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Solve the given problems by finding the appropriate derivatives.During a television commercial, a rectangular image appeared, increased in size, and then disappeared. If the length $l$ (in $\mathrm{cm}$ ) of the image, as a function of time $t$ (in s) was $l=6-t,$ and the width $w(\text { in } \mathrm{cm})$ of the image was $w=t^{2}+4,$ find the rate of change of the area of the rectangle with respect to time when $t=5.00 \mathrm{s}$.

Calculus 1 / AB

Chapter 23

The Derivative

Section 6

Derivatives of Products and Quotients of Functions

Derivatives

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All right. The theme in this problem is that they want you to evaluate D. A. D. T. Um or D. A. D. X. I guess now it's T. Uh And all that you're given is the length and the width and we know areas length times width. So it makes sense then that you substitute. I don't know if this makes sense, but it does to me the length with its quantity six minus T. And you replace W with its quantity T squared plus four. And what they want you to do is to utilize that product rule because it's a product in here. So just a reminder with the product rule, um you have four spaces in here. This is how I do it where I tell students to take the derivative of one side. So that's the derivative of that is negative one. You leave the other piece alone and then leave the six minus T alone and multiply by the derivative of the right side. Which would be to T. Now I will circle this because some teachers will let you leave your answer like that. I think that's all we have to do in this problem. Oh no. Uh That's not it. We also have to evaluate D. A. D. T. At T equals five. It's kind of our notation. So everywhere that we have a. T. In the problem I need to replace it with five. So I'm gonna do some of the math in my head, five squared is 25 plus four would be 29. And then over here six minus five would be one and two times five is 10. So it looks like I have negative 29 plus 10 to give me an answer of negative 19. So as long as I didn't make any mistakes, this should be the right answer. As far as units go. Uh Let's see talks about centimeters. I don't know if your teacher area is in square centimeters, but it's a rate. Um, so I don't know what the time is, what T. Stands for in time. Uh, it looks like there's s so centimeter squared per second is my guess if I read the question more carefully, but negative 19 is the correct numerical answer.

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