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I. What's related about these rates? (Or, "How to recognize a Related Rates problem.")

Related rates problems will always give you the rate of one quantity that's changing, and ask you to find the rate of something else that's changing as a result. Here are three common problem-scenarios to illustrate:

1. Expanding Circle Example. Given: The radius of a particular circle increases at 1 millimeter each second. As a result, its area changes. Question: How fast is its area changing [at some particular instant]?
   Play video
2. Sliding Ladder Example. Given: A 10-foot ladder leans against a wall. The base of the ladder slides horizontally away from the wall at 2 feet per second. As a result, the top of the ladder moves down the wall. Question: How quickly is the top of the ladder sliding down the wall [at some particular instant]?
   Play video
3. Water Leaving a Cone Example. Given: A large cone of given size is being drained of water at the constant rate of 15 cm3 each second. The water's surface level in the cone falls as a result. Question: At what rate is the water level falling [at a particular instant]? (We'll solve this problem from start to finish in our next post.)

In each case you're given the rate at which one quantity is changing. That is, you're given the value of the derivative with respect to time of that quantity:

1. "The radius . . . increases at 1 millimeter each second" means the radius changes at the rate of 𝑑𝑟𝑑𝑡 =1 mm/s.
2. "The base of the ladder slides horizontally away from the wall at 2 feet per second" means the ladder-base's x-position changes at the rate of 𝑑𝑥𝑑𝑡 =2 ft/s.

3. When a quantity is decreasing, we have to make the rate negative.

   "A cone is being drained of water at the constant rate of 15 cm3 each second" means the volume of water in cone changes at the rate of 𝑑𝑉𝑑𝑡 = −15 cm3/s. Notice that we have to make the rate negative to capture that the water's volume is decreasing. To do so, we insert the negative sign "by hand"—we just stick it in there.

The upshot: Related rates problems will always tell you about the rate at which one quantity is changing (or maybe the rates at which two quantities are changing), often in units of distance/time, area/time, or volume/time. The question will then be how fast. . . how quickly. . . at what rate. . . ...does another quantity change as a result? The rate you're after is related to the rate(s) you're given. Your job is to find that relationship.

II. Your Job: Develop a Relationship between Quantities

This is where students often initially get stuck when doing Related Rates problems. (But after working lots of these problems you won't anymore; they'll be totally routine for you.) The skill you have to practice is developing a relationship between

• the quantity you're after, and
• the quantity (or quantities) you've been given.

There's just a tiny handful of approaches you'll ever use.

You'll take a derivative in a minute to get a relationship between the rates. Before then, though, we have to write a relationship between the quantities themselves. Fortunately, there's just a tiny handful of approaches you'll ever use. We'll illustrate two of the most common using our first two examples above. (We'll analyze the third example in our next post.)

1. Expanding Circle Example. We need a relationship between the circle's area, and its radius. This one's easy, and just draws upon our everyday geometry-knowledge for a circle:𝐴=𝜋𝑟2 This expression captures the relationship between
   • the quantity you're after (something about the circle's area, A), and
   • the quantity you've been given (something about the circle's radius, r).
2. Sliding Ladder Example. This example illustrates one of the two approaches you'll use most often: the Pythagorean theorem. Let's call the horizontal distance from the wall to the ladder's bottom x, and the vertical distance from the ground to the ladder's top y. Then since the ladder is 10 feet long, at any instant we know𝑥2+𝑦2=102 This expression captures the relationship between
   • the quantity you're after (something about the ladder-top's vertical position, y), and
   • the quantity you've been given (something about the ladder-bottom's horizontal position, x).

The upshot: The new skill you'll probably need to practice is developing the relationship between the quantity whose rate you're after, and the quantity (or quantities) whose rate(s) you know about. There are only a few approaches you'll typically use; we've listed them all in our Problem Solving Strategy box below.

III. Take the Derivative with Respect to Time

Related Rates questions always ask about how two (or more) rates are related, so you'll always take the derivative of the equation you've developed with respect to time. That is, take 𝑑𝑑𝑡 of both sides of your equation. Be sure to remember the Chain Rule!

Let's apply this step to the equations we developed in our two examples above:

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1. Expanding Circle Example.

𝐴=𝜋𝑟2𝑑𝑑𝑡[𝐴]=𝑑𝑑𝑡[𝜋𝑟2]𝑑𝐴𝑑𝑡=2𝜋𝑟𝑑𝑟𝑑𝑡

Why is that dr/dt there? Open for an explanation.

Jump ahead to 4:05 in the video to see the relevant discussion.

Play video

Are you wondering why that 𝑑𝑟𝑑𝑡 appears? The answer is the Chain Rule.

While the derivative of 𝑟2 with respect to r is 𝑑𝑑𝑟𝑟2 =2𝑟, the derivative of 𝑟2 with respect to time t is 𝑑𝑑𝑡𝑟2 =2𝑟𝑑𝑟𝑑𝑡.

Remember that r is a function of time t: the radius grows as time passes. We could have captured this time-dependence explicitly by writing our relation as 𝐴(𝑡) =𝜋[𝑟(𝑡)]2 to remind ourselves that both A and r are functions of time t. Then when we take the derivative, 𝑑𝐴(𝑡)𝑑𝑡=2𝜋𝑟(𝑡)[𝑑𝑑𝑡𝑟(𝑡)] =2𝜋𝑟(𝑡)[𝑑𝑟𝑑𝑡]

[Recall 𝑑𝑟𝑑𝑡 =1 mms in this problem.]

Most people find that writing the explicit time-dependence A(t) and r(t) annoying, and so just write A and r instead. Regardless, you must remember that r depends on t, and so when you take the derivative with respect to time the Chain Rule applies and you have the 𝑑𝑟𝑑𝑡 term.

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That last expression gives us the relation we need between the unknown 𝑑𝐴𝑑𝑡 and the known 𝑑𝑟𝑑𝑡 =1 mms, and so we're essentially finished. The problem will always specify the other conditions of the moment you're interested in. For instance, this problem might have asked:

Question: How fast is the circle's area changing when its radius is 20 mm?

In that case we have 𝑟 =20 mm and 𝑑𝑟𝑑𝑡 =1 mms, and so 𝑑𝐴𝑑𝑡=2𝜋𝑟𝑑𝑟𝑑𝑡  =2𝜋(20)(1)=40𝜋mm2s✓ The problem could have specified any value for r; it doesn't matter. At this point, we're just substituting in values.

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2. Sliding Ladder Example.

𝑥2+𝑦2=102𝑑𝑑𝑡[𝑥2+𝑦2]=𝑑𝑑𝑡[102]2𝑥𝑑𝑥𝑑𝑡+2𝑦𝑑𝑦𝑑𝑡=0 [*]

The last expression gives us the relation we need between the unknown 𝑑𝑦𝑑𝑡 and the known 𝑑𝑥𝑑𝑡 =2 fts, so again the problem is essentially finished. If it had asked, for instance,

Question: How quickly is the top of the ladder sliding down the wall when the base is 8 feet from the wall?

then we would have x = 8 ft and 𝑑𝑥𝑑𝑡 =2 ft/s, and we're looking for 𝑑𝑦𝑑𝑡. But then in this case we still have one unknown: we don't know y.

So we have to stop and do a quick subproblem: our original Pythagorean statement specifies y when 𝑥 =8.

Begin subproblem.



Find 𝑦 when 𝑥 =8:

𝑥2+𝑦2=102 82+𝑦2=100 𝑦2=100−64=36 𝑦=6 ft 

End subproblem.

So now we have 𝑥 =8 ft, 𝑦 =6 ft, and 𝑑𝑥𝑑𝑡 =2 ft/s. We're solving for 𝑑𝑦𝑑𝑡.



Starting with the equation marked [*] above: 2𝑥𝑑𝑥𝑑𝑡+2𝑦𝑑𝑦𝑑𝑡=02𝑦𝑑𝑦𝑑𝑡=−2𝑥𝑑𝑥𝑑𝑡𝑑𝑦𝑑𝑡=−𝑥𝑦𝑑𝑥𝑑𝑡=−86(2)=−83fts✓

Again, the problem could have given us any value for x, and then we'd find the associated y and substitute those values in. (Or the problem could specify a value for y and then we'd have to find the associated x. Whatever.) At this point we're just substituting in values.

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3. Water Leaving a Cone Example.

To see the complete solution to this problem, please visit Part 2 of this blog post on how to solve related rates problems.

The upshot: Take the derivative with respect to time of the equation you developed earlier. Then substitute the values you've been given to find the quantity you're after. (You might first have to complete a little subproblem to determine another quantity at the moment of interest, like we did in the Ladder Example.)

RELATED MATERIAL
• Related Rates Problems & Complete Solutions, all for your practice and for free so you'll be ready for your exam.

IV. 4-Step Problem Solving Strategy

To summarize, here are the four steps that will help you solve very-nearly any Related Rates problem (an image, so you can easily save it):



As promised, in the next post we'll complete the "Water Leaving A Cone" example, which will illustrate the common use of similar triangles in solving Related Rates problems.

If you'd like more example problems with complete solutions, please visit our Related Rates page.

Over to you:

• What tips do you have to share about solving Related Rates problems?
• Or what questions do you have? Related rates problems can be tricky to start, and we're happy to help!
• Or how can we make posts such as this one more useful to you?
Please post on the Forum and let us know!