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Definite Integrals As Net Change - Problem 2 3,013 views

Teacher/Instructor Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

The definite integral can be used to solve for net change, when given a function for the rate of change. Remember that the rate of change will be some quantity over a period of time. For example, if given a rate of change function F'(t) representing velocity, or distance traveled per second, of a skydiver, integrate this function to find F(t) from a to b. Since F'(t) is height fallen per second, F(t) is simply the height of the skydiver. Therefore, F(b) - F(a) is the net change in height during time b and time a.

Let's do another problem that involves the net change theorem. Recall that net change theorem says; that the integral from a to b of F'(x) equals F(b) minus F(a). This is a net change, and we can calculate the net change by integrating the rate of change.

So let's see how that works in this problem. A skydiver jumps out of an airplane 5000 feet above ground. Her velocity, t seconds later, is given by v(t) equals 160e to the -0.2t minus 160. This is in feet per second. She opens her parachute after 30 seconds. What is her height above ground at that time?

Well, so this is a net change theorem problem, because, we want to know what her final height is. We're given information about the rate of change of that height. Recall that; velocity, v(t) is the rate of change of position; so s'(t). Of course the function that we're given is 16e to the -0.2t minus 160

Now what do we know about position? Well, we know that the initial height s(0) is 5000. We're asked to find the height after 30 seconds. So that's s(30). Now we can relate s(30) to a net change. The net change, and the height from 0 to 30. So it's really important when you're using the net change theorem, that you're given sum value of the height or whatever quantity you're working with. One of these values is known so that you can find the other.

So this one is 5000, and this net change, by the net change theorem is going to equal the integral from 0 to 30 of the derivative of this function; s'(t). Of course that's our function up here. So let fill that in. 0 to 30; the integral of 160e to the -0.2t minus 160 dt. S(0) is 5000. I'm looking for s(30).

So let's evaluate this integral. I'm going to need an antiderivative, The antiderivative for 160e to the -0.2t well, I remember that when I differentiate e to the -0.2t, I get -0.2e to the -0.2t. When I integrate it, I get e to the -0.2t divided by this constant. So my antiderivative is 160 divided by -0.2e to the -0.2t.

For -160, the antiderivative I need is -160t. So I'm going to take this from 0 to 30. By the way what is this number; 160 over 0.2? Let me do a little calculation here. 160 over -0.2 is the same as 1600 over -2. You multiply the top, and bottom by 10, this is -800. So this is -800 times e to the negative. Then if I plug in 30 for t, 30 times 0.2 is 6. So it's -6, minus 160 times 30. So this is what I get when I plug 30 into my antiderivative.

What do I get when I plug in 0? Well, when I plug 0 into this guy, I will get 0, but when I plug 0 in here remember that e to the 0 is 1, so I'm going to get my -800. So what does this give me? 16 times 3 is 48 so this is going to be 4800, and this is -800, I'll evaluate this in a second on my calculator, minus 4800 minus, minus plus 800. So that's -400e, -800e to the -6.

Now I'll erase this. Let me just copy down. This is s(30) minus 5000. So if I add 5000 to both sides, I'll get 1000 minus 800 e to the -6. That will give me my s(30). I'm going to write that over here. S(30) equals 1000 minus 800 e to the -6. So I'll need a calculator for this. 1000 minus 800e to the -6. It's approximately 998. This will be in feet.

So let's figure out what just happened. Going back to the beginning here, I knew that I can use the net change theorem to find the change in the height; s(30) minus s(0). To do that, I have to integrate from 0 to 30. Now that change in height, the integral gave me -4000 minus 800e to the -6. So whatever value this is, this is how much the height changed from t equals 0 to t equals 30. It makes sense that this number should be negative, because her height is actually decreasing.

Now, s(30) minus 500 equals that amount. So when I add 5000, I get this quantity. This is her actual height at 30 seconds; 998 feet; her final height when the parachute opens.

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