We are open Saturday and Sunday!

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

← Previous 1 2 3 4 5 Next →

Example Question #31 : Rate Of Flow

Water is spilling from a cone at $2cm^{3}/min$ . The cone has a radius of 16cm at the top and is 25cm high. At what rate is the depth of the water changing when the water reaches a height of 12cm ?

Possible Answers:

$\frac{dh}{dt}=\frac{1}{8\pi}$

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

$\frac{dh}{dt}=\frac{25}{12\pi}$

$\frac{dh}{dt}=\frac{625}{12288\pi}$

$\frac{dh}{dt}=\frac{25}{16\pi}$

Correct answer:

$\frac{dh}{dt}=\frac{625}{12288\pi}$

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 .

$\frac{r}{h}=\frac{16}{25}$

$r=\frac{16}{25}h$

The volume equation is now

$V=\frac{\pi }{3}\left ( \frac{16}{25}h \right )^{2}h=\frac{256\pi }{1875}h^{3}$

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.

$\frac{dV}{dt}=\frac{d}{dt}\left [\frac{256\pi }{1875}h^{3} \right ]=\frac{256\pi }{1875}*3h^{2}\frac{dh}{dt}$

$\frac{dV}{dt}=\frac{256\pi }{625}h^{2}\frac{dh}{dt}$

We are given the following parameters

$\frac{dV}{dt}=2cm^{3}/min$ and we need to find $\frac{dh}{dt}$ at h=12cm .

Subsituting this information into the first derivative

$\frac{dV}{dt}=\frac{256\pi }{625}h^{2}\frac{dh}{dt}$

$3=\frac{256\pi }{625}(12)^{2}\frac{dh}{dt}$

$3=\frac{36864\pi }{625}\frac{dh}{dt}$

$\frac{dh}{dt}=3*\frac{625}{36864\pi}=\frac{625}{12288\pi}$

Report an Error

Example Question #32 : Rate Of Flow

Water is poured into a cone shaped cup at a rate of 1 m/s. The radius of the cone is 5 m. What is the rate of change of the height of the water in the cone?

Possible Answers:

$\frac{1}{9\pi}$

$\frac{1}{25}$

$\frac{1}{25\pi}$

$\frac{1}{9}$

Correct answer:

$\frac{1}{25\pi}$

Explanation:

The equation for the volume of a cone is $V=\pi r^2h$ . To find the rate of change of height we must differentiate with respect to h and treat r as a constant.

$V=(5)^2\pi h$

$\frac{dV}{dt}=25\pi \frac{dh}{dt}$

We can plug in the flow rate of 1 and solve for the rate of change of height.

$1=25\pi \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{1}{25\pi}$

Report an Error

Example Question #33 : Rate Of Flow

A cylindrical water tank has a 2 feet radius and is 5 feet long. If the tank is halfway full of water that weighs 25 pounds per cubic feet, what is the force by the water acting on one of the tank's ends?

Possible Answers:

$\frac{200}{3}$

$\frac{800}{3}$

$\frac{400}{3}$

$\frac{50}{3}$

$\frac{500}{3}$

Correct answer:

$\frac{400}{3}$

Explanation:

For this problem, the length of the tank is irrelevant.

Write the formula for the force exerted by a fluid.

$F= \int_{a}^{b} \rho h(y)L(y)\:dy$

We will need to find the equation of the of circle given the radius.

The equation of a circle is: x^2+y^2 =r^2

The center of the circle is (0,0) .

Substitute the radius and simplify: x^2+y^2 =4

Rewrite the equation in terms of .

x^2=4-y^2

$x=\sqrt{4-y^2}$

The length of the horizontal rectangular strips are twice the length of :

$L(y) = 2\sqrt{4-y^2}$

The height h(y) , or the depth, is assuming that the downward direction of the y-axis is positive.

The pressure $\rho$ is $25\: \frac{lbs}{ft^3}$ .

The bounds are from the center of the circle to the outer rim of the tank, which is a distance of two. The bounds are from zero to two. Write the integral and solve.

$F= \int_{0}^{2} (25\: \frac{lbs}{ft^3}) (y)(2\sqrt{4-y^2})\:dy$

$F= \int_{0}^{2} 50y\sqrt{4-y^2}\:dy$

Use substitution to solve.

u=4-y^2

$du=-2y \:dy$

$-\frac{1}{2}du=y \:dy$

Replace the variables of to in terms of in order to solve the integral. Pull out all constants out in front of the integral.

$F= 50\cdot -\frac{1}{2}\int \sqrt{u}\:du = -25\times \frac{2}{3}u^{\frac{3}{2}}$

Re-substitute u=4-y^2 .

$F=-\frac{50}{3}(4-y^2)^{\frac{3}{2}}|_{0}^{2} = 0-[(-\frac{50}{3})(8)] = \frac{400}{3}$

Report an Error

Example Question #34 : Rate Of Flow

A cone-shaped funnel with a base diameter of and a height of is initially full of water, though water begins to drain from the tip at a rate of . How fast does the water level fall when it's at a height of ?

Possible Answers:

$\frac{12}{5}$

$\frac{12}{5\pi}$

$\frac{5}{12\pi}$

$\frac{5\pi}{12}$

$\frac{12\pi}{5}$

Correct answer:

$\frac{5}{12\pi}$

Explanation:

It might be assumed intuitively that the water level will decrease faster as it lowers towards the tip of the cone. To model this, relate the radius of the plane of the water to the height of the water:

d=2r

$\frac{2r}{h}=\frac{6}{5}$

$r=\frac{3}{5}h$

Now, since the volume of water at a point in time is approximately a cone shape, it can be written as:

$V=\frac{1}{3}\pi r^2 h$

$V=\frac{1}{3}\pi (\frac{3}{5}h)^2 h$

$V=\frac{3}{5}\pi h^3$

To find the rate of change, take the derivative with respect to time:

$\frac{dV}{dt}=\frac{9}{5}\pi h^2 \frac{dh}{dt}$

The term $\frac{dh}{dt}$ is unknown and is what we're solving for, but we're told that $\frac{dV}{dt}=3$ .

At height h=2 :

$3=\frac{9}{5}\pi 2^2 \frac{dh}{dt}$

$3=\frac{36}{5}\pi \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{5}{12\pi}$

Report an Error

Example Question #35 : Rate Of Flow

A cone-shaped funnel with a base diameter of and a height of is initially full of water, though water begins to drain from the tip at a rate of . How fast does the radius of the water's surface diminish when it's at a height of ?

Possible Answers:

$\frac{15\pi}{196}$

$\frac{60}{49\pi}$

$\frac{15}{196\pi}$

$\frac{15}{98\pi}$

$\frac{3}{5\pi}$

Correct answer:

$\frac{15}{98\pi}$

Explanation:

Itmay be guessed that the radius of the water's surface will decrease faster as it lowers towards the tip of the cone, i.e it's not constant. To model this, relate the radius of the plane of the water to the height of the water:

d=2r

$\frac{h}{2r}=\frac{10}{8}$

$h=\frac{5}{2}r$

Now, since the volume of water at a point in time is approximately a cone shape, it can be written as:

$V=\frac{1}{3}\pi r^2 h$

$V=\frac{1}{3}\pi r^2 (\frac{5}{2}r)$

$V=\frac{5}{6}\pi r^3$

To find the rate of change, take the derivative with respect to time:

$\frac{dV}{dt}=\frac{5}{2}\pi r^2 \frac{dr}{dt}$

The term $\frac{dr}{dt}$ is unknown and is what we're solving for, but we're told that $\frac{dV}{dt}=3$ .

At height h=7 :

$r=\frac{2}{5}h=\frac{14}{5}$

$\frac{dV}{dt}=\frac{5}{2}\pi r^2 \frac{dr}{dt}$

$3=\frac{5}{2}\pi (\frac{14}{5})^2 \frac{dr}{dt}$

$3=\frac{98}{5}\pi \frac{dr}{dt}$

$\frac{dr}{dt}=\frac{15}{98\pi}$

Report an Error

Example Question #36 : Rate Of Flow

A cone-shaped funnel with a base radius of and a height of is initially full of water, though water begins to drain from the tip at a rate of . How fast does the water level fall when it's at a height of 0.5 ?

Possible Answers:

$\frac{1}{16\pi}$

$\frac{\pi}{4}$

$4\pi$

$\frac{4}{\pi}$

$\frac{1}{4\pi}$

Correct answer:

$\frac{4}{\pi}$

Explanation:

It might be assumed intuitively that the water level will decrease faster as it lowers towards the tip of the cone.

Cone drain

To model this, relate the radius of the plane of the water to the height of the water:

$\frac{r}{h}=\frac{12}{4}$

r=3h

Now, since the volume of water at a point in time is approximately a cone shape, it can be written as:

$V=\frac{1}{3}\pi r^2 h$

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

$V=3\pi h^3$

To find the rate of change, take the derivative with respect to time:

$\frac{dV}{dt}=9\pi h^2 \frac{dh}{dt}$

The term $\frac{dh}{dt}$ is unknown and is what we're solving for, but we're told that $\frac{dV}{dt}=9$ .

At height h=0.5 :

$\frac{dV}{dt}=9\pi h^2 \frac{dh}{dt}$

$9=9\pi 0.5^2 \frac{dh}{dt}$

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

Report an Error

Example Question #37 : Rate Of Flow

A cone-shaped funnel with a base circumference of $10\pi$ and a height of is initially full of water, though water begins to drain from the tip at a rate of $32\pi$ . How fast does the water level fall when it's at a height of ?

Possible Answers:

$4\pi$

$2\pi$

$\frac{4}{\pi}$

Correct answer:

Explanation:

It might be assumed intuitively that the water level will decrease faster as it lowers towards the tip of the cone.

Cone drain

To model this, relate the radius of the plane of the water to the height of the water:

$C=2\pi r$

$\frac{2\pi r}{h}=\frac{10\pi}{5}$

r=h

Now, since the volume of water at a point in time is approximately a cone shape, it can be written as:

$V=\frac{1}{3}\pi r^2 h$

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

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

To find the rate of change, take the derivative with respect to time:

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

The term $\frac{dh}{dt}$ is unknown and is what we're solving for, but we're told that $\frac{dV}{dt}=32\pi$ .

At height h=4 :

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

$32\pi=\pi 4^2 \frac{dh}{dt}$

$\frac{dh}{dt}=2$

Report an Error

Example Question #38 : Rate Of Flow

Possible Answers:

$\frac{\pi}{5}$

$25\pi$

$\frac{5}{\pi}$

$\frac{1}{5\pi}$

$5\pi$

Correct answer:

$\frac{1}{5\pi}$

Explanation:

It might be assumed intuitively that the water level will decrease faster as it lowers towards the tip of the cone.

Cone drain

To model this, relate the radius of the plane of the water to the height of the water:

d=2r

$\frac{2r}{h}=\frac{20}{2}$

r=5h

Now, since the volume of water at a point in time is approximately a cone shape, it can be written as:

$V=\frac{1}{3}\pi r^2 h$

$V=\frac{1}{3}\pi (5h)^2 h$

$V=\frac{25}{3}\pi h^3$

To find the rate of change, take the derivative with respect to time:

$\frac{dV}{dt}=25\pi h^2 \frac{dh}{dt}$

The term $\frac{dh}{dt}$ is unknown and is what we're solving for, but we're told that $\frac{dV}{dt}=5$ .

At height h=1 :

$5=25\pi 1^2 \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{1}{5\pi}$

Report an Error

Example Question #39 : Rate Of Flow

Water is being added to an empty pool using a hose that sprays out water at a constant rate of flow. The pool holds 500 gallons of water, and took 3 hours to finish filling up.

Find an expression for the total amount of water in the pool between t = 0 and t = 180 minutes.

(Hint: Find the rate in which the water was flowing out of the hose and into the pool)

Possible Answers:

4.12t

278t+C

278t

$\frac{t-500}{3}$

3.56t+C

Correct answer:

278t

Explanation:

The rate of flow for the pool can be found by taking the total amount of water that flowed from the hose (the 500 gallons) divided by the amount of time it took to give those 500 gallons (3 hours or 180 minutes). Therefore, the rate of flow of the hose is determined to be 500/180 = 2.78 gallons/min. Afterwards, in order to develop an expression for the amount of water in the pool at a given time, you can take the integral of the flow. This integral will sum all the water that has been added up to that point and give you the total amount of water in the pool.

$f(t) = \int v(t)dt = \int_{0}^{t}2.78dt= 2.78t$

This makes sense because 2.78 gallons are being added every minutes, so the total number of minutes passed multipled by the 2.78 gallons gives the total amount of gallons in the pool. Notice that the C constant was not included, this is because the pool is originally empty. If the pool was not originally empty, the intial amount of water in the pool would have been set as the value of C.

Report an Error

Example Question #40 : Rate Of Flow

Determine the rate of flow of a liquid at t=5 given that it's flow function is determined by:

p(t)=-t^3+4t^2-6t+10

Possible Answers:

Correct answer:

Explanation:

To determine rate of flow, we can take the first derivative of our function by using the power rule:

p(t)=-t^3+4t^2-6t+10

p'(t)=-3t^2+8t-6

At t=5 ,

p'(5)=-3(5)^2+8*5-6=-41

Report an Error

← Previous 1 2 3 4 5 Next →

View Calculus Tutors

Raphael
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Computer Science.

View Calculus Tutors

Johnathan
Certified Tutor

Central Connecticut State University, Bachelor of Science, Geology. Montana Tech of the University of Montana, Master of Scie...

View Calculus Tutors

Yili
Certified Tutor

Shanghai University of Finance and Economics, Master of Science, Economics.

All Calculus 1 Resources

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

Popular Subjects

Chemistry Tutors in Atlanta, MCAT Tutors in Denver, Statistics Tutors in San Diego, MCAT Tutors in Phoenix, GMAT Tutors in New York City, Biology Tutors in Boston, MCAT Tutors in Atlanta, Calculus Tutors in Boston, MCAT Tutors in New York City, Chemistry Tutors in San Francisco-Bay Area

Popular Courses & Classes

Spanish Courses & Classes in Boston, SAT Courses & Classes in Phoenix, MCAT Courses & Classes in Denver, GMAT Courses & Classes in Atlanta, MCAT Courses & Classes in Phoenix, SAT Courses & Classes in Chicago, Spanish Courses & Classes in San Francisco-Bay Area, ACT Courses & Classes in Chicago, SSAT Courses & Classes in New York City, GRE Courses & Classes in Washington DC

Popular Test Prep

MCAT Test Prep in Philadelphia, GMAT Test Prep in Seattle, LSAT Test Prep in Atlanta, GMAT Test Prep in Chicago, ISEE Test Prep in Philadelphia, ACT Test Prep in Miami, GRE Test Prep in Washington DC, ACT Test Prep in Denver, LSAT Test Prep in Boston, GMAT Test Prep in Atlanta

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 #31 : Rate Of Flow

Example Question #32 : Rate Of Flow

Example Question #33 : Rate Of Flow

Example Question #34 : Rate Of Flow

Example Question #35 : Rate Of Flow

Example Question #36 : Rate Of Flow

Example Question #37 : Rate Of Flow

Example Question #38 : Rate Of Flow

Example Question #39 : Rate Of Flow

Example Question #40 : Rate Of Flow

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: