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.