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((20 Marks) The length of a rectangular is decreasing at the rate 2 cm/sec, that is, 4 ~2 cm/sec while the width w is increasing at the rate cmlsec_ When 12 cm and ...

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((20 Marks) The length of a rectangular is decreasing at the rate 2 cm/sec, that is, 4 ~2 cm/sec while the width w is increasing at the rate cmlsec_ When 12 cm and W = 5 cm, find the rate of change of The area The perimeter Length of the diagonals of the rectangle. Which of those quantities are decreasing; and which are increasing?(20 Marks) A balloon is rising vertically above = level, straight road at constant rate of = Just when the balloon is 65 ft above the ground bicycle moving at= constan

((20 Marks) The length of a rectangular is decreasing at the rate 2 cm/sec, that is, 4 ~2 cm/sec while the width w is increasing at the rate cmlsec_ When 12 cm and W = 5 cm, find the rate of change of The area The perimeter Length of the diagonals of the rectangle. Which of those quantities are decreasing; and which are increasing? (20 Marks) A balloon is rising vertically above = level, straight road at constant rate of = Just when the balloon is 65 ft above the ground bicycle moving at= constant ft/sec. President University, Jababeka Educalion Park Cikarang Baru; Bekasi 17550 _ Indonesia JI: Ki Hajar Dewantara, Kota Jababeka angolaren epaesidenLecia, Wan presidertac d 8910 9762-63_ Fax (021) 8910 9768. Email



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The length $x$ of a rectangle is decreasing at the rate of $5 \mathrm{~cm} /$ minute and the width $y$ is increasing at the rate of $4 \mathrm{~cm} /$ minute. When $x=8 \mathrm{~cm}$ and $y=6 \mathrm{~cm}$, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.

To find the rates of change of area for the quest. But you're seeing the length they left. Let l. S decreasing decreasing and scare it's and they read or soon centimetre divided by seeking to centimeter per second, then the width and with W. Is increasing is increasing is increasing. It's two centimeter based second as well. Two hands. And so you have why should a relative of air with respect to partial derivative of T. To be called a negative too centimetre happy second. Then crush a divisive for W. If we spoke to T. It's also equal to two sentiments up second. So our area to the area E. It's equals out W. So to derive the rates that's a special divisive of aid with respect to see will be your passion you received with respect the city of El. I'll talk to you and this is a vehicle to the products through here. So why should there be some of L with respect to see you keep w constant less. Keep l constant pressure divinity for W. With respect to see. So then this will be equal to negative two to have negative to here. W. For this parts plus L. Sure. So since we need the risk of change when L it's equal to Hell. Yeah, it's equal to 12 centimetre BMW is equal to five cm. We can substitute that so that we have negative two Size W. Which is five year. We have centimeters great face again. Let's this would give us 24. So solve two. You have centimeters square big second, then this thing. It's equal So Mhm. This will be equal to have negative sale centimetres Great a second plus 24 cm great their second which is equal to foods in centimeters great seeking. So there is why should they receive both? The area with respect should see this is rate of change the fire area. So this is because since the area. So you realize that area area of area of the giving race angle is increasing. It's increasing here. Then for the 2nd part B. For me, B. A perimeter P will be called to two L. Plus two W. So now we derive with respect to T. To obtain the risk of change of perimeter pasha. David E. P. With respect to T. Will be equal to is a respected. See we have two L. Plus two W. And this is going to give us two Russia derivative of air with respect to T Plus two Pasha derivative of W. with respect to see. And this is equal to So you have to this was negative too last two. Which is this was also to us. So so you have negative four plus four which is equal to zero. So then this implies that the rates of change of our perimeter. Mhm rates of change of our perimeter here. It's equal to zero. So there Mhm. So this implies that our perimeter very made style. It's a constance it's of course that's and they take parts let's see foot see dead bugs. See you see is the formula for the length of the dia gonna of direct tangle. Which is given by D. To be go to square roots of all squared platz, W. Squared. So to find a rich of change of length of dia gonna will be equal to one divided by you have soon that's a square roots of L. Squared glass. The blue square and we have to L. A partial derivative of L. With whispered to see last two. W. Russia David steve of W. With respect to T. So this is going to give us one divided by square roots of those squared bloods W. Squared. We have L. So we have L. Here name two. Let's W. So so this day is going to be cold. So we have L. L. So you have negative to L. This was negative. It gets you to kill plus two W. To C. W. Divided by square root soft L squared plus W. Squared. So since we need a risk of change we know L. It's called. It's wolf. W. It's equal to five. So we substitutes for that. If he substitutes you have you have this to be equal to negative two times so of last five times two divided by square roots of Swath squared plus five squared. And this will be equal to food scene divided by state scene Since he needs up 2nd. So then our rates of change of the dia gonna I should be able to deal with respect to T it's equal to 14 negative for the same year, Negative negative, or simply divided by 13 centimetre, their second as uh resorts.

In this situation, we have a rectangle and you could maybe imagine that it's expanding, uh, outwards from every corner and inwards, depending on how x y and L are all interacting here and the way that we could find out the x p t d y d t d l d t is by setting out Pythagorean there x rayed What's y squared equals z squared. And then we're gonna differentiate with respect to t And so, using the power rule for each of these, we'll get two x the Axe City plus two. Why? What d y d t bless her sari is equal to two l the old ET. Now, if any of these was fixed, the let's say, for instance, just supposing that like X was was a fixed blank than that DX DT would end up being zero, for instance. Um and then that's for part A. And then for part B, we want to know very specific. Ah, situation here when X is three will go ahead and start substituting in so well. We'll do that underneath, so we've got to Times X is three DX DT is increasing and it is increasing at a rate of one half half a foot each second plus two times why in this situation is for and then that's multiplied by D Y d t which that, uh, y side it's actually decreasing battery of 1/4 of a foot. So what? Actually, this will be negative 1/4 and then don't forget the right side that is gonna equal to. And then as you concede the red there, we've got a 34 and then it's gonna be a 345 triangle. So l would be five with Exxon wire three and forth and then we want to solve or D l B t. And then looking at the left side here, all of this is just going to equal six up six, divided by two Rather. So that's actually just gonna be three. Uh, these to cancel. And so this will give us minus two. And so, really, the left side is equal to one. And then we're going to divide that by this 10 year. So the Elby T in the end is 10 1 10 feet. Hurry. Second, since its positive, we know that the deal or that a diagonal ill is actually increasing if it was negative would be decreasing and so we could also add that this is increasing at a very slow rate compared to the other ones, but still increasing nonetheless.

Section 3.8 looking at problem number 21. So here we're dealing with changing dimensions of a rectangle, so you know that a rectangle is going to have a length times a with okay. And what they're telling us is this is that length of the rectangle is decreasing two centimeters per second, so the the length is changing. So the change D l d t of the land is gonna be minus two centimeters for a second. So keep in mind when they tell you something is changing decreases a negative. The width is increasing, so D W D t is increasing at a rate of two centimeters for a second. So we're lengthening the width and we're shortening. Ah, the length at the same rate. Okay, they're asking us when when the length is 12 and the width is five. Uh, tell us what is happening to the area, the perimeter and the length of the diagonal. So let's start with party. We're gonna figure out how does this affect the area? The area of a rectangle is the length times the wit. So over time, the change in area is going to be l, uh D W D W D t plus w d l e t. So how is this happening at these points under under these conditions that you see here. So under those conditions, I'm gonna find that d a d t is going to be equal to 12. And then d w t t is going to be two plus, the width is five. And the change of the length is minus two. So this turns out to be, um um 14 season 24 24 minus 10. So this is 14. So the area is changing 14 centimeters per second. Okay, And it's increasing. So we see an increase in this situation. Let me right here in crease, because is a positive number. So I can see that if I shorten the length and increase the width Ah, at the is opposing speeds. I net effect is I'm changing the area, increasing the area 14 square centimeters per second. So Part B says let's do this same problem. But instead of area, let's focus on the perimeter. So if I have the same rectangle l and W, the perimeter is going to be to l plus two w so over time. How does the perimeter change so D l b d d d p d t is going to be So I take the derivative that's going to d l d t plus two d w d t So in this particular case to evaluate under the conditions And what was that we had? Ah, we had l equal to 12 w equal to five, and we were increasing. What is a d l d t minus two d w d t ah, in this case was to let me double check that So d w was to d l minus two. Ellis 12 w is five. So this becomes too DLD t, which is minus two plus two d w t t, which is to this becomes zero. So what does it mean if the derivative is zero? That means that P is not changing. So the perimeter is constant and that makes sense. I mean, if you are stretching out the length at a certain rate, but you're decreasing and compressing the width at the same rate. The total trip around that around that rectangle should not change. And that's what the math is. Bearing that out So then we moved to part C and then part see of this equation. They're saying, How does Thea length of the diagonals of the rectangle change? Okay, so the length of the diagonal. So let's go back to our rectangle. We had l times w Okay. And so if you look at the Dag No. There. So what is the length of that diagonal that's going to be the square root of l squared plus W square Just using the Pythagorean theorem. So if you look at it, um, the length of our diagonal is going to be the square root of l squared plus w squared. So how does that change? So D d d t is going to be one over two l squared plus w squared times the derivative what is under ah, that radical. That's gonna be to L d l d T plus two w d w d t This simplifies a little bit so d the D t is equal to sew the two's all go away, so that becomes a factor that is out of there. So all we end up with here is, um, one over the square root off l squared plus w squared times l DLD t plus w do you w d t. Okay, so let's evaluate it on the conditions that they gave us. So they gave us the conditions that what l is equal to Let's go back. Ellis 12 w is five. So this is 12. This is five. And we knew that d l was minus two d w is too so DLD t so d l d t minus two d w d t is equal to two centimeters for second. So let's evaluate all of this. So we get one over, we're gonna have the square root of five squared plus 12 squared times. L d l d T. That's gonna be 12. I was negative two plus W d W D T. There's gonna be five times two. So we know from looking at this this is a 5 12 13 triangle, so that's gonna evaluate one over 13. So I'm gonna get one over 13 and then if I look back, I've got negative. 24 plus 10. So negative. 24 plus 10. This is negative. 14 over 13. And so the length of that I am murders. This is changing in centimeters per second. Okay, so it tells me I get a negative value for that, right? Negative words. So it tells me that this is a decrease. So it tells me as I stretch this triangle according to what they said, Um, I'm seeing the length of the diagnose decreasing negative 14 13 centimeters per second.

This problem states that the width of a rectangle is increasing at two centimeters per second and its length is decreasing at two centimeters per second. Find the rates of change of the area perimeter and diagonal of the rectangle when it's linked. This 12 and it's with us five centimeters, which quantities are increasing and which quantities are decreasing. So let's start with the picture. So this is our rectangle. We know that this side is with in the side is length and the area for a rectangle is given by with times, length or length, times width. So to answer part A, we get that d a d t is equal to w times d l d t plus l times d W D T And we know that the link this 12 and the with this five and then we also know that the width of the rectangle is increasing at two centimeters per second and this length is decreasing at two centimeters per second. So we can use that information here. So we get that de a D t is equal to the wits, which is five times d l d t, which is positive to our know the length is decreasing at two centimeters, it's decreasing. So we get that. That's minus to D L. D T is minus two plus l, which is 12 times D W D t, which was two. And that's equal to minus 10 plus 24 which is positive. 14. So we get that D A D T is equal to 14 centimeters squared per second. And because this is positive, we know that that's increasing. And also let's go ahead and write the information that's given in the problem. So but with those increasing at two centimeters per second so we know that D W D T is to and the length is decreasing at two centimeters per second. So D L D t is minus two and then lastly, the link this 12 and the with this five for the rectangle. And so for part B, we want the change in perimeter. What's respected time. So we want DP DT. We know that the equation for perimeter is just two times the with plus two times the length. So we get that DP DT is equal to two times dw DT plus two times d d l d T again D W D. T is two and the L. D. T is minus two. So we get that that is two times positive, too, plus two times minus two. And that's just four minus for So we get that that zero, which is constant. So DP DT is equal to zero centimetres per second and then we can write Constant next to DP DT and then for part C. We want to know how the diagnosis changing what's respected time. So this blue line is the length of the diagonal, and we know that we'll label that as capital D. We know that d is equal. Teoh Ah, the square root of w squared plus elsewhere. But we can also rewrite this as D squared is equal to W squared plus l screen, which will help make the the process of differentiating both sides a little bit easier. So we get D squared is equal to W squared plus elsewhere, which means that two times d d d d t is equal to two times w d W D T plus two times l d l d t and we get that two times d well again D is just a square root of w squared plus l squared and L squared is 12 squared, which is 1 44 and W squared is by squared, which is 25. So we get the square root of 144 plus 25 which is thes square root of 169. So we get that d is 13 so we get to Times 13 which is 26 d d d t is equal to two times W, which is 10 multiplied by D W D t, which is to plus two times l, which is two times 12. So that's 24 and we know that d l d t is minus you. So that means that D D d t is equal to 20 minus 48 which is minus 28 all over 26 which is equal to minus 14 over 13. And that IHS centimeters for a second. And because this number is negative, we know that d d d t is decreasing, and that completes the problem


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