2})=\frac{2}{3t^{\frac{1}{3}}}$

Report an Error

Example Question #1811 : Functions

Water is poured in a cylindrical container at a rate of $4 \txtup{ L/s}$ . If the radius is meters how fast is the water rising.

Possible Answers:

$9\pi$

$\frac{1}{2\pi}$

None of these

$\frac{1}{\pi}$

Correct answer:

$\frac{1}{\pi}$

Explanation:

The volume of a cylinder is $V=\pi r^{2}h$ . By implicit differentiation while holding the radius constant,

$\frac{dV}{dt}=\pi r^2\frac{dh}{dt}$

The rate of volume change is 4 and the radius is 2 so

$4=\pi*4\frac{dh}{dt}$

$\frac{dh}{dt}=\frac{1}{\pi}$

Report an Error

Example Question #23 : Rate

Water is poured into a square pyramid shaped container at a rate of 10 L/s . The container has a side length of . What is the rate of change of the height?

Possible Answers:

None of these

Correct answer:

Explanation:

The equation for volume for the container is $V=\frac{s^{2}h}{3}$ . Implicit differentiation while holding side length constant gives

$\frac{dV}{dt}=\frac{s^2}{3}\frac{dh}{dt}$

With a side length of 3 and a flow rate of 10

$10=\frac{9}{3}\frac{dh}{dt}$

$\frac{dh}{dt}=30$

Report an Error

Example Question #1812 : Functions

A cubic tank with sides of 10 feet is being filled at a rate of $Flow=10\frac{ft^3}{s}+3t^2\frac{ft^3}{s^3}$ .

If the water level in the tank is initially one foot from the bottom, how long will it tank for the tank to fill?

Possible Answers:

9.31s

10s

17.22s

5.23s

9.67s

Correct answer:

9.31s

Explanation:

Begin by finding the capacity of the tank:

$V_{max}=(10ft)^3=1000ft^2$

Now, determine the amount of water initially in the tank:

V_0=1ft(10ft)^2=100ft^3

From here, integrate the flow equation to determine the amount of water at any point in time:

$V(t)=\int Flow=\int (10\frac{ft^3}{s}+3t^2\frac{ft^3}{s^3})dt$

$V(t)=10t\frac{ft^3}{s}+t^3\frac{ft^3}{s^3}+C$

This constant of integration can be found by using the initial condition:

$V(0)=V_0=10(0)\frac{ft^3}{s}+0^3\frac{ft^3}{s^3}+C$

C=100ft^3

$V(t)=10t\frac{ft^3}{s}+t^3\frac{ft^3}{s^3}+100ft^3$

Now, to find when the tank is full, set the equation equal to the max capacity of the tank:

$V(t_f)=10t_f\frac{ft^3}{s}+t_f^3\frac{ft^3}{s^3}+100ft^3=1000ft^3$

$10t_f\frac{ft^3}{s}+t_f^3\frac{ft^3}{s^3}=900ft^3$

t_f=9.31s

Report an Error

Example Question #1813 : Functions

Neo loses money such that the amount of dollars in his bank account at time (in hours) is m(t)=5000-0.0001e^t .

What is the net rate of flow of money out of Neo's bank account when t=15 ?

Possible Answers:

$-\$326.90/\textup{hr}$

$\$4673.10/\textup{hr}$

$\$326.90/\textup{hr}$

$-\$4673.10/\textup{hr}$

Correct answer:

$\$326.90/\textup{hr}$

Explanation:

The flow of money out of the bank account is the negative of the rate of change (increase) in the amount of money.

Thus, the value we are looking for is

$-m'(15)=-(-0.0001e^{15})=0.0001\cdot3268984.4=326.90$ .

So the net rate of flow out of the bank account is $326.90/hr.

Report an Error

Example Question #1821 : Functions

A -gallon tank begins to drain at a rate of $0.3t^2\frac{gallon}{s^3}$ , where is a time in seconds.

How much water will remain in the tank after five seconds?

Possible Answers:

1.3gallons

13.8gallons

27.5gallons

2.5gallons

32.5gallons

Correct answer:

27.5gallons

Explanation:

To begin this problem, note that the total amount of water drained over a span of time can be found by integrating the rate function with respect to time:

$V(t)=\int Rdt$

$V(t)=\int -0.3t^2 \frac{gallon}{s^3}dt$

$V(t)=-0.1t^3\frac{gallon}{s^3}+C$

Note that the rate is treated as negative, since it is flow out of the tank. To find the value of C, the constant of integration, use the initial condition:

$V(t)=40gallons=-0.1(0^3)\frac{gallon}{s^3}+C$

C=40 gallons

Therefore the integral is:

$V(t)=40gallons -0.1t^3\frac{gallon}{s^3}$

$V(5s)=40gallons -0.1(5s)^3\frac{gallon}{s^3}$

V(5s)=27.5gallons

Report an Error

Example Question #1822 : Functions

A -gallon tank begins to drain at a rate of $0.3t^2\frac{gallon}{s^3}$ , where is a time in seconds.

How much time will it take for the tank to be completely drained?

Possible Answers:

23.29s

12.67s

11.18s

7.37s

6.23s

Correct answer:

7.37s

Explanation:

To begin this problem, note that the total amount of water drained over a span of time can be found by integrating the rate function with respect to time:

$V(t)=\int Rdt$

$V(t)=\int -0.3t^2 \frac{gallon}{s^3}dt$

$V(t)=-0.1t^3\frac{gallon}{s^3}+C$

Note that the rate is treated as negative, since it is flow out of the tank. To find the value of C, the constant of integration, use the initial condition:

$V(t)=40gallons=-0.1(0^3)\frac{gallon}{s^3}+C$

C=40 gallons

Therefore the integral is:

$V(t)=40gallons -0.1t^3\frac{gallon}{s^3}$

Now to find when the tank is drained, solve for when the equation is zero:

$V(t_f)=0=40gallons-0.1t_f^3\frac{gallon}{s^3}$

t_f^3=400s^3

t_f=7.37s

Report an Error

Example Question #1823 : Functions

Water flows into a cylindrical container at a rate of $3t^2\frac{ft^3}{s^3}+6t\frac{ft^3}{s^2}\textup{}$ .

If the cylinder has a radius of ten feet, a height of twenty feet, and is empty initially, how long will it take to fill completely?

Possible Answers:

9.37s

17.51s

44.29s

21.38s

10.88s

Correct answer:

17.51s

Explanation:

Begin by finding the volume of the tank:

$V_{tank}=\pi r^2h$

$V_{tank}=2000\pi ft^3$

Now, to find the amount the tank has filled over a given time, integrating the flow function with respect to time:

$V(t)=\int (3t^2\frac{ft^3}{s^3}+6t\frac{ft^3}{s^2})dt$

$V(t)=t^3\frac{ft^3}{s^3}+3t^2\frac{ft^3}{s^2}+C$

To find the constant of integration, use the initial condition of an empty tank:

$V(0)=0=0^3\frac{ft^3}{s^3}+3(0)^2\frac{ft^3}{s^2}+C$

C=0

$V(t)=t^3\frac{ft^3}{s^3}+3t^2\frac{ft^3}{s^2}$

Now, to find the time when the tank is full, set the equation equal to the max volume of the tank:

$V(t_f)=t_f^3\frac{ft^3}{s^3}+3t_f^2\frac{ft^3}{s^2}=V_{tank}=2000\pi ft^3$

t_f=17.51s

Report an Error

Example Question #1824 : Functions

The equation for the volume of fluid over time is given by

$Q=\frac{1}{3}t^3+2.5t^2+4t+7$ , where is measured in m^3 and is measured in seconds .

Determine the rate of flow at t=4s .

Possible Answers:

$40 \frac{m^3}{s}$

$-40 \frac{m^3}{s}$

$32 \frac{m^3}{s}$

$200\frac{m^3}{s}$

Correct answer:

$40 \frac{m^3}{s}$

Explanation:

Since we're given volume and need to calculate flow, we take the derivative of volume with respect to time .

$Q=\frac{1}{3}t^3+2.5t^2+4t+7$

To find the derivative of Q{} with respect to t{} , we need to use the power rule.

Recall that the power rule for differentiation is given as

$\frac{\mathrm{d} }{\mathrm{d} t} (At^n)=nAt^{n-1}{}$ , where A,n{} are constants and t{} is a variable.

Using this rule for

$Q=\frac{1}{3}t^3+2.5t^2+4t+7{}$

$\frac{dQ}{dt}=t^2+5t+4$

At t=4,{}

$\frac{dQ(4)}{dt}=4^2+5*4+4= 40 \frac{m^3}{s}{}$ .

We know the units are in $\frac{m^3}{s}{}$ because we determined the rate of flow, which is volume of fluid moving per unit time.

Report an Error

Example Question #2851 : Calculus

Water is passing through a cone-shaped filter at $1cm^{3}/min$ . The cone has a radius of 4cm at the top and is 8cm high. At what rate is the depth of the water changing when the water reaches a height of 3cm ?

Possible Answers:

$\frac{dh}{dt}=\frac{475 }{921\pi}$

$\frac{dh}{dt}=\frac{4 }{9\pi}$

$\frac{dh}{dt}=\frac{3}{8\pi}$

$\frac{dh}{dt}=\frac{4 }{8\pi}$

$\frac{dh}{dt}=\pi$

Correct answer:

$\frac{dh}{dt}=\frac{4 }{9\pi}$

Explanation:

We will use the equation for the volume of a cone and its derivative to solve this problem.

The equation for the volume of a cone is

$V=\frac{\pi }{3}r^{2}h$

where is the radius of the cone at the top and is the height of the cone.

Since we do not know $\frac{dr}{dt}$ , we must eliminate from the volume equation before we differentiate.

We use the ratio of height to radius to eliminate .

$\frac{r}{h}=\frac{4}{8}=\frac{1}{2}$

$r=\frac{1}{2}h$

The volume equation is now

$V=\frac{\pi }{3}\left (\frac{1}{2}h \right )^{2}h=\frac{\pi }{12}h^{3}$

We will now take the derivative of the equation with respect to time or . We will use the power and chain rules to find the derivative.

$\frac{dV}{dt}=\frac{d}{dt}\left [\frac{\pi }{12}h^{3}\right ]=\frac{\pi }{12}*3h^{2}\frac{dh}{dt}$

$\frac{dV}{dt}=\frac{\pi }{4}h^{2}\frac{dh}{dt}$

We are given the following parameters

$\frac{dV}{dt}=1cm^{3}/min$ and we need to find $\frac{dh}{dt}$ at h=3cm .

Subsituting this information into the first derivative

$\frac{dV}{dt}=\frac{\pi }{4}h^{2}\frac{dh}{dt}$

$1=\frac{\pi }{4}(3)^{2}\frac{dh}{dt}$

$1=\frac{9\pi }{4}\frac{dh}{dt}$

$\frac{dh}{dt}=\frac{4 }{9\pi}$

Report an Error

View Calculus Tutors

Andrew
Certified Tutor

University of Wisconsin-Stevens Point, Bachelor of Science, Mathematics. University of Wisconsin-Oshkosh, Master of Science, ...

View Calculus Tutors

Magdy
Certified Tutor

Cairo University, Egypt, Bachelor of Science, Electrical Engineering. New Mexico State University-Main Campus, Doctor of Phil...

View Calculus Tutors

Michael
Certified Tutor

Cleveland State University, Bachelor of Science, Mathematics.

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Calculus Tutors in Philadelphia, Algebra Tutors in Miami, Calculus Tutors in Dallas Fort Worth, MCAT Tutors in San Francisco-Bay Area, ISEE Tutors in Miami, ISEE Tutors in San Francisco-Bay Area, Spanish Tutors in Boston, SAT Tutors in Miami, SAT Tutors in Washington DC, French Tutors in Philadelphia

Popular Courses & Classes

SSAT Courses & Classes in Los Angeles, Spanish Courses & Classes in San Francisco-Bay Area, ACT Courses & Classes in Miami, ISEE Courses & Classes in Philadelphia, LSAT Courses & Classes in Denver, GMAT Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in San Diego, MCAT Courses & Classes in New York City, GRE Courses & Classes in Denver, LSAT Courses & Classes in Miami

Popular Test Prep

SSAT Test Prep in Dallas Fort Worth, SSAT Test Prep in San Diego, ISEE Test Prep in Philadelphia, SSAT Test Prep in Philadelphia, MCAT Test Prep in Denver, ISEE Test Prep in Denver, SSAT Test Prep in Denver, SSAT Test Prep in Miami, LSAT Test Prep in Miami, SSAT 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 1 : How to find rate of flow

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #21 : Rate

Example Question #1811 : Functions

Example Question #23 : Rate

Example Question #1812 : Functions

Example Question #1813 : Functions

Example Question #1821 : Functions

Example Question #1822 : Functions

Example Question #1823 : Functions

Example Question #1824 : Functions

Example Question #2851 : Calculus

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: