Do They Get It? The Instantaneous Rate of Change Exactly

Today in calculus I wanted to check if students really understood what they were doing when they were finding the instantaneous rate of change. (We haven’t learned the word derivative yet, but this is the formal definition of the derivative.)

So I handed out this worked out problem.

And I had them next to each of the letters write a note answering the following individually (not as a group):

A: write what the expression represents graphically and conceptually

B: write what the notation \lim_{h\rightarrow0} actually means. Why does it need to be there to calculate the instantaneous rate of change. (Be sure to address with h means.)

C: write what mathematical simplification is happening, and why were are allowed to do that

D: write what the reasoning is behind why were are allowed to make this mathematical move

E: explain what this number (-1) means, both conceptually and graphically

It was a great activity. I had them do it individually, but I should have had students (after completing it) discuss in groups before we went to the whole group context. Next time…

Anyway, the answers I was looking for (written more drawn out):

A: the expressions represents the average rate of change between two points, one fixed, and the other one defined in relation to that first point. The average rate of change is the constant rate the function would have to go at to start at one point and end up at the second. Graphically, it is the slope of the secant line going through those two points.

B: the \lim_{h\rightarrow0} is simply a fancy way to say we want to bring h closer and closer and closer to zero (infinitely close) but not equal zero. That’s all. The expression that comes after it is the average rate of change between two points. As h gets closer and closer to 0, the two points get closer and closer to each other. We learned that if we take the average rate of change of two points super close to each other, that will be a good approximation for the instantaneous rate of change. If the two points are infinitely close to each other, then we are going to get an exact instantaneous rate of change!

C: we see that \frac{h}{h} is actually 1. We normally would not be allowed to say that, because there is the possibility that h is 0, and then the expression wouldn’t simplify to 1. However we know from the limit that h is really close to 0, but not equal to 0. Thus we can say with mathematical certainty that \frac{h}{h}=1

D: as we bring h closer and closer to 0, we see that h-1 gets closer and closer to -1. Thus if we bring h infinitely close to 0, we see that h-1 gets infinitely close to -1.

E: the -1 represents the instantaneous rate of change of x^2-5x+1 at x=2. This is how fast the function is changing at that instant/point. It is graphically understood as the slope of the tangent line drawn at x=2.

I loved doing this because if a student were able to properly answer each of the questions, they really truly understand what is going on.

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7 thoughts on “Do They Get It? The Instantaneous Rate of Change Exactly

  1. This is awesome. I feel like I should do activities like this more often with my students. Seems like a great way to assess whether or not they know exactly what is going on. I think activities like this are also a great way to de-emphasize “getting the right answer,” which seems to be the only thing many students care about.

  2. Wow, wow, wow. Seeing this, daughter brings to a higher register your prowess as a math teacher. Mesmerized at how colloquial and down to earth how you communicate your process. Formulas are pretty, aren’t they. Make we want to turn a problem in to a wall art poem.

  3. Very nice. Would be interested in what some of the responses looked like.

    A question I like a lot is to give a few expression that look like the limit definition and ask which will work, which won’t work, and graphically what each is measuring.

    f(3 + 2h) – f(3) /2h ; f(3 + h) – f(3 – h) / 2h ; f(3) – f(h) / h f(x) – f(3) /x-3 etc.
    (with limit notation of h -> 0 or x -> 3)

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