In each exercise, a function and an interval of its independen...

Question

In each exercise, a function and an interval of its independent variable are given. The endpoints of the interval are associated with points $P$ and $Q$ on the graph of the function.

a. Sketch a graph of the function and the secant line through $P$ and $Q$ .

b. Find the slope of the secant line in part (a), and interpret your answer in terms of an average rate of change over the interval. Include units in your answer.

After $t$ seconds, an object dropped from rest falls a distance $d=16t^2$ , where $d$ is measured in feet and $2\leq{t}\leq{5}$ .

Accepted Answer

Welcome back, everyone. In this problem, a balloon is being filled with air and its volume V after T minutes is given by V equals 4 T2, where V is measured in liters. Draw the graph of the function and the C line passing through A and B. The end points of the interval of T being between 1 and 3 correspond to A and B. Then calculate the slope of the equant line and interpret it in terms of an average rate of change over the interval. Here in our answers, we have a diagram and a description of our our average rate of change over the interval, where A says that the balloon's volume increases by 16 L per minute on average. Here is the graph of B and the answer is -16 L per minute, which means the balloon's volume decreases by 16 L per minute on average. See Says that the answer is 8 L per minute, which means the balloon's volume increases by 8 L per minute on average, OK. And D says that it's -8 L per minute, where the balloon's volume decreases by 8 L per minute on average. Now, I also have a graph, OK, that we're going to use to plot V of T, OK? So this is going to be a graph, OK, of the volume as a function of time, and that is going to help us figure out what the slope of the C line is going to be and the average rate of change over the interval. So, let's try to draw our graph here where we know that our function V of T is equal to 4T2. Now, on our graph, notice that on our X-axis, we go by one unit. Yes, we go up by 1 unit each time and on our Y axis, we go up by 5 units each time. So let's try to plot 3 points that can help us to draw or curve, OK? Because we know we're interested in interval where T is between 1 and 3. Now, when T equals 0, OK, then V is going to be equal to 4 multiplied by 0 squared, which equals 0. So that means that 00 is on our graph, 00. Next, when T equals 1, V is going to be equal to 4 multiplied by 1 squared, which is 4. So 14 is also on our graph. So we can plot that here. Next, when T equals 2, then V 2 equals 4 multiplied by 2 squared, as 4 multiplied by 416. So 216 is on our graph 216. And when T equals 3. Then V is going to be equal to 4 multiplied by 3 square, which is 36. So 336, OK, 336 is also on our graph, OK? And now we have enough points to plot a curve. I'm going to try and draw it as smooth as possible, but I think it would look something like that. No, remember that we said the graph has a function that has a C can line passing through A and B, where the endpoints of the interval of T being between 1 and 3 correspond to A and B. So, in other words, what we're saying is that on our graph, OK, this point where X equals 1 is going to be a. But then red. This point where X equals 3 is going to be B and thus our C can line, fix that here, or C can line passes through these points. So no C can line is probably gonna look something like this, OK? We're gonna look something like that. And no, so that's at 3:36 and 14. OK. So now to get the slope of our second line. The slope of AB, OK. slope of the CA line at AB. Let's call that. Well, it's a slope. We know it's going to be the change in Y coordinates divided by the change in X coordinates. So that's 36 minus 4 divided by 3 minus 1, 36 minus 4 equals 32, 3 minus 1 equals 2 and 32 divided by 2 is going to be equal to 16. Remember that our volume is in liters and our time is in minutes, so our slope is going to be 16 L per minute. Now what does this mean? Well, since our slope is positive, this means that, OK, so let's make a note here. Since our slope is positive. That means the volume of the balloon, OK, the volume of the balloon increases because it's positive. So it increases. OK, I don't think I spelled increase, right? Let me fix that. At an average rate. Of 16 L per minute, OK. So this is what our graph would look like. This is what we know about our slope, and if we compare this result to our answer choices, then we can tell that answer choice A would have been the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Secant Line

Average Rate of Change

Graphing Functions

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