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Car X and Car Y are parked at the same place At exactly 3 OOpm, Car X starts moving east at a constant rate of 40 miles per hour. At exactly 5.00pm, Car Y starts mo...

Question

Car X and Car Y are parked at the same place At exactly 3 OOpm, Car X starts moving east at a constant rate of 40 miles per hour. At exactly 5.00pm, Car Y starts moving north at a constant rate of 40 miles per hour.At what rate Is the distance between Car X and Car increasing at exactly 6.OOpm? (Round to two decimal places)15 pointsType ndte eaorch

Car X and Car Y are parked at the same place At exactly 3 OOpm, Car X starts moving east at a constant rate of 40 miles per hour. At exactly 5.00pm, Car Y starts moving north at a constant rate of 40 miles per hour. At what rate Is the distance between Car X and Car increasing at exactly 6.OOpm? (Round to two decimal places) 15 points Type ndte eaorch



Answers

Two cars leave an intersection at the same time. One travels north at 40 miles per hour, and the other travels east at 30 miles per hour. If $z$ represents the distance between the cars, (a) describe how the rate at which the cars are separating is related to the rates at which they are traveling. (b) At what rate is the distance between the two cars changing at the end of two hours?

Okay, so we have a problem here. One car leaves a given point, travelling north at thirty miles per hour. So let's draw a picture. So here's her point. One car leaves going thirty miles per hour. Then another car leaves the same point in tramples the west at forty miles per hour. And then we want to know what rate is the distance between the two cars changing at the instant when the cars have traveled to ours. So if the second car is out here in the first cars out here, here's the distance between them. Okay, So call it s And this is why I'LL say this is Axel wise. The distance that the first cars traveled North X is the distance that the second cars traveled traveling west, as is the distance between them. And so the relationship between ex twine s is, of course, western and northern perpendicular. So we have the part that I'm going to tear my ex career plus life squared is s squared. And, of course, thirty and forty are the rates of change of, uh, X and y said d x d t It's party the variety his thirty and So if we differentiate this with respect to TV A to X t X T plus two I divided team Nichols to S t s t t. Ok, And so what we need, I guess we have the extra weight density divided tur looking for D s tt We need X and y and s well, it says after two hours. Okay, so after two hours, So after two hours? Well, ex so that this second car has been traveling forty miles per hour. So it's gone eighty miles. Why? Has been traveling thirty miles for two hours, said sixty miles. And then if we come up here in just two eighty squared plus sixty squared is s squared Well, you do eighty square plus sixty squared. You've been and gets, uh, see ten thousand insolence for ten thousand one hundred. So s is one hundred. This is one hundred miles. That's nice. So then we have two times eighty times forty plus two times sixty. That's thirty equals two times, one hundred times. Well, we're looking for OK, And so everything has to everything has two zeros, So in forty, that's gonna give us one hundred times dividing my hundred and see what a real actors were left with AIDS times for plus six times. Three Nichols D s Dean team. And what is that? Well, this is thirty two was eighteen. So, fifteen, it is? Yes, into that's course miles per hour. Okay, so that was a party and part of being okay. So now we suppose that the second car, the one traveling west left one hour later than the first car. Well, everything is the same. So we still have this relationship. So this relationship here relating the race have changed. The race of change is still the same. So we still have X dx DT and divided by two. Because every term as a factor to s to t t. But what's different now? Well, now, after a cease after two hours, at what rate is the distance between the two cars changing An incident of the second car has travelled one hour. Well, now, instead of ex being eighty now, X is just going to be forty. Because when why has traveled for two hours, ex waited an hour, sit in trouble, anything anywhere and then traveled forty miles. So now X is eighty and why is still so excess forty? And why's sixty? So that means s changes. So we have. So what is s we know? Forty squared plus sixty squared now is s squared. It's part B. So then what is s well s is going to be about seventy two point one Now if we put everything in, we have forty times eighty plus a sixty Sturdy eagle seventy two point one one. Yes, Titi. And you just saw his calculator yesterday. T is about forty seven point one five miles per hour.

Here. I have two cars. They leave the same spot at the same time. Okay, The first car is going to travel north at 30 MPH, and the second car is going to travel west at 40 MPH. What we want to know is how fast is the distance between them changing after two hours? Okay, so what we're looking at is this distance here. Well, we see we have a right triangle there, okay? And what we're trying to find if I label my triangle A B and C. Okay, I'm gonna make this side a This one's be. And this is See, I'm trying to find d c d t. All right, So they told me after two hours. Well, after two hours, B is going to be 60 and a is going to be 80. So let's go ahead a and set up her Pythagorean theorem and take the derivative with respect to time. I would have to a d a d t plus to be d b d t equals two c d CDT. All right, so I need to plug in things that I know. So over to here, 80 de a T t we know is 40. You have to times 60 times 30 equals to what is C? Okay, so I'm gonna have to do another Pythagorean theorem to figure out. See? Wouldn't come down here. And I must say, Well, 80 squared plus 60 squared equals C squared. Okay, so that's gonna give me 10,000 equals C squared or C equals 100 so I can go ahead and plug 100 in here, and I'm gonna solve for D C D t. So this side over here is gonna give me 10,000 is equal to 200 d c d t. So d C D t is going to equal 15 MPH. So the distance between the two cars is changing at the rate of 50 MPH. Okay, so let's change it up a little bit. Let's say, Well, that second car car a left. One hour later, we're going the same speeds, but is one hour later. So that changes my a So 40. Okay, be is still going to be 60 right? So we can go ahead and take this derivative that we already had. Okay, We can plug things in here, so we're gonna have to times 40 times 40 plus two times 60 times 30 equals two times. Well, what is? See, we have to figure that out again. This is gonna be D C D t. Right. So, another thuggery and zero, I'm gonna have 40 squared Los 60 squared equals C square. It's gonna give me 50. 200 equals C square and C is equal to 72 so I could put 72 in here. So this side of my equation left side evaluates to 6800 is equal to 1 44 Do you see DT and D C D t is equal Teoh 47.2 MPH.

We have one car traveling south at 60 MPH and another car travelling west at 25 MPH. And if we consider the car going south to be traveling in the Y direction and the car traveling west to be traveling in the extraction, then we can express this distance between them, using the Pythagorean theory as the square root of X Squared plus y squared, where X and Y are the current positions of the cars. And so if we consider the distance between them to be what say our thin our equals, the square root of X squared plus y squared and the derivative of our with respect to time will be equal to 1/2 the quantity X squared plus y squared to the negative 1/2 times two x dx DT plus two y do I didn t and we want to know what D R D t is two hours after the cars have left their location. So after two hours, the car traveling 25 miles an hour we'll have gone 50 miles, and the car traveling 60 miles an hour will have gone 120 miles, so we have d r t t equals and we'll just plug in our values for X y dx DT and do I D. T. So. 1/2 times 50 squared plus 120 squared to the negative. 1/2 times two times 50 times 25 plus two times 1 20 times 60 And if we start simplifying, things will get 50 squared plus 120 squared to the 1/2 is 130 so we'll have 11 over 260 times 2500 plus 14,400. And if we add those together we get 16,900 over to 60 which is equal to 65. So we have 65 MPH as the rate of increase of the distance between the two cars.

I'm going to draw a diagram of what's going on here. We've got two cars, one is driving south and going away from an intersection and the other one is getting going toward the intersection and moving west. So yeah, have trouble drawing points on here. So I'm going to call that the intersection um south away from the intersection. So this will be Car A. And then Carby is moving to the west toward the intersection. Don't know why things disappear here. Okay. Again. Um Makes a triangle just like in the previous question. And I'm going to call this X. And I'm going to call this why? And so I'm also going to call that L. So L squared equals X squared plus Y squared according to the Pythagorean theorem. I'll take the derivative with respect to time two L. D. L. Over D. T. Equals two X. Dx over D. T. Two. Why? Dy over A. D. T. Now let's see if we have all the information we need. Oh let's also think about positive and negative. So the way that I set up X. And why um and the direction that car A. Is going then uh the rate of A. Is positive. But the rate of B. Is negative because it's causing uh X. To get smaller. So let's see what we got here. Car A. Is 40 miles away and traveling at 55 MPH. So that would go here 40 miles away at 55 MPH. B. Is 30 miles away at 45 MPH. 30 miles away negative 45. All right so l. Is just the square root of X. Squared plus Y squared. So I'm going to write that in there. Yeah let's see the distances are 30 and 40. Okay I'm gonna put that into a calculator two times 30 times negative. 45 with two times 40 times 55. All of that divided by two. And that square root which ends up being 50 in the end. That gives me the L. Over D. T. Is 17. Yeah. MPH. Yeah. And since this is positive, the cars are getting farther away. Okay?


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