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Studying for a test? Prepare with these 5 lessons on Applications of derivatives.
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Video transcript
It's late at night, and some type of nocturnal predatory bird, maybe this is an owl, is diving for its dinner. So this right over here is a mouse. And it's diving straight down near a street light. And let's get some information about what's going on. So the street light right over here is 20 feet high. So this is a 20 foot high street lamp. And right at this moment, the-- and I haven't drawn it completely to scale-- the owl is 15 feet above the mouse. So this distance right over here is 15 feet. And the mouse itself is 10 feet from the base of the lamp. Let me draw that. So the mouse is 10 feet from the base of the lamp. And we also know-- we have our little radar gun out-- we know that this owl is diving straight down. And right now, it is going 20 feet per second. So right now, this is going down at 20 feet per second. Now what we're curious about is we have the light over here. Light is coming from the street lamp in every direction. And it creates a shadow of the owl. So right now the shadow is out here. And as the owl goes further and further down, the shadow's going to move to the left like that. And so, given everything that we've set up right over here, the question is, at what rate is the shadow moving? So let's think about what we know and what we don't know. And to do that, let's set up some variables. So let me draw the same thing a little bit more geometrically. So let's say that this right over here is the street light. That is 20 feet tall. And then this right over here is the height of the owl right at this moment, so this is 15 feet. The distance between the base of the lamp and where the owl is going, where that mouse is right now. This is 10 feet. And if I were to think about where the shadow is, well, the light's from right over here. And so the owl blocks the light right over there, so the shadow is going to be right over there. So if you just draw a straight line from the source of light through the owl and you just keep going and you hit the ground, you are going to figure out where the shadow is. So the shadow is going to be right over here. It's going to be right over there. And we need to figure out how quickly is that moving. And it's going to be moving in the leftward direction. So let's set up some variables over here. So let's say-- so what's changing? Well, we know that the height of the owl is changing. So let's call that y. Right at this moment, it's equal to 15, but it is actually changing. And let's call the distance between the shadow and the mouse x. Now given this set up, can we come up with a relationship between x and y? And then using that relationship, what we're really trying to come up with is, what is the rate at which x is changing with respect to time? We know what y is right at this moment. We know what dy/dt is right at this moment. Can we come up the relationship between x and y and maybe take the derivative with respect to t so we can figure out what dx/dt is at a given moment in time? Well, both of these triangles-- and when I say both of these triangles, let me be clear what I'm talking about. This triangle right over here, the smaller triangle in green, is a similar triangle to the larger triangle. This is a similar triangle to this larger triangle that I am tracing in blue. It's similar to this larger one. How do I know that? Well, they both have a right angle, right over here. They both share this angle. So all three-- they have two angles in common-- then all three angles must be in common. So they are similar triangles, which means the ratio between corresponding sides must be the same. So we know that the ratio of x to y must be the ratio of this entire base, which is x plus 10, to the height of the larger triangle, to 20. And right there, we have a relationship between x and y. And if we take the derivative of both sides with respect to t, we're probably doing pretty well. Now before taking the derivative with respect to t-- I could do it right over here-- just to simplify things a little bit, let me just cross multiply. So let me multiply both sides of this equation by 20 and y, just so that I don't have as many things in the denominator. So on the left hand side it simplifies to 20x. I don't want to write over it, well, I'll just write-- 20x. And on the left hand side is 20 x. And then on the right hand side-- let's see this cancels with that-- we have xy plus 10y. And now let me take the derivative of both sides with respect to time. So the derivative of 20 times something with respect to time is going to be the derivative of 20 times something with respect to the something, which is just 20. That's the derivative of 20x with respect to x, times dx or the derivative of x with respect to t, is equal to. Now over here we're going have to break out a little bit of the product rule. So first we want to figure out the derivative of x with respect to time. So the derivative of the first thing times the second thing, times y. Plus just the first thing times the derivative of the second thing. So derivative of y with respect to t is just dy/dt. And then finally, right over here. The derivative of 10y with respect to t is the derivative of 10y with respect to y, which is just 10, times the derivative of y with respect to t, which is dy/dt. And there you have it. You have your relationship between dx/dt, dy/dt, and x and y. So let's just make sure we have everything. This is what we're trying to solve for, dx/dt. And let's see here, we have another dx/dt here. We're going to try to solve for that. We know what y is. y is equal to 15. We know what dy/dt is, dy/dt. If we make the convention since y is decreasing, we can say it's -20. So we know what this is. And so if we just know what, so we know what this is. So if we just know what x is we can solve for dx/dt. So what is x right at this moment? Well, we can use this first equation-- we could actually use this one up here, but this one is simplified a little bit-- to actually solve for x. So let's do that, and then we'll substitute back into this thing where we've taken the derivative. So we get 20 times x is equal to x times y. y is 15. And just remember, I could have used this equation, but this is just one step further. We've already crossed multiplied. So it's x times y. y is 15. So it's x times 15, plus 10 times y. Plus 10 times 15. Did I do that right? 20x is equal to x times 15 plus 10 times 15. So let's see if you subtract. So just this is 20x is equal to 15x, plus 150. Subtract 15x from both sides, you get 5x is equal to 30-- sorry, 5x is equal to 150, my brain is getting ahead. 5x is equal to 150. Divide both sides by 5. You get x is equal to 30 feet. x is equal to 30 feet right at this moment. So this distance, just going back to our original diagram. This distance right over here is 30 feet. So let's substitute all the values we know back into this equation to actually solve for dx/dt. So we have-- so let me do it right over here-- we have 20 times dx/dt. I'll do that in orange, we'll solve for that. Actually no, I already used orange. So let's say dx/dt-- I'll use this pink-- 20 times dx/dt is equal to dx/dt times y. y right now is 15 feet. So times 15, times-- I didn't want to do that color. Times 15 plus x, we already know that x is 30. Plus 30, times dy/dt. What is dy/dt? dy/dt, we could say is -20 feet per second. y is decreasing, the bird is the bird is diving down to get its dinner. So times negative-- well, times 20 feet per second. So that's that right over there. Plus 10 times dy/dt. So plus 10 times -20 feet per second. And now we just solve for dx/dt. So let's see, what do we have? We have 20 times-- let's see, let me subtract 15 dx/dt from both sides of this equation-- and we get 5dx/dt's. I just subtracted this from both sides of the equation. This is 15dx/dt this is 20. So that we have 5dx/dt's, is equal to this is, this part right over here is -600. And this part right over here is -200. So it's equal to -800 feet per second-- or -800, and actually this will actually be in feet per second. And so dx/dt is equal to, dividing both sides by 5, 5 times 16 is 80. So this is -160 feet per second. And we're done. And we see the shadow is moving very, very, very, very fast to the left. x is decreasing, and we see that. That's why we have this negative sign here. The value of x is decreasing. It is moving to the left at quite a nice speed here.