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(Continuation of previous exercise.) Suppose that the average price of a VCR during the same time period was governed by the rule $p=820-50 t$.(a) Find the price and total revenue from sales for $1990,1994,$ and 2000 .(b) Find the rate of change of total revenue for $1990,1994,$ and 2000 .

(a)(i) 820,620,320(ii) $410,15190 / 3,9680 / 3$(b) $4690,-11019,-8155 / 18$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 10

Related Rates

Derivatives

Missouri State University

Campbell University

University of Nottingham

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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for this problem. We're referring back to problem number 17. What I've done is I've made a little chart with our answers from number 17. Now, if you have any questions about where these came from, go back and look at number 17. Go through that problem, then come back because we will be referring to our answers from that problem as we go through problem 18. Okay, so for this same time period were told that the average price of a VCR during this time period follows this formula. P equals 820 minus 50 t. So there's my price of a VCR. So let's find the price and the revenue for each of these, um, three years we looked at. So first of all, price, we went price at year zero, price at year four and price at year 10. Well, that's just plugging that right into our problem here. If t is zero, that means I have 820. If he is four, that's a 20 minus 200 or 620. And if t is 10, that means I'm subtracting 500 which gives me 320. So those are my prices Now let's look at revenue or recall our revenue function. Revenue is the number that sold times the price. Okay, sometimes we call that number X. But this case N is the number. And we found that in our last problem. So in order to find our revenue at each of these times, well, multiply the n value that we got for that year times the p value we got for that year. So when are when t is zero? That means I'm multiplying one half times 820 the p and N values for that year. And that gives us 410 okay, of R equals four were multiplying 49 6 times 620 our revenue revenue is 15,190 divided by three. And if t equals 10 are of 10 were multiplying 121 12 times 320. It gives us 9680 over three. Okay, that's the first part of this problem. The price and revenue numbers. Now, our last thing is, we want to find the change of total revenue. Okay, so the change in revenue means I'm going to take my revenue function Take a new color here. The change is the derivative change in revenue with respect to time. So for this problem I have a product, okay? And times P product rule says it's the first times the derivative of the second plus the second times the derivative of the first. So I'm gonna want to find the change at each of these. So at time equaling zero, let's start there. I'm just gonna plug in all the pieces. I know, I know N That's from our table from our last problem. So we know that one. I know P. We just found those d n d t is also from the last one thing. Only thing we don't have is DP DT. So let's look at that for a second. And I'm just gonna come right here. I've got a little bit of room. Here is my price function that we were given. If I take the derivative to find DP DT, it's a constant negative 50 which makes sense because this is a linear equation, so it makes sense. It's gonna have ah constant slope. So dp DT isn't negative 50 and I'll be able to plug that right in there. Okay, So when I put these in, if T equals zero, I'm gonna find the corresponding number for each of these values. Plug those in and I get a value of 4690. Well, what about the next case when T is 4 1994. Plug those in and I get a value of negative 220 over 18. And if I take my last case when t equals 10 and I plugged that in, we get negative. 8155 over 18.

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