17Calculus Derivatives - Related Rates Involving Volumes
This page covers related rates problems specifically involving volumes where the shape of the volume is described by an equation and is involved in the solution. These types of problems involve cylinders (often called right-circular cylinders), spheres and troughs (tanks) with a regular geometric shape.
Related rates involving cones are covered on a separate page.
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Overview
When solving these types of problems, you first draw of picture and pick out the type of geometric figure involved. The key here is to write down as many equations as you can think of and then use only the ones that seem to apply. Sometimes, especially with cones, you need to come up with relationships that might not be very intuitive. We will show you some things to watch for. Once you have a figure with all the parts labeled, you can write down the equations involved.
On the main related rates page, we discuss two ways to work these problems based on when you take the derivative, either before combining equations or after. Especially with volume problems, it is often better NOT to combine equations too soon, since the equations can get quite messy very quickly.
What To Do With Constants In Related Rates Problems
What do you do with constants that are given in the problem? First of all, you never want to just go in and plug in all your constants before you take the derivative.
Safe Answer - - Wait and plug in your constants only after you have the derivative. So, you would label all distances with variables, take the derivative with respect to t and then plug in all your given constants. This is what you need to do when you first start learning to work related rates problems. After you have some experience, you can go on to the more experienced technique.
Experienced Answer - - Once you learn the basics of related rates problems, you will have a feel for which constants you can plug in right away and which ones you can't. The difference you need to look for is
- if the variable is NOT changing, then you can substitute the constant in before taking the derivative;
- but, if the variable is changing over time, then you must wait until after you take the derivative before you can substitute the constant into the equation.
At this point, it will just confuse you more if we write down a bunch of theory on how to work these problems. You need to actually see one, then work many in order to see patterns. If you haven't already, read the top section (above the resources section) of the main related rates page. Then, come back here, watch a video or two and try your hand at a few problems.
Cylinders
Problems involving cylinders (also called right-circular cylinders) and cylindrical tanks are usually pretty straightforward. One of the major variations is the orientation of the tank.
- If the tank is on end, then the volume equation is just the area of the circular cross-section times the height of the tank or the depth of the fluid in the tank, i.e. \(V = \pi r^2 h\).
- The other orientation you may come across is a tank on it's side. In this case, you have to be careful to note if the fluid in the tank is above or below the center line. The equations are different in each case.
Probably, if the problem statement does not say that the tank is on it's side, you can usually assume that it is sitting on end. However, check with your instructor to confirm this.
Practice
A cylindrical tank with a radius of 10 meters is being drained. The fluid is draining from the tank at a rate of 1.2m3/min. At what rate is the fluid level inside the tank changing?
Problem Statement |
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A cylindrical tank with a radius of 10 meters is being drained. The fluid is draining from the tank at a rate of 1.2m3/min. At what rate is the fluid level inside the tank changing?
Solution |
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3733 video
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Water is draining out of a cylindrical tank at a rate of 5ft3/min. The diameter of the tank is 8ft and the tank is 10ft tall. How fast is the water level falling when the water is 6ft deep?
Problem Statement |
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Water is draining out of a cylindrical tank at a rate of 5ft3/min. The diameter of the tank is 8ft and the tank is 10ft tall. How fast is the water level falling when the water is 6ft deep?
Solution |
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video by Krista King Math |
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Water is being poured into a cylindrical drum at a rate of 0.010m3/sec. If the radius of the drum is 1.2 meters, how fast is the water level rising?
Problem Statement |
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Water is being poured into a cylindrical drum at a rate of 0.010m3/sec. If the radius of the drum is 1.2 meters, how fast is the water level rising?
Final Answer |
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\(1/(144\pi)\) m/sec
Problem Statement |
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Water is being poured into a cylindrical drum at a rate of 0.010m3/sec. If the radius of the drum is 1.2 meters, how fast is the water level rising?
Solution |
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2248 video
video by Michel vanBiezen |
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Final Answer |
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\(1/(144\pi)\) m/sec |
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Spheres
Sphere problems show up in three main types.
1. If you have a sphere, like a balloon that is a complete sphere. In this case, you use the equation of the volume of a sphere with radius \(r\) as \(V = 4\pi r^3 / 3\).
2. If you have a spherically shaped tank and it is only partially full of fluid. In this case, you need have to notice if the tank level is above or below the half-way point. The equations are different in each case.
3. For the surface area of a sphere, the equation you need is \( A = 4 \pi r^2 \). [Note: Although these problems use the word 'area', we include them on this page since the shape is 3-dimensional.]
Practice
The surface area of a snowball decreases at a rate of 6ft2/hr. How fast is the diameter changing when the radius is 2ft?
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The surface area of a snowball decreases at a rate of 6ft2/hr. How fast is the diameter changing when the radius is 2ft?
Solution |
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3722 video
video by The Organic Chemistry Tutor |
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A spherical balloon is inflated with gas at a rate of 900cm3/min. How fast is the radius of the balloon changing when the radius is 12cm?
Problem Statement |
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A spherical balloon is inflated with gas at a rate of 900cm3/min. How fast is the radius of the balloon changing when the radius is 12cm?
Solution |
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3724 video
video by The Organic Chemistry Tutor |
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Air is being pumped into a spherical balloon at 10cm3/min. Calculate the rate at which the radius of the balloon is increasing when the diameter is 15cm.
Problem Statement |
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Air is being pumped into a spherical balloon at 10cm3/min. Calculate the rate at which the radius of the balloon is increasing when the diameter is 15cm.
Solution |
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video by Krista King Math |
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A spherical balloon is being filled with air so that the radius is increasing at the rate of 0.5 inches per second. How fast is the volume changing when the radius is 2.5 inches?
Problem Statement |
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A spherical balloon is being filled with air so that the radius is increasing at the rate of 0.5 inches per second. How fast is the volume changing when the radius is 2.5 inches?
Solution |
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Air is being pumped into a spherical balloon so that its volume is increasing by 400 cm3/sec. How fast is the radius increasing when the radius is 100 cm?
Problem Statement |
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Air is being pumped into a spherical balloon so that its volume is increasing by 400 cm3/sec. How fast is the radius increasing when the radius is 100 cm?
Final Answer |
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\(dr/dt=1/(100\pi)\) cm/sec
Problem Statement |
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Air is being pumped into a spherical balloon so that its volume is increasing by 400 cm3/sec. How fast is the radius increasing when the radius is 100 cm?
Solution |
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1999 video
video by Krista King Math |
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Final Answer |
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\(dr/dt=1/(100\pi)\) cm/sec |
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A spherical balloon is being filled with air so that its volume is increasing at a rate of 3ft3/min. How fast is the radius increasing when the radius equals 1ft?
Problem Statement |
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A spherical balloon is being filled with air so that its volume is increasing at a rate of 3ft3/min. How fast is the radius increasing when the radius equals 1ft?
Final Answer |
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\( 3/(4\pi) \) ft/min
Problem Statement |
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A spherical balloon is being filled with air so that its volume is increasing at a rate of 3ft3/min. How fast is the radius increasing when the radius equals 1ft?
Solution |
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2243 video
video by Michel vanBiezen |
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Final Answer |
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\( 3/(4\pi) \) ft/min |
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If we have a sphere whose radius is expanding at a rate of 2cm/sec, how fast is the volume changing when the radius is 10cm?
Problem Statement |
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If we have a sphere whose radius is expanding at a rate of 2cm/sec, how fast is the volume changing when the radius is 10cm?
Final Answer |
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\(800\pi\) cm3/sec
Problem Statement |
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If we have a sphere whose radius is expanding at a rate of 2cm/sec, how fast is the volume changing when the radius is 10cm?
Solution |
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2249 video
video by Michel vanBiezen |
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Final Answer |
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\(800\pi\) cm3/sec |
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We have a semi-spherical bowl (a sphere cut in half) that we are filling with water at a rate of 1L/sec. The radius, R, of the sphere is 40cm. How fast is the water level rising when the height is R/2?
Problem Statement |
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We have a semi-spherical bowl (a sphere cut in half) that we are filling with water at a rate of 1L/sec. The radius, R, of the sphere is 40cm. How fast is the water level rising when the height is R/2?
Hint |
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The volume of water in the bowl is given by the equation \(V_w = \pi R h^2-(\pi/3) h^3\) where h is the height of the water.
Also, 1L = 12252 cm3.
Problem Statement |
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We have a semi-spherical bowl (a sphere cut in half) that we are filling with water at a rate of 1L/sec. The radius, R, of the sphere is 40cm. How fast is the water level rising when the height is R/2?
Final Answer |
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\(5/(6\pi)\) cm/sec
Problem Statement |
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We have a semi-spherical bowl (a sphere cut in half) that we are filling with water at a rate of 1L/sec. The radius, R, of the sphere is 40cm. How fast is the water level rising when the height is R/2?
Hint |
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The volume of water in the bowl is given by the equation \(V_w = \pi R h^2-(\pi/3) h^3\) where h is the height of the water.
Also, 1L = 12252 cm3.
Solution |
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2252 video
video by Michel vanBiezen |
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Final Answer |
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\(5/(6\pi)\) cm/sec |
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Troughs and Other Volumes
These types of problems include troughs (tanks) that have some kind of geometric shape that is the same for the entire length of the trough when you look at the cross-section. The volume equation for these types of tanks is just the area of the cross-section times the length.
Key - It is important to notice that the cross-section of the tank is exactly the same no matter where along the tank you look at the cross-section.
If you have a triangular cross-section, you will end up with similar triangles. See the precalculus similar triangles page for a review on how to set up the ratios for use in related rates problems.
Practice
We have a 10 meter long trough that we are filling with water at a rate of 0.3 cubic meters per minute. The bottom of the trough is 30 cm wide, the top of the trough is 80 cm wide and it is 50cm tall. What is the rate at which the water level rises when the depth of the water is 40 cm?
Problem Statement |
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We have a 10 meter long trough that we are filling with water at a rate of 0.3 cubic meters per minute. The bottom of the trough is 30 cm wide, the top of the trough is 80 cm wide and it is 50cm tall. What is the rate at which the water level rises when the depth of the water is 40 cm?
Final Answer |
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30/7 cm/min
Problem Statement |
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We have a 10 meter long trough that we are filling with water at a rate of 0.3 cubic meters per minute. The bottom of the trough is 30 cm wide, the top of the trough is 80 cm wide and it is 50cm tall. What is the rate at which the water level rises when the depth of the water is 40 cm?
Solution |
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2220 video
video by Michel vanBiezen |
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Final Answer |
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30/7 cm/min |
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A trough is 12ft long, 4ft wide and 2ft high. The ends of the trough are isosceles triangles. What is poured into the trough at 36 ft3/min. How fast is the water level changing when the water is 1ft deep?
Problem Statement |
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A trough is 12ft long, 4ft wide and 2ft high. The ends of the trough are isosceles triangles. What is poured into the trough at 36 ft3/min. How fast is the water level changing when the water is 1ft deep?
Solution |
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video by The Organic Chemistry Tutor |
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The sides of a cube are increasing at a rate of 5cm/s. How fast are the surface area and volume increasing when the side length is 10cm?
Problem Statement |
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The sides of a cube are increasing at a rate of 5cm/s. How fast are the surface area and volume increasing when the side length is 10cm?
Solution |
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3725 video
video by The Organic Chemistry Tutor |
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You CAN Ace Calculus
Topics You Need To Understand For This Page
For related rates problems involving similar triangles, it may help you to review how to set up the ratios. You can find a discussion of this on the similar triangles precalculus page. |
Related Topics and Links
related topics on other pages |
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external links you may find helpful |
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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