Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Related rates. The easiest way to do related rates is to draw a picture. Name the variables and the constants. Write down any numerical information. Write down what you're asked to find, write an equation that relates the very the variables, differentiate in terms of time, and then evaluate. So we're going to look at some examples. The first example is going to say the surface area. Suppose that the radius R and surface area S equal 4 Pi. R-squared of a sphere are differentiable functions of T. Write an equation that relates the change of the surface area in time to the change of radius in time. So if we know that we have's equals 4 Pi R-squared, we're going to take the derivative in terms of time of the surface area equaling the derivative in terms of time of the four Pi R-squared where the radius is going to be changing. So DSDT is going to equal the derivative of four Pi R-squared using our power rule bring down the two. So we get 8 * π * r the R goes to 1 less power DRDT. What that really says is that the surface area changes at some rate as the radius changes. If you think about blowing up a balloon, when you have a small balloon, you don't have a very big surface area. But as the radius gets bigger, the surface area gets bigger. If we wanted to do the same thing, looking at the volume, volume is 4 thirds π R cubed. So taking the derivative of the volume in terms of time and the derivative of the Four Thirds π R cubed in terms of the radius. So the left side would be DVDT, the right side would bring down the three and take it to the one less power in that exponent. So we'd get 4 Pi R-squared DRDT. So the surface area changing in terms of the radius, surface area in terms of time changing as radius changes in terms of time, and the volume changing in terms of time as the radius changes in terms of time. So let's look at a different example. We're given that the volume R and height H of a right circular cone are related to the cone's volume by the equation V equal 1/3 Pi r ^2 H. So we draw our picture right cone and we know that we're going to take the derivative in terms of time. The first part we're going to say let's let the R be constant. So if we take the derivative in terms of V, we get DVDT. If R is constant, the only variable that's changing would be the H. So we'd have 1/3 Pi R-squared because that R is constant and the H is changing, so it'd be DHDT. The derivative of H is one, but we need to say what variable is changing in terms of time. So the derivative of H in terms of derivative derivative of HDT. So Part B says let H now be a constant going back to the original equation. So if H is a constant, the only variable that's changing besides the volume would be the radius. So the derivative of that R-squared is going to be two RDRDT. So when we do this, we bring down the two from the exponent, raise it to the one less power. So we'd get 2 thirds π RHDRDT. The last part says both of them are changing. So if they're both changing, we're going to have to use the product rule. If we use the product rule, we're going to take the derivative in terms of time of the volume equaling the derivative in terms of time of the whole thing. Well, the derivative of the H was DHDT, so it's going to be the derivative of H times the whole rest of it by the product rule, plus the derivative of the other variable, in this case the R-squared times the entire rest. So part C is really just parts A&B added together. So if you thought about one variable being a constant, the other variable being a constant, adding them together, you really have the product rule. Looking at some more examples, if we look at heating a plate, when a circular plate of metal is heated in an oven, its radiant radius increases at the rate of .01cm per minute. At what rate is the plate area increasing when the radius is 50 centimeters? So the first thing we have to do is we have to realize that we're talking about the area of a plate. So the area of a circle area equal π R-squared. So we're going to figure out how things are changing in relationship to time, and we're going to take the derivative of each side. So the derivative in terms the derivative in terms of time of A is going to equal the derivative in terms of time of π. R-squared or DADT equal 2π RDRDT. Now we were told that DRDT is .01 and we want to know what's happening when the radius is 50. So literally we're going to stick in 50 for R and .01 for DRDT. When we compute this, we can say that the area is changing by π. If we look at another example, this one's going to be two commercial airplanes are flying at an altitude of 40,000 feet along straight line courses that intersect at right angles. Plane A is approaching the intersection point at a speed of 442 knots. Speed. Speed means derivatives. Speed is the absolute value of velocity. Speed in this case is going to be. The velocity is going to be negative because we're getting closer and closer and closer. Or the C, the distance between our two planes is getting smaller and smaller and smaller. So our DADT is going to be -442 plane B is approaching the intersection at 481 knots. So once again it's going to be a negative because they're getting closer together. So -481 knots. At what rate is the distance between the planes changing when A equal 5 nautical miles from the intersection point and B equal 12 nautical miles from the intersection point? Well, if we know A&B by Pythagorean theorem, we can find C Since A is five, we have 5 ^2 + 12 ^2 = C ^2 or C is 13. So we want to find our DCDT. We know the relationship by Pythagorean theorem of a ^2 + b ^2 = C ^2. If we take the derivative of all of these in terms of time, we'd get 2A, the derivative of A in terms of time plus 2BD BDT equaling 2, CDCDT two, our distance A was given us 5. We knew our DADT is -442 + 2. We knew our B is 12. Our DBDT was negative 481 equaling our two. We found our C given our A and our BA13, and we're trying to find DCDT. Now it's just a matter of plugging and chugging. So we eventually find DCDT is -614 knots. Still another example is going to involve draining a conical reservoir. Water is flowing at the rate of 50 meters cube per minute from a shallow concrete conical reservoir. If it's flowing, it's going out, the volume is changing and it's getting smaller. So we're going to have -50m cube per minute from a shallow concrete conical reservoir vertex down of a base radius 45 meters and a height of six. So the original conical had a radius of 45 meters and a height of 6 meters. Obviously the pictures not to scale. The thing we need to understand here is that we have similar triangles because the angles are going to be the same as it's decreasing the radius and the height will be changing. But we still have similar triangles. So the original radius to the original height is going to equal the new radius to the new height. 45 / 6 = r / H or we could think of R equaling 15 halves H We know the formula for the volume of a cone is 1/3 Pi r ^2 H, so 1/3 Pi. Instead of the R, I'm going to put in 15 halves H and square it times H So doing some math and simplifying, we're going to get the volume equaling 75 fourths Pi H cubed. Now we're told to find the change in the volume as the height changes. So the DVDT is going to equal 225 force Pi H ^2 DHDT. We're told in the original that the water was flowing out at 50m cube per minute. So -50 we want to figure out what happens when the water level falling is at a height of five. So we're going to put five in for our H and we want to figure out how fast the height is falling. So DHDT, doing some plugging and chugging here. We're going to get -8 / 225 Pi meters per minute. Well, that's kind of an interesting number, and it doesn't have a lot of meanings, so let's try to change that into something smaller, like centimeters per minute. If we want to do conversion, we know that there are 100 centimeters in 1m, so the meters top and bottom are going to cancel. And if we plug this into our calculator, we get approximately -1.13cm per minute. Then we're asked to find how fast is the radius of the water surface changing then. And the nice thing about this is we could go back to our relationship between our radius and our height. If we know R equal 15 halves H, then we can show that DRDT is just equaling 15 halves DHDT. And we found DHDT a moment ago as -1.13, so 15 halves times -1.13 is -8.49 centimeters per minute. And that concludes that problem. Thank you and have a wonderful day.