University of North Dakota
From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.
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University of North Dakota
From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.
Related rates; they are an incredibly useful topic. Very, very useful. For example let?s say an engine designer working for a car company wants to design a new engine. Well, part of the engine has a lot of moving parts in it. For example the piston is moving up and down. Well, if you want to design an engine that?s efficient, yet light and cheap to make, what you have to do, is make sure you?re using just the right amount of material.
To do that, you have to know what the accelerations are of the pistons going up and down, so you know how strong to make them. Related rates can help an engineer figure that out. If you?re watching this episode right now, you should have looked at Implicit Differentiation in episode 7 first. Because Implicit Differentiation is the heart of how this works.
I?m going to say this 1, 2, 3, 4, probably 40 times. The key mistake students make on related rates is often doing the derivative before, or after they?ve substituted in a variable that doesn?t change. Let me say that again. If you?re deriving something, you shouldn?t have substituted in anything that can change. Wait or not until after the derivative.
So there?s a number of common problem types we?re going to through each of the types. We?ll either do a full problem, or at least the set up for a problem. And you?ll find these all on the bonus materials. Common problem types; there are spheres. Those are usual problem where the radius is changing with respect to the volume. Cylinder; often it?s water tank problems like these. Similar triangles and cones, Pythagorean Theorem, and Trig functions. Trig function ones are usually when an angle is changing while some kind of a length is changing.
Spheres are a really common problem type, very common. With a sphere problem, you?re generally trying to relate how fast the radius is changing, with how fast the volume is changing. You might think wait a minute radius double, shouldn?t the volume just double? No, it doesn?t work that way.
This is the problem we?re going to be doing, but let me show you again with what I?m talking about about the radius changing. What I?m going to do is blow two breaths of air into this balloon, in other words a fixed volume and see how the radius changes. So the radius went up a certain amount; quite a bit actually. Now I?m going to put two more breaths into it. Observe how much it changes. Same amount of volume, but the radius didn?t get that much bigger, one more time. Now the radius is barely changing with two breaths. Well, that?s because as it gets bigger and bigger, it takes more and more gas to fill in the larger and larger shells effectively that you have expanding here.
So the rate at which the radius changes, and the rate at which the volume changes, they vary.
So we?re going to find the rate at which the radius of the balloon is changing at the instant its radius of 5cm, if it?s being inflated at a rate of 10 square centimeters per second. That should be cubic centimeters. Our goal here is to find the rate at which the radius of the balloon is increasing at a certain time at that instant in time that it?s radius is 5cm.
When it was smaller, the volume was decreasing at a different rate, or the radius increasing at different rate. When it?s larger, the rate changes again. Remember the bigger it gets, the slower the radius changes. Now I notice a little mistake here. It says it?s being inflated at a rate of 10 square centimeters per second.
That should be cubic centimeters per second. I?ve set up a data chart here. I really recommend doing this. These data charts are ones which really help you organize your information. It helps you to realize what your question is. And you can fill in the numbers as you extract them from the parabola. Because there?s a lot of numbers you usually have to get to set one of these up.
Dr over dt, that?s the rate at which the radius is changing. That?s what we?re being asked to find. So put a question mark there. Dv over dt, that?s the rate of change for the volume. We?re given that one. That one is 10 cubic centimeters per second. Now the radius is 5, but I?m going to put that in quotation marks, because that only happens at a certain instant. You can?t use that one until near the end of the problem.
Next, you have to come up with your base equation. The base equation always has to relate the two things that the problem is asking you to relate. We have to relate these two rates which means we have to have something that relates volume to radius. It?s a balloon. A balloon is basically a sphere and the volume formula for a sphere is 4/3 pi r³. Volume is 4/3 pi r³.
You heard this one last in Geometry. You should probably memorize it. There is a number of basic volume formulas that you should probably know before you go into the AP test. This is one of them. Another one is the volume formula for a cylinder. Another one is the volume formula for a prism etcetera.
Let?s continue on with the problem. So here is out formula. I have v and I have r in it, but I?m supposed to have dr and dv. That?s where the Implicit Differentiation comes in.
Now you?ll remember in the Implicit Differentiation episode, we are always doing d over dy to both sides, because that?s trying to find a regular slope. We have to find dt, the rate at which this is changing with respect to time. Most of the related rates problems you?re doing are going to be with respect to time.
So on both sides we?re going to do d over dt, d over dt, d over dt. Now d over dt of this, the derivative of a single variable. If this was t, its derivative will just be 1, if not just t. So remember Implicit Differentiation is really like Chain Rule. So I have to have that dv over dt factor, because I don?t really know how t and v are related yet. Yeah I know you have this over here, but don?t use it. It will make it a mess.
Notice I haven?t put the 5 in yet, even though it?s 5. That?s very important. I?m going to say this a million times. Anything that?s changing shouldn?t be substituted in until later. You could put in 3.14 in for pi, because that never changes. On this side we will have 3 times 4/3 that is 4. I?m doing the derivative here. Pi is a constant factor, so it just stays in there. The derivative r³ I?m still continuing that. Remember it was 3r³; the 3 turned the 4/3 into a 4. So we have the r² factor and I?m going to write this as times. It?s a Chain Rule factor, which in this case is dr over dt. We are getting close to done, just about finished.
Now we?re going to be able to do a substitution on this one next. We can substitute anything in that?s changing at this point. So I can put the 10 in for the dv over dt. I can put in the 5 for r² because now that we have a rate, we can substitute in the things that are changing. So we have 4 pi times 5² times dr over dt. We?re almost done. Our goal was to find the rate at which the radius is changing with respect to time. So I?ll solve for that, and I?m done. So our final result is dr over dt equals over here I?ve got 4 times 25, that?s a 100. Dividing both sides by a 100 gives you 1/10 over here. I flipped the side of course. I still have the pi factor, and we?re done.
Cylinders are one of the common problem types. You know cylinders your old friends that look like this. Well, as far as cylinders go what you need to know is the formula for this. The volume is the area of its base that?s pi r² times the height of the cylinder. Basic formula; you?ve got to memorize that one. You may have a cylinder problem where the water level is just moving steadily up within the tank like so.
It?s a relatively easy problem. So we?re not going to done of those. You may have one like this, but I think it?s very unlikely where the cylinder is on its side. As it fills those up, you have rectangles of volumes increasing up there. We?re going to not do that one either though this one is very unlikely for you to see on the AP test. It?s quite complex.
We?re going to go onto another one that?s a really common problem type, the conical reservoirs. So everything we just talked about with reservoirs as the level is increasing are built into this type of problem. If you can do this one, you can do the others.
So here is a little example. You?ve got a conical water tank 10 meters in diameter at the top. It?s 20 meters high and being it?s being drained at a rate of 5 cubic meters per minutes. Drained, watch out for that word. How fast is the depth of the water dropping at the instant that it?s 15 meters deep?
So in this picture, this large cone is the reservoir itself. This smaller cone down here, that?s the water. Remember that we?re going to have to actually treat this as if it?s two different cones; a cone for the water and a cone for the tank. We?ll get back to that later.
Let?s fill in the information in here and let?s see. Dh over dt well, that?s what we?re being asked to find out. How fast is the depth dropping? Depth is the height. Dv over dt, that?s the rate of change of volume. Cubic meters that?s volume, but it?s decreasing. You have to have -5 here. The height, well that?s 15, but it?s one of those momentary instant ones. We?re not going to use that number until almost at the end of the problem. So here is our data ready to go. Quotation marks you?re not using that until the end of the problem.
Now remember that in a related rate problem, you?ll always have to have an equation that relates the two things, the two quantities that you?re trying to compare. In this case volume and height. I?ve got a little bit of a problem though. Look at volume, we?ve got height, but the volume formula involves another variable.
We have to get rid of that one somehow. You?ve got to get rid of. And we want to do that before we do the derivative. Well, the way you get rid of it is something you might remember from Geometry. Here is how it works.
If you look at this overall shape, you see you?ve got the cylinder. Within the cylinder, there?s a right triangle. That right triangle has a height of 20. I?m going to call that capital H. It?s got a length here in the triangle. I?m going to call that capital R. So this right triangle has a known radius and a known height.
Down here, here is another right triangle. This right triangle is one that is related to the water that?s in the tank. Now, if you look at the two triangles, they?re both right triangles, and they both have this same angle down here. Well, if two triangles have two matching angles, they?re similar to each other. That means the side lengths are proportional. That?s where this came from. So the height over radius ratio for the small triangle has to exactly the same as the height over radius ratio for the big triangle.
Let?s fill in the number. We don?t know the height or radius of the little guy, but we do know that the height of big one is 20. We do know that the radius of the big one, this triangle side is 5. So now we have something that relates h and r. Remember that the variable we had to get rid of was r for this problem, because the problem was asking us to find the rate of change of height. So I?m going to solve this for r.
R, let?s see you cross multiply you get 5h equals 20r. That means r is dividing both sides by 20. The radius is ¼ of the height. So we just found out that the radius is ¼ of the height. Remember the reason we were doing that is, we don?t want to do the derivative, until we?ve gotten down to the two rates; the two variables that we?re supposed to relate. That was h and v. So we have to get rid of the r. Let?s substitute.
So we have volume equals 1/3 of pi times the radius is ¼ of the height. Substitute that. ¼ of the height and there?s a square that went with the radius, time the height on the end. A little bit simplifying ¼² is 1/16. 1/16 times 1/3 is 1/48. 1/48 pi we have h² times another h that?s h³. That equals the volume.
Well, now we?re going to have to do the derivative. To take care of this derivative, with respect to time, d over dt on both side. We have d over dt of V equals, I?m going to write the d over dt behind this, but technically I really shouldn?t. I just ran out of a little bit of room. So doing the derivative of this with respect to time gives us 1/48 pi.
The derivative of h³ is 3h², times 3h². But h is not t, so we have to have that Chain Rule factor; the dh over dt. Times dh over dt and that equals over here, dv over dt. We are almost done. We?ve done the derivative, so now we can substitute in quantities that are changing. The quantities that we're changing were this. We can substitute in the height. We can substitute in for that.
Let?s write that out again. A little more room here d over dt on both sides. So we?ve got dv over dt equals 1/16h²dh over dt.
Now we can go and put in our numbers. So our goal was to find the rate at which the height is changing. We don?t have that, but we do know that the volume was changing. It was losing 5m³ every minute. I have 1/16 and the problem was at the instant that the height is 15 meter deep. So I have 15² times dh over dt. A little bit of simplifying, solving for dh over dt, and you got the answer. I?ll leave that for you to do it. It?s just a little bit of Arithmetic.
Pythagorean Theorem shows up all the time in Related Rates problems. Let?s have a look at one. So here is an example where you?ve got a ladder it?s sliding down the wall. The typical ladder is sliding down the ladder at a fixed rate. The question is how fast is the base moving away from the wall at the instant the ladder?s end is 16 feet from the ground. Well, let?s fill in our table.
We are supposed to find out how fast the base is moving away. That means x is changing, so dx over dt is our unknown. Dh over dt though that?s how fast this quantity is changing and the tip of the ladder is the thing that?s moving so we do have the rate for that one. It?s 3, but don?t forget that since it?s getting smaller, you have to call it -3.
X, that?s the distance from the tip of the ladder to the base of the wall, 16 feet. But that?s one of those little guys that I?m going to put in quotation marks, because that is a change in quantity and it can?t be substituted until the end. Length of the ladder though, 20 feet. That never changes. We?ve be able to use that one earlier on.
So in this slide, you?ll see we have a Pythagorean Theorem set up; h²+ x² =l². That?s the ladder distance from the base, the height from the base. Since this is not changing, we can put the 20 in right now. So we have 20² there over in this side. H² and r² are changing so we can?t substitute it for either one yet. H² plus x². Now we?re ready to do the derivative with respect to time; d over dt on both sides. D over dt over here, d over dt over here.
Derivative of this, we have derivative of h² that?s 2h. But since h is not t we have to have the dh over dt Chain rule factor times dh over dt. Derivative of x² is 2x, but since x is not t, we have to have the dx over dt Chain Rule factor. On the other side, the derivative of 20² it?s a constant, that?s 0.
Here is just where we left off. We had 2h, dh over dt, plus 2x, dx over dt. We?ve done our derivative now, so we?re free to substitute in things that change. The things that change, that?s the 16 for x. Put the 16 in and I can save us a little bit of trouble. Notice that it?s 2 times that, 2 times that. I can divide both sides by 2 to make those factors go away, to save us a little trouble.
The height, we actually do need the height of the ladder. That we can finally do though. It?s not on the chart, but it turned out we needed it. We knew this was always 20. At the instant we?re doing this problem, x is 16. Well, we have to use the Pythagorean Theorem again, but this time with a couple of numbers. At the instant there?s a 16 that squared plus 16² equals 20². The height works out to be 12 just at that instant. Put that number in; 12 for h.
Dh over dt that?s -3 we?re going to put that in. You get a 3 for that and dx over dt that is our unknown. Dx over dt equals 0 we?re almost done. So the 2 is divided away. H gets replaced with 12, dh over dt with -3, x with 16. If we do a little bit of Algebra on this, we can solve for dx over dt. Our final result is dx over dt equals 9/4. It?s going to need to put the units in as well. Since this is feet, this should be feet per second.
The last kind of problem type you?ll see will be trigonometric problems. They can involve sines or cosines, or tangents. You really don?t know until you start the set up. So I?m going to base this on the problem we were just doing, the sliding ladder. But this time what we?re going to look for is how fast the angel its base makes with the ground. That?s this one right here, theta.
So as you go along, the ladder slides down and down. That angle gets smaller and smaller. So the rate at which that angle changes, is going to vary depending on this rate right. So now we?ve got to identify the things we need.
We have to have theta in here, because that?s for the rate in which the angle changes. We?re going to have to have h in there, because that?s the rate we?re relating it to. So now we?ve got to decide are we?re going to use sine or cosine or tangent? What you want to do, is if there?s something that doesn?t change, you want to build it into the trig equation. X is changing don?t use it.
Use the 20. 20 doesn?t change. If you use this you?d have too many variables, but if you use this one, the sine of theta is h over 20. Sine theta h over 20 and now we?ve got what we need. We have a base equation that relates the two variables we?re supposed to relate just those two. So we won?t even have to do any substitutions this time. To go on from here of course you d over dt on both sides. There?s nothing I can substitute in yet, because these variables change.
So d over dt to both of these right there. I?m going to have to write this behind. It?s technically not correct, but I want to write it to let you know that I?m doing the derivative of both sides. So if you do the derivative of both sides on this, this one will give you the cosine of theta times d theta over dt. This side here let?s see, derivative of h is 1, so the derivative of h over 20 is 1/120. 1/120 dh over dt.
We?ve done the derivative and now we are free to substitute in anything that changes. I?m going to leave the problem here and you?ll find it in the bonus materials along with all of the steps for you to try to finish yourself. You are going to need to find the cosine and theta, but nothing could be easier. Because you know this is 20 and you?re doing it at the instant that h is 16 so you can work out the cosine just by using Pythagorean Theorem to find x, nothing simpler.
In this episode we talked about related rates. Relating how one quantity is changing in respect to another. We started by talking about the different problem types with examples of each. Remember, don?t substitute anything that?s changing until after you do that derivative.
So there?s one other thing I have to caution you. You?ll often find related rate problems on the AP test. You?ll also see some other problems that talk about rates, yet they?re not Related Rate problems. You?ll find a lot of those in the AP Trick episodes.