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In the shadow problem, how was "12 fect away from the lamp post " used t0 solve the 'problem? A) Extrancous info B) Set initial distance C) Show ra...

Question

In the shadow problem, how was "12 fect away from the lamp post " used t0 solve the 'problem? A) Extrancous info B) Set initial distance C) Show rate of change D) Show Similar Triangles

In the shadow problem, how was "12 fect away from the lamp post " used t0 solve the 'problem? A) Extrancous info B) Set initial distance C) Show rate of change D) Show Similar Triangles



Answers

Sketch the situation if necessary and used related rates to solve for the quantities.

Using the previous problem, what is the rate at which the tip of the shadow moves away from the person when the person is 10 ft from the pole?

We have a pipe lit tall person walking away from a wall at a rate of two people second and a spotlight is located on the ground 40 feet away. We want to determine at what rate is this person shadow changing? All right, so let's go ahead and see how to do. So we have a triangle here inside of another triangle. We look at it, and because of that we can use the fact that triangles like this are congruent with each other. So what I'm going to do is I'm going to call this distance between the person, the spotlight, X. And since I know the distance from him to the spotlight is 40 that would mean the rest of this over here should be 40 minus X. It was important that we use 49 sex as opposed to 10 because remember, the person is closer to it over here. And then he's moving over until he hits the 10 foot mark. So we need to make sure that we're using that. So the other thing is, we still need to figure out the height out here. Well, we don't necessarily know how high his shadow is. But we could just go ahead and call it. Why for now And what we want to do is find what d y by DT iss. So let's go ahead and first set this up using our congruence. So the first triangle are smaller, triangle will be This will be five X and then our larger triangle will be why and then 40 Because I would be the total distance. So you would end up with why over 40 is equal to five over X and then we can multiply each side by 40. And in doing that, we get why is equal to 200 over X. And now let's actually move this up top here because if we take the derivative of this with respect to time, that's going to give us d y by D. T is equal to 200 over. So remember this is really X to the negative first power since its one over X. So we would need to use Powerball to take this derivative. That would be negative one and then we subtract one so it would be excellent negative second power. So we should have 200 over X square and then we're going to multiply this by D x d t. Because we are implicitly differentiate right now, the only thing we need to figure out from this is what is x of what is D exp i t t So, in the case of what he's 10 feet away, that means this distance here should be 30. So we have exes, he go to 30 was right that off on the side over here and then the rate at which excess changing well, the persons moving away from the pool of two feet per second. So you might want to say that the X by d. T is equal to to eat per second. But the more the person walks this way, the shorter X becomes so. This should be negative two feet per second so we can go ahead and just plug everything. Didn't know to find what the rate of change of our shadow is. So we would have negative 200 over, so it would be 30 squared, which is 900 then times negative too E for a second. So our negatives here cancels out in 200 over 900 should be two nights, so this year should simplify down too, or night beat her second, so the rate of change all his shadow would be four nights feet per second.

In this given question ticket under Deasy Quarto five b as if pretending at the white for second now by using up a party of similar climates Taken right s I B Porto six my weapon. The work of my life. Not so. I have been like this. We get, I guess, is the photo s. That's groovy. Yes, Don't be s a week or two to my really now opened up a little bit. Report a five picture. We get people to buy three until five p in the rectory.

Were given a problem where a six with person is walking away. Herman eaten. Put Lamport. The person is currently 12 feet away from the lamppost and moving away out of rid of two feet per second. We were asked to find a redoubt which the length of the shadow is changing. We were given a hint to help solve the problem. Andi, that Hank is basically saying The proportions of the triangle formed by the lamp post has equal the proportion of the smaller triangle formed by the individual under shadow. So since I'm interested in finding go the rate at which the shadow is changing, I will first simplify this equation so that it is sold for us. So we can cross, multiply and are that will give Times X Plus s pastor equal to 18 s solving for s. We find that uh 12 s is equal to it's. And if we take the derivative of this, then what we end up with is that 12 s prying is equal to six. I'm specs bright. Oh, that s prime is equal to 1/2 expert. So the redoubt was just Shadow is changing. Can be flown by ticking How, um, to think per second, which is one or the length of the shadow is changing at a rate of one put for a second.

If this problem. We're looking at a ball, dropping in front of a light and looking What happens to the ball shadow anytime you have a related rates problem, if you can draw a picture, that's a really good place to start. So I'm going to draw a very quick sketch of what we have. Okay, we have a light on. We're told that that is 50 ft tall and that number is not changing, so I could put that on my picture here. Next we have a ball that's at the same height. So right about here, I've got a ball and it is 30 ft away, and that number is not changing either. And at a given time, we're going to drop the ball so you can imagine the light will strike the ball and then there'll be a shadow as the ball drops. That shadow is going to move, so I'm gonna call that piece X because that's a variable distance. And my question that I have here is how fast is the shadow of the ball moving on the ground? So I am looking for DX DT. That's what I want to know. How fast is that shadow's length changing. Okay, What else do I know that I could put on my picture? Well, I know that I'm going to be looking at a snapshot when t equals one half of a second. I'm not putting that on my picture, because time is changing. That's just my snapshot moment. We also know that I am a certain height above the ground at any given time. I'm just going to call that h, because again, that is a variable distance. Okay? And I'm gonna put one other thing here. We're told that the ball falls at 16. T squared feet in two seconds, so I'm just gonna put that here. We know that as well. Okay, because this is a related rate problem. We have our picture now. We need our actual formula that relates the rates to each other. So I have nested triangles here Anytime you have these overlapping triangles, a small triangle nested inside a large one. Comparing the sides of the triangle is usually a good place to start. So let's look at the big triangle. The height is 50 and the length is 30 plus x. The small nested triangle has a height of H and a length of X. Okay, so let's see if we can rewrite this a little bit. If I cross multiply, I get 50 X equals 30 h plus h X. Now I'm trying to solve for X. So I'm going to combine all my exes together. I have 50 x minus h X equals 30 h factor my ex out and solve for X. That gives me 30 h over 50 minus h now. Yeah, I don't know exactly what H is, but I can put it in terms of tea because I know that this distance that I have dropped is going to be equal to 16. T squared because I know my ball falls 16 t square feet every t seconds. So that's the distance I've fallen. Which means that my H has to be 50 by full distance, minus the distance. It's dropped so I could make that substitution for H. I could say the X is 30 times H, which is 50 minus 16 t squared over 50 minus eight, which again is 50 minus 16 t squared, bro. I got a little more room. Let's try to simplify. This is much as we can. Before we take the derivative, my numerator is going to be 1500 minus 480. T squared the denominator the fifties were going to cancel on. I'm just left with 16 t squared, simplifying that out as much as I possibly can. I am gonna end up with 375 over four times t to the negative, too. And I put that up there. Just when I go to take the derivative, I find it easier if I have a negative exponents versus a denominator, if possible. And then for the second term 480 divided by 16, that's going to be 30 and the T Square's canceled. So now I have, ah, formula that relates X and T to each other. Let's take the derivative DX DT. Well, I'm gonna multiply 375 over four times negative, too, and I subtract one from the exponents. I have t to the negative third derivative of 30 0, so I could just drop that off and let's evaluate. This is my derivative and I want t to be one half second. That's I was told. That's what I'm going to be looking at. I want one half the negative third power, and when I evaluate all of this together, I end up with a negative 1500. So that is the rate at which my shadow is changing its negative because it's getting smaller, the closer the ball gets to the ground, the smaller that shadow is going to be. The negative tells me that that number is shrinking at this point. So for this ball dropping, the change of the shadow with respect to time is negative. 1500.


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