Calculus starts out with two main branches. Finding slopes of curvy lines (aka finding the derivative or differentiating) and finding areas and volumes. It turns out that these two different sorts of questions are inverses to one another (like subtraction undoes addition, and logarithms undo exponents), but that's not at all obvious at first.
Tutoring Artemis has continued to be a joy. We've been exploring all sorts of things, not going in any set direction. Today we started out not knowing what we wanted to do, and played with Guess My Function for awhile. It was fun, but I think we both wanted something that would move us along more. We've done derivatives, so I figured it might be time for the inverse.
I didn't say any of that, I just said I had a problem he might enjoy*. I drew one of our favorite functions, y = x 2, and said, "I want to design a part for a toy car for my son. The area under this curve, from x=0 to x=3, is a good shape for the part, but I need to know how heavy it's going to be." We decided to estimate it with rectangles and triangles.
So 1 + 4 (for rectangles) + 1/2 + 3/2 + 5/2 (for triangles) = 9 1/2.
I said the squares were square centimeters, and the metal was 1 gram per square centimeter (is that a reasonable density?), so we knew the part would wiegh (mass) less than 9.5 grams. The curve is always below the stright line segments that the triangles make. So the estimate must be higher than the true value. I told him we wanted to make sure it was under 9.3 grams, so we might want a better estimate. He was loving it. We did an estimate with rectangles and triangles 1/2 centimeter wide, and got 9.125. That took a while. Lots of fraction practice.
I'm thinking I should have claimed we needed it lighter, like 9.1 grams. But Artemis immediately says, "I want to know how much it will be exactly, let's take the limit." Sure, let's do that... ;^)
So we did, his way. When I teach this, I always use just rectangles. But they don't brush up against the curve nearly so nicely as rectangles with triangles on top, and he insisted on including the triangles. I was worried it would get too messy, but it wasn't bad. It was interesting to see the area contributed by the triangles drop out when we took the limit at the end. Maybe I'll teach it differently from now on.
When we got to the sum of a bunch of squared terms, he immediately wrote down what the sum would be! It's in the lower right corner in the diagram to the right, that fraction with a mess over 6. Good thing, because I would not have known how to lead him to that - I don't even have that memorized. We would have taken a long, leisurely detour here, and ended with the problem halfway done.
We found that the car part would weigh 9 grams. I'm so glad it worked out to a nice number; he really appreciated the beauty of that.
As he walked to the car, he was telling his mom about finding the weight of a toy car part. I had almost made the mistake of thinking the concrete context I had given it didn't matter to him. It did. He can zoom into abstraction like a rocket, but he does like a good launching pad.
Next week, we can do another problem like this, and look for patterns. I think I can lead us though the Fundamental Theorem without ever 'teaching' it.
[Edit: Next post in the series here.]
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*If my description of our work goes above your head, and you'd like to understand, please ask. I'd love to help you make sense out of it.
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"1 gram per square centimeter (is that a reasonable density?)"
ReplyDeleteis not valid because density is mass per unit volume. 1 gram per cubic centimeter is the density of water, and metal is quite a bit denser than that (roughly an order of magnitude).
Otherwise, what you are doing sounds very neat! I think it probably makes sense to think about antiderivatives (indefinite integrals) before getting to the whole riemann sum = definite integral bit, but it sounds like you are headed that way and tying together the ideas.
Dang it, I knew density would be the wrong word. But what I'm asking is whether that's too few grams for the area (assume it's whatever thickness needed for a toy car, so can it be thin enough to make sense).
ReplyDeleteI think putting anti-derivatives first points students to think anti-derivative too quickly in the area problem. (Especially given the notational overlap, where the integration symbol means anti-derivative for indefinite integrals, and does not mean anti-derivative for definite integrals - it means area then.)
Aluminum is roughly 3 g/cm^3 (among friends), so to have an areal density of 1 g/cm^2, the part would need to be around 1/3 of a centimeter thick. Steel is closer to 8 g/cm^3, so it would need to be around 1/8 cm thick.
ReplyDeleteYou could always switch back and forth between Newton and Liebniz notation and include some history in the lesson. :)
ReplyDeleteI'm going to have to look that up before Monday! I'm sure I know both notations, but I have no idea which one goes with which person.
ReplyDeleteI've got to come up with some story about why we'd want to find the area out to t (x is in use already). Maybe we can afford a bit more weight and we want to see how long we can make the part.
He's going to discover that it's the anti-derivative without me ever mentioning them. That's my hope.
(And thanks, Hao. I don't usually prepare at all. Having this conversation is helpful.)
ReplyDeleteSue, thank you for sharing the story. I especially appreciated the pictures.
ReplyDelete