 This lesson is on related rate problems. When you do a related rate problem, you are computing the rate of change of one quantity in terms of the rate of another. There are certain strategies that I have already written up for you, but let's go over them once again. The first one is to read the problem carefully, not once, not twice, but many, many times until you begin to understand what is going on. And as you begin to read it over and over, you should draw a diagram and write down what you know and what you have to find and also when you have to find it. And we will be doing lots of problems on this and we will follow this pattern every single time. The next thing you need to do is take the derivative with respect to the independent variable, which is usually t. And then you substitute in the given values and solve. For every problem you are given, you really should follow these strategies because they will help you solve the problem correctly. Our first problem is the soap bubble or balloon problem. And it reads, air is expanding a soap bubble so this volume is increasing at the rate of 10 centimeters cubed per second. How fast is the radius changing when it is 10 centimeters? And that means the radius is 10 centimeters in this particular problem. Well, what do we know? We know that we need the volume of a sphere and we will get that formula in a minute. And what else do we know? Well, we know the change in volume, which we write as dv dt, and that's equal to 10 centimeters cubed per second. The volume of a sphere is volume equals four thirds high r cubed. What do we need to find out? We need to find out the change in rate over the change in time. So that's dr dt when r is equal to 10. So we know that the volume formula is four thirds pi r cubed. We take this derivative and dv dt, remember we're taking with respect to t, becomes four thirds pi times 3 r squared dr dt. Again, the derivative is taken with respect to t. Cleaning this up, we get 4 pi r squared dr dt. Now dr dt becomes dv dt over 4 pi r squared. We know dv dt is 10. We know that r is 10 and we're filling those into our formula. And eventually we can reduce this and get 1 over 40 pi. So it's approximately equal to 0.008 centimeters per second, which is pretty slow. So this is how you would solve a balloon type problem. The next problem is a cylindrical tank problem. And it reads, how fast is water level dropping when the radius is 30 centimeters in a cylindrical tank when it is being drained at the rate of 3,000 centimeters cubed per second? What do we know? Well, we know the volume of a cylinder and we need that. So the volume of a cylinder is volume equals pi r squared h. And we also know that dv dt, the change in volume of the change in time, is negative because it's dropping negative 3,000 centimeters cubed per second. What do we need to find? Well, again we need to find the change in height over the change in time and again we're given when the radius is 30 centimeters. Of course, you know with a cylinder the radius will always remain the same. So let's go for our solution. Well, let's draw a diagram first. We know that that radius is 30. So when we develop our equation, volume equals pi r squared h, we are filling in for that radius because it is constant throughout the problem. So that is equal to 30 squared pi h or 900 pi h. When we take the derivative dv dt, we have 900 pi. When we take the derivative on h, it becomes dh dt. And we know what dv dt is. It's negative 3,000. So we divide that by the 900 pi and we get dh dt. So dh dt is equal to 30 over 9 or 10 over 3 pi centimeters per second. These first two problems are relatively simple because we've just taken the derivative with the formulas that we have had. Let's go on to something that might be a little bit more difficult. And this is a right triangle type of problem. A boat is being pulled to a dock at the rate of 2 feet per second. The pulley is 8 feet from the water and the rope is tied to the boat at 1 foot above the water. How fast is the boat approaching the dock when it is 20 feet away? We have a little animation on this, which we can show you. The solution to this right triangle type of problem is to first draw a right triangle. So let's draw a right triangle. We know certain things. We know that the distance from the rope to the top of the pier, where the pulley is pulling it in, is 7 feet. We don't know the distance from the boat to the pier. And of course, we don't know how far the rope is out. So we'll call the first distance x and the second distance z. So now we can use Pythagorean theorem on this because it is a simple right triangle. So we'll have x squared plus 7 squared is equal to z squared. Take the derivative on this and we get 2x dx dt. Again, everything taken in a relationship to t. 7 is a constant. And when we try to take that derivative, we know it's 0, so let's not even write it in. And then we'll have equal to 2z dz dt. We can cross out the 2s and solve for dx dt and get z dz dt over x. Now we can fill in for the numbers we know. We know that in this particular problem, x is going to be 20. We know that dz dt is equal to 2. Right now we do not know z, but we can find it by just putting numbers into our right triangle. We know that x is equal to 20 when we are trying to find our answer for dx dt. So now we have a right triangle with two sides known and one side not known. And using Pythagorean theorem, we find out that z is equal to the square root of 449. So we can fill that in the square root of 449. So now we get an answer of, this can be reduced, the square root of 449 over 10, and that's feet per second, which we can change to its equivalent decimal form in three decimals, 2.119 feet per second. That is a right triangle problem. Let's go on and do another type of right triangle problem. This one reads, two trucks leave a depot at the same time. Truck A travels east at 40 miles per hour, and truck B travels north, so they're traveling away from each other at 30 miles per hour. How fast is the distance between the trucks changing six minutes later when A is four miles from the depot and B is three miles from the depot? Let's put down what we know in this right triangle type of problem. We know that we have a right triangle, and it looks like the Pythagorean theorem is going to be involved in it. We also know that DADT is equal to 40 miles, and it's going in the eastern direction, and DBDT is 30 miles, and it's going north, and we need to find out DSDT, which is the change in the distance between them when A is four and B is three. So let's do our triangle. We have A going east, and we have B going north. So we have that DADT equal to 40. We have the DBDT is equal to 30, and we want to find DSDT when A is equal to four, and B is equal to three. Again, Pythagorean theorem, A squared plus B squared is equal to S squared. Take the derivative. Take the derivative on all three components, so we have 2A DADT plus 2B DBDT equals 2S DSDT. Again, canceling out those twos, we have this formula in which DSDT is standing alone. In this one, I'm just going to fill in the numbers for A, B, and S, DA, DT, DBDT, and then solve for DSDT. In the former problems, I actually worked out the DSDT before I filled in the numbers, but sometimes it's just easier to fill in the numbers first. So A is four, and we know DADT is 40. We know that B is three, and DBDT is 30, and we just need to determine what S is, because what we are looking for is DSDT. So if we fill in A is four, and B is three, we realize we have a three, four, five triangle, so that makes S equal to five. Solving for DSDT, we get 50 miles per hour. And if you want, you can do the math on your own and check this out. So this is the second type of right triangle where we have movement away or even eventually towards each other and some sort of triangle is being formed. Let's go on. This time we're going to deal with a conical tank. So this is a cone type problem, and the problem reads, a water tank in the shape of right circular cone has a radius of three meters and a height of six meters. If water is being pumped into the tank at the rate of four meters cubed per minute, find the rate at which the water level is rising when the water is four meters deep. Well, what do we know? We know the volume of a cone, and we all know that the volume of a cone is one-third pi R squared H. We know the relationship between the base and height of the cone, because we had said that the base of the cone is three meters, this is three meters, and this is six meters. So the relationship between the radius in the height is three, is to six. So the R to H is three to six. And we also know that the change in volume of the change in time is four meters cubed per minute. What do we need to find out? Well, we need to find out the change in height of the change in time when the height is four meters. In our solution, let's go back and rewrite that R to H is equal to three over six, which equals one-half. And the volume we know to be one-third pi R squared H. What we are looking for is dH dt. So our formula should be in H's, but we have this R here. So somehow we have to replace that R with something else. And it comes from the R over H formula, that relationship. So we can say R over H is equal to one-half, or R is equal to one-half H. And we can substitute that in for R in our volume formula. So our volume formula now becomes one-third pi one-half H quantity squared times H. And if we simplify this a little bit, we get one over twelve pi H cubed. Now we can take the derivative very readily. So we'll take dV dt is equal to three-twelfths of one-fourth pi H squared dH dt. Now we are looking for dH dt, that right there. So we need to substitute for the H and for the dV. And we find that dV is four. So we write that in and then copy down to one-fourth pi. And the H is four. So that's four squared. And then we have dH dt. Solving for dH dt, which isn't very difficult to do in this one, because I put a lot of fours in it, we get one over pi. And that's meters per minute. And if we want to evaluate that, we get approximately 0.318 meters per minute. Let's go to the street light problem. This time we have to deal with similar triangles. The problem reads a 6.5 foot man, six and a half foot man walks away from an 18 foot light pole at the rate of seven feet per second. We want to find two things. The first one is at what rate is the length of the shadow moving? And the second one is at what rate is the tip of the shadow moving? The first one is at what rate is the length of the shadow moving? And we are going to make the triangle for that one. If we create our triangle, it will look like this. We have this pole that's 18 feet tall with some light flashing on it. We have a man who is 6.5 feet tall, and he's walking away from that light pole. We also know that he's walking at the rate of seven feet per second. So if we call this x, then we know that dx dt is equal to seven feet per second. We are trying to find how fast this piece is changing, so we'll call that y. So we need to find dy dt. In looking at this triangle, we see we have a large triangle, and we also have a smaller triangle. So if we set up a ratio between the large triangle and the small triangle, we can get all our x's and y's in there and then take our derivative. So we can set up 18 over x plus y is equal to 6.5 over y. And now since we want dy dt, let's solve this for y. So we have 18y equals 6.5x plus 6.5y, or 11.5y equals 6.5x, and then y is equal to 6.5x over 11.5. Now we can take the derivative. So we have dy dt is equal to 6.5 over 11.5 dx dt. We know dx dt is seven, so we can put a seven in for that dx dt, and we find out the answer is 3.957 feet per second. This problem actually worked with the length of the shadow. And again, set up the right triangle, set up the ratio, and then solve for what you need. Now the next part of the problem is the tip of the shadow. The tip of the shadow is a little bit different when we set up our triangle. Again, we'll have the man at 6.5 feet, the pole at 18 feet, a little light on the pole. This is x and this is y. The tip of the shadow is actually the sum of x plus y, and that's important to know when you're doing these types of problems. That's your tip. We still know that the x dt, that hasn't changed, of course, is seven feet per second. What do we need to know? Well, we need to know the derivative of x plus y with respect to t. Now this problem becomes quite simple because of the fact that really the derivative of x plus y is the derivative of x plus the derivative of y. So we only have to add what we've already calculated together. We know dx dt is seven, and dy dt is equal to 3.957, and we add those together and we get 10.957 feet per second. So once you know the length of the shadow, so if you're given this type of problem, you do the length of the shadow first, you can easily find the tip of the shadow. And again, going back and remembering that in this type of problem, you look for ratios. So it doesn't have to be a streetlight problem or a length of a shadow problem to work with ratios. You should always consider that as you're working with different problems. Well, let's go to our final problem, which is an angle problem. This one's stated as Mario is standing 25 feet from a launch pad of a hot air balloon. If the balloon is rising at the rate of 20 feet per minute, how will the angle between Mario's eye and the balloon be changing after five minutes? This one's a little bit different because we have to use trig. So what do we know? We know that Mario is 25 feet from the launch pad. So we know x is equal to 25. We don't know y. We don't know z. But we need to find this angle here, which is theta. So we need to find d theta dt when t equals 5. We also know dy dt is equal to 20 feet per minute. So this is what we know that x is equal to 25 all the time. We also know the change in y over the change in t is 20. We're trying to find theta dt at a certain time when t is equal to 5. Let's go on to the solution. Well, the first thing we were given was when t is equal to 5. So let's translate that to something we know. Well, when t is equal to 5 and the balloon is going up at 20 feet per minute, then in five minutes y will be at 100 feet. So with our triangle, we had a 25 here. We had a y there. We were looking for theta. We need some sort of trig function that deals with opposite and adjacent. And of course, that's tangent. So we say tangent theta is equal to y over 25. We can take the derivative on this now. So we get secant squared theta, d theta dt, don't forget that, is equal to 1 over 25 dy dt. Now this time again, I'm going to solve for d theta dt and show you why. d theta dt is equal to, this would be 1 over secant squared on the right-hand side, but 1 over secant squared is cosine squared. So let's just write it as cosine squared theta. So we have cosine squared theta over 25 dy dt. Well, let's fill in what we know. We know dy dt is equal to 20, and we have our 25 sitting here. Now we need to find cosine theta. Well, at the point in question, we know, again, x is 25, y is 100, and we just have to find out what the hypotenuse is because cosine of theta is adjacent over hypotenuse. So the hypotenuse is the value 25 square root of 17. Why don't you calculate 100 squared plus 25 squared under the square root? So this becomes 25 divided by 25 square root of 17 quantity squared, and that evaluates out to 4 over 85, and since this is an angle, we do this in radians per minute, and that becomes in decimal form 0.047 radians per minute. This concludes your lesson on related rate problems. Good luck with doing them.