Estimating rate of change 2011 AP Calculus AB free response 2a
About this video
Sal solves 2011 AP Calculus AB free response 2a: estimating how fast a pot of tea is cooling at t = 3.5 minutes from a table of temperatures. He sketches the data, then uses the slope between the nearest data points, t = 2 and t = 5, to get (52 - 60)/(5 - 2) = -8/3 degrees Celsius per minute.
You'll be able to
- explain that a rate of change at a time is the slope of the graph at that time.
- choose the data points on either side of a time to estimate a derivative from a table.
- compute and give units for an estimated rate of change.
Key words
- Rate of change
- How fast a quantity is changing; here, degrees Celsius per minute.
- Difference quotient
- (H(b) - H(a))/(b - a), the slope between two points, used to estimate a derivative.
- Differentiable function
- A function that has a derivative (a defined slope) at each point in its domain.
Transcript
- [Instructor] As a pot of tea cools, the temperature of the tea is modeled by differentiable function H for zero is less than or equal to t is less than equal to 10, where time t is measured in minutes and temperature H of t is measured in degrees Celsius. Values of H of t at selected values of time t are shown in the table above, so that's right over there.
Use the data in the table to approximate the rate at which the temperature of the tea is changing at time t equals 3.5. Show the computations that lead to your answer. So they give us a bunch of points. So let's just graph this just so we can visualize what this data is telling us a little bit, and I suspect that this might be useful.
So this right over here is our temperature axis. This is our temperature at any moment in degrees Celsius, and then this is our time axis. And they gave us the time 0, 2, 5, 9 and 10, or they give us a temperature at those times. So this is zero, two. Five's a little bit further.
5, 9, and then 10. And at time zero, the temperature is 66 degrees Celsius. 66, so it's right over there. At time 2, it is 60 degrees Celsius. At time 2, it is 60 degrees Celsius. At time 5, it is 52 degrees Celsius. So time five, maybe it's right over here. This is 52 degrees Celsius.
At time nine, or after nine minutes I should say, it's at 44 degrees Celsius. So this is 44 degrees Celsius. And after 10 minutes it's at 43 degrees Celsius. So this is really just a graph showing how it cools off over those 10 minutes, over those 10 minutes. That's what the data is telling us.
It's kind of a curve like this, and we've sampled it at these points. So going back to part A, use the data in the table to approximate the rate at which the temperature of the tea is changing at time t equals 3.5. So the rate of change is really just the slope of this curve at time t is equal to 3.5.
So really we just want to find the slope at that point right over there. We want to find the slope, and we don't know the actual function, so the best way to approximate the slope at that point is really just find the slope between time, between minute five and minute two.
So the rate of change, so we could say that the rate is going to be approximately our change in temperature over those three minutes. So H of five minus H of two, and this is obviously going to be in degrees Celsius, over the number of minutes that changed. So from five minutes, or we end at five minutes and we started at two minutes.
And so the rate, the rate is going to be approximately... At 5 minutes, our temperature is 52 degrees. 52 degrees. At 2 minutes, our temperature is 60 degrees. 60 degrees. And then this is over change in three minutes, five minus two is three. So this gets us, let me scroll to the right a little bit.
This gives us negative eight. Negative eight degrees Celsius over three minutes. Or the rate at three and a half minutes is going to be approximately negative eight thirds degrees Celsius per minute, so that is part A. Part A.
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