solve-the-problems-in-related-rates-a-rectangular-image-400-in-high-on-a-computer-screen-is-wideni-2

Question

Solve the problems in related rates.
A rectangular image 4.00 in. high on a computer screen is widening at the rate of 0.25 in./s. Find the rate at which the diagonal is increasing when the width is 6.50 in.

Daphne Pusey · Accepted Answer

here we have a rectangle with dimensions two by four. I&#x27;m just gonna label to as w for width and l for length for the four. We&#x27;re also told that each side is going to begin increasing in length at a rate of one centimeter per second. So D W duty is one centimeter per second and the L D. T is one centimeter per second. What we want to know is how quickly the area is increasing after 20 seconds. Okay, so let&#x27;s start with our area formula. We know area is length, times wet. I&#x27;m gonna take the derivative of both sides with respect to t. So de a t t is equal to We have a multiplication problem here. So we&#x27;re gonna use our product rule first times the derivative on the second, plus the second times, the derivative of the first. All right, so let&#x27;s plug in what we know. So de a t t is what we&#x27;re trying to find out. L we Joan. Exactly. No, l Yet we know DW DT is one, and we know d l d t is one. So we gotta find l and w. So if we think about this a second. We know side links are increasing at one centimeter per second. Ok, well, if the length starts at four after 20 seconds, it will have increased by 20 centimeters. Which means L is 24. W started at two after 20 seconds. If it increased one centimeter per second, it would have increased 20 centimeters. So two plus 20 is 22. Right? So we&#x27;re just gonna go ahead and add these together. De A T T is equal to 46 centimeters squared for a second.

Seth Bullwinkle · Answer

Hello, everybody. For this problem, we are given a rectangle that at time zero initially measures two centimeters in width and four centimeters in length were also given that the rate of change of the length with respect to time is one centimeter Her second, which is also the same as the rate of change of the with The WTT were asked to find the rate of change of the area a d t. That&#x27;s our missing right there. And when t equals 20 seconds. Okay, so the area of a rectangle was given by length times with, so you&#x27;d be tempted to just differentiate with these, but you would need the product rule. The reason being is that this length is a function of time and this width is a function of time. So this is the product of two functions. So I&#x27;ve gone ahead, written out a generic form of the product rule so we can begin our differentiation here. So we differentiate with respect to the left side the a d d. Which is what we&#x27;re looking for now We&#x27;re gonna follow this product will hear you is l R length with respect the time he is W R with with respect to time. So you So we just get l right there. Take the derivative of R B function, which is W in this case, That&#x27;s D W D t. Who, us? We&#x27;re gonna take RV function w multiplied by the derivative of the U function l of tea right here. So the l d t. Okay, so we can we can put in some of our knowns right now, So that&#x27;s L times One centimeter per second. Sack hand plus w times are ready to change over here again. One centimeter per second. But now many this l and this w So we know that these lengths have been increasing in. These wits have been increasing one centimeter per second and 20 centimeters. 20 seconds. Excuse me, have passed. So that&#x27;s a change of 20. Both these numbers. So after 20 seconds, the length will equal 24 centimeters time. Still, I haven&#x27;t dealt with that yet. One centimeter, her second. Who? Us 22 Now for the With that it&#x27;s centimeters in our rate. One centimeter per second. So a little bit more clean up here when I move over to this one. So the A B t right then again was equal to one times 24 Still 24 centimeters of centimeters square centimeter her 2nd 24 arm square centimetres per second. Plus the same thing&#x27;s gonna happen 22 square centimetres per second. So we&#x27;ll clean this up even more notice. Our units are nice and the same there. What we&#x27;re looking for square centimetres because we&#x27;re in area 22 plus 24 is indeed what we get. 46 46 square centimeters. Her second is our rate of change of the area after 20 seconds.

Ashley Volpe · Answer

All right, So this is just a Pythagorean Theorem question. And so we know a squared plus weeks where cease for if our dimensions air 20 by 25 those are legs were looking for the diagonal, which would be our hot pot news. So C squared equals 1025 and then square rooting. That means that this is going to be a 32 inch TV.

Melvin Adkins · Answer

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&#x27;s linked. This 12 and it&#x27;s with us five centimeters, which quantities are increasing and which quantities are decreasing. So let&#x27;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&#x27;s decreasing. So we get that. That&#x27;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&#x27;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&#x27;s increasing. And also let&#x27;s go ahead and write the information that&#x27;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&#x27;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&#x27;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&#x27;s respected time. So this blue line is the length of the diagonal, and we know that we&#x27;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&#x27;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

Robert Daugherty · Answer

section 3.8 looking at problem number 21. So here we&#x27;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&#x27;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&#x27;re lengthening the width and we&#x27;re shortening. Ah, the length at the same rate. Okay, they&#x27;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&#x27;s start with party. We&#x27;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&#x27;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&#x27;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&#x27;m changing the area, increasing the area 14 square centimeters per second. So Part B says let&#x27;s do this same problem. But instead of area, let&#x27;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&#x27;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&#x27;re decreasing and compressing the width at the same rate. The total trip around that around that rectangle should not change. And that&#x27;s what the math is. Bearing that out So then we moved to part C and then part see of this equation. They&#x27;re saying, How does Thea length of the diagonals of the rectangle change? Okay, so the length of the diagonal. So let&#x27;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&#x27;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&#x27;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&#x27;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&#x27;s evaluate it on the conditions that they gave us. So they gave us the conditions that what l is equal to Let&#x27;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&#x27;s evaluate all of this. So we get one over, we&#x27;re gonna have the square root of five squared plus 12 squared times. L d l d T. That&#x27;s gonna be 12. I was negative two plus W d W D T. There&#x27;s gonna be five times two. So we know from looking at this this is a 5 12 13 triangle, so that&#x27;s gonna evaluate one over 13. So I&#x27;m gonna get one over 13 and then if I look back, I&#x27;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&#x27;m seeing the length of the diagnose decreasing negative 14 13 centimeters per second.

Problem

Solve the problems in related rates. A metal cube…

Problem 22 Easy Difficulty

Solve the problems in related rates.
A rectangular image 4.00 in. high on a computer screen is widening at the rate of 0.25 in./s. Find the rate at which the diagonal is increasing when the width is 6.50 in.

Answer

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Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus