###### 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.

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# Instantaneous Rate of Change - Problem 1

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.

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To estimate the instantaneous rate of change of an object, calculate the average rate of change over smaller and smaller time intervals. Recall that the average rate of change is the change in some quantity divided by the change in time. Eventually, these values will get closer to some point, which you can call the estimated instantaneous rate of change. When the rate of change is positive, that means that quantity is increasing. When the rate of change is negative, the quantity is decreasing.

We’re talking about instantaneous rate of change and I have a problem here. The amount A(t) in milligrams of pain reliever in a patients system after t minutes is given by this function. A(t) equals 8 times t times e to the (-t over 50). What I want to do first is complete this table of average rates of change.

When you look at this expression here, this is an average rate of change, 100 plus delta t, 100. I’m calculating the change in the amount of medication in the patient’s system over time. So let’s start with the increments. Delta t, when delta t is 100, 100 plus delta t is 200. Remember t is in minutes, 200 minutes after the medication’s been taken.

I want to calculate A(200) minus A(100) and I want to divide that by delta t which is 100. That’s an average rate of change. I’ll do that on my calculator. I get -0.7896. I want to do the same thing with smaller and smaller increments of time. So delta t equals 10, 100 plus delta t is 110, so I’m going to have A(110) minus A(100) over 10. Again using my calculator I get -1.0761. And I’m going to keep doing this for smaller and smaller values and see if I get some sense of what value this number is approaching.

I’ll do it for delta t equals1, I get -1.0826, and I’ll do it again for 0.1, I get -1.0827. It looks like it’s getting pretty close to converging as my values of delta t are getting smaller and smaller. So part b asks me to estimate the instantaneous rate of change of A(t), at t equals 100. That’s the instantaneous change of the amount of drug in this patient’s system.

Now, these values are getting successively closer and closer, to some magical value. And it looks like it's -1.083, and that’s my instantaneous rate. What are the units? Remember, the amount of drug in the system was in milligrams and time was in minutes, so this is milligrams per minute.

That means that, at this point in time, the amount of drugs is actually decreasing at this rate. We represent this rate on the graph. If this is a graph of my function remember my function was 80 times e to the -t over 50. There’s a surge in the beginning and then at t equals 50, the amount of drugs starts to decrease. Here’s where we are right now. The rate of decrease, this rate actually gives me the slope of the curve at this point. That’s a way of thinking about instantaneous rate of change. It’s the slope of the curve at that point. So the slope would be -1.083 milligrams per minute.