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Solve the problems in related rates.The shadow of a $24-\mathrm{m}$ high building is increasing at the rate of $18 \mathrm{cm} / \mathrm{min}$ when the shadow is $18 \mathrm{m}$ long. How fast is the distance from the top of the building to the end of the shadow increasing?

Calculus 1 / AB

Chapter 24

Applications of the Derivative

Section 4

Related Rates

Derivatives

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either. So in this problem, we have, uh the problems actually illustrated for us. So let's copy that picture down. First of all, we have a five meter lamp. Oh, so this is always five. You have a man whose height is 1.8. So this is always 1.8, and the lamppost is casting a shadow on the ground here as he's walking. So as the book, the book is calling this distance X with this red distance here because calling X and then this blue part, which is the shadow, This is the part we care about. We're calling that. Why? So we have a right triangle. Um good. And the other thing we know is that he's walking at a speed of 1.2 meters per second. Mr, In other words, of this red side here is increasing at 1.2 meters per second, so we can actually translate that into math. If he is walking that quickly, that means his distance away from the lamppost X is changing at that rate. So what they've actually given us is DX DT. And it's positive 1.2 because, uh, because X is growing at 1.2 meters per second. So they have told us that. Okay. What we want to get is how fast the shadows growing. We want d y t t the rate of growth of side. Why? Okay, what's our equation? This is trickier than a lot of them because it's not a Pythagorean theorem. Ah, problem. We don't know much about the high pot news at all here. So what's the one thing we do know? Well, the one thing we know is that these triangles are similar triangles. The small triangle here a lot lining green into the small triangle and the large triangle, right? They're similar triangles. So what do you know about similar triangles? Well, don't think what we know about the side lengths, but we know that five is Thio 1.8. As you know, you're you have to be a little careful. What is the overall bottom? While the bottom of the big triangle is X plus y? And the bottom of the small triangle is just why? So make sure you can see this. This is just similar Triangles have proportionate sides. That's all we used here. Okay, so now at this point. Is there anything that we can do to make the math easier before we take derivatives? Um, I would say so. So let's see. Why don't we actually multiply both sides by 1.8 and then multiply both sides by why? Or if you prefer to stay cross products So right five times y equals 1.8 x close 1.8. Why, okay. Or, in other words, you can do even better. And if we combine our I like terms five minus 1.8 gives us 3.2. Why equals 1.8 x? Okay, so we have a nice equation relating X and y. Now, when we take the derivative so we'll take both sides derivative with respect to t and see what we get. So you have the derivative of 3.2. Why, it's just 3.2. Then we need to multiply radio I d. T on the right. Likewise, the derivative of 1.8 x is 1.8, but we need two more fly by DX DT. And since we want to solve for D Y e t, all we have to do is divide by 3.2 and we'll get d Y t t by itself. 1.8 over three points. You d x t t Remember, we know that's 1.2. So in the end, we have it. We just have to calculate this out. So we need to d'oh! 1.8 times one point to you and then divide that answer by 3.2, we should get 0.675 Now, let's just make sure we have the right units. We're measuring distance and this problem distances of meters and a unit of time for our rate in this problem. This seconds. So we get points 675 meters per second. Kind of a tricky problem. Hopefully, this helped

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