Cartoon Calculus: Ch 14 (Physics)

Quick links to Front matter, Back matter, and:
Part One: Ch 1: Introduction, Ch 2: Speed, Ch 3: Area, Ch 4: Fundamental theorem, Ch 5: Limits
Part Two: Ch 6: Derivatives, Ch 7: Toolkit, Ch 8: Extreme, Ch 9: Optimization, Ch 10: Economics
Part Three: Ch 11: Hard way, Ch 12: Easy way, Ch 13: Revisited, Ch 14: Physics, Ch 15: Conclusion

Page 173: What does it mean that acceleration…

Pages 174-175: It’s not immediately obvious why Leibniz… // Newton’s laws of motion and gravitation…

Pages 176-177: Newton’s second law of motion… // That’s because we need limits to define acceleration.

RH: Newton’s 2nd law correctly stated says that the NET force on an object equals its mass x its acceleration. This is an important distinction from just saying F = m x a. Here’s why: A book sitting still on a table is not accelerating (a = 0) but it has two forces on it, neither one of them being zero. So, if we say F = m x a (which is NOT true) and (a = 0) this somehow implies there’s no forces acting on the object. But there are! So adding the word “net” is critical here (and takes nothing away from your story). YB: Thanks for this suggestion; we will ponder.

RH: You make this claim: “…AND ACCELERATION IS HARD TO UNDERSTAND WITHOUT CALCULUS.” I actually disagree with this but I think it IS fair to say that “acceleration is easier to understand WITH calculus.” YB: I think we may just have to agree to disagree, but note (relative to the next paragraphs) that when the book talks about things like velocity and acceleration it’s referencing something instantaneous, not something averaged over time.

RH: You state “…SO WILL OUR VELOCITY INCREASE BY 9.8 METERS PER SECOND DURING THE NEXT SECOND?” and then answer “NOT QUITE.” But that IS what the 9.8 m/s2 means! Given that the acceleration due to gravity near the surface of the earth is 9.8 m/s2, any object in free fall will increase its velocity (in the negative, downward direction, mind you) by 9.8 m/s for every second of flight, regardless of its initial velocity, be it zero, upwards, or downwards, or even any combination of those. All that the limiting process does here is zero in on the instantaneous acceleration. But in the case of free fall the acceleration is constant over time so the limit process doesn’t nearly have the same impact here as it does for speed since the instantaneous acceleration is always equal to an object’s average acceleration. YB: Good point, we can try to clarify this (as we do in Ch 2) by saying, e.g., that your velocity might not increase by 9.8 meters per second during the next second if you open your parachute, or if you hit the ground, etc. This is what we did in Ch 2, and hopefully we have the space to do it here as well.

RH: You also bring this issue up in CH2 Speed. When I teach physics I bring up the idea that there are different ways to measure speed: The AVERAGE SPEED of something during a given time interval is the slope of its position-time graph between two points on the graph which bookend the time interval. But if we wish to know something’s speed RIGHT NOW (a specific moment in tie) we need to find its INSTANTANEOUS SPEED. And that is where the limit ∆t 0 comes into play. [I always thought this would be a good joke in a book like yours, particularly so if you showed it with the associated position-time graph: Police Officer to a driver he just pulled over: “Your instantaneous speed just now was way above the speed limit.” DRIVER: “But officer, my AVERAGE speed was UNDER the speed limit.” Both of them can be right!] YB: Okay, but we’re focusing just on the instantaneous speed… so the joke will have to wait for a different book.

MB: First off, it seems as if you are referring to forces other than gravity when you say the measured acceleration after one second won’t be = g. That turns out not to be your point, but I am not sure what your point is. Yes, acceleration is defined as a derivative, i. e., a limit, but if it is constant over a second, you will indeed see delta v = 9.8 m/s. Is it important at this level to say anything else? Couldn’t you just draw a graph of x[t] and fit it by little segments that get smaller and smaller? YB: This is similar to RH’s comments; clearly we need to address this point.

RH: Page 177: Again, you don’t NEED limits (as stated at the top of P5) to understand acceleration but they certainly can help. YB: Discussed above.

Pages 178-179: In the language of mathematics… // Luckily, the math itself is pretty easy.

RH: I like the sequence of equations you show for the Flying Apple Example. One thought here: What if you used a small font to label what each number means? For example, for the f(t) equation the number 2 represents the initial position (in meters), the +19.6 is the initial speed (in m/s) and the – 4.9 is half the acceleration (in m/s2). These same meanings carry through the rest of the equations, too. YB: Good ideas. I kind of doubt that we have the space, but we can try.

Pages 180-181: Of course, the pull of earth’s gravity isn’t constant. // Figuring out how to get an apple to an astronaut…

RH: This line: “NEWTON’S LAWS TELL US THAT IT CHANGES WITH ALTITUDE.” When people refer to “Newton’s Laws” they mean his three laws of motion, not usually his Universal Law of Gravitation, to which you refer here. Yes, it is “a law” that Newton authored but I think you should restate it is a follows “NEWTON’S LAW OF GRAVITATION TELL US THAT IT CHANGES WITH ALTITUDE.” YB: Good point. We try to finesse this on page 3 by referencing “Newton’s laws of motion and gravitation”, and the question is whether the extra verbiage provides extra clarity or just gets in the way. We will ponder.

RH: I think you should add the units (meters per second per second or m/s2) to the vertical axis here. It looks wrong (to me) to have units on one axis but not the other. YB: Excellent point. We will definitely fix this. (We have a bunch of work to do on the graphs in the book more generally.)

RH: Near the top of the page you say you are trying “to get it (the apple) up into orbit.” Our scenario: There is an astronaut 2.5 x 107 meters away from the center of the earth and you are trying to “throw” an apple up to him. This is a perfectly good problem and you guys solved it correctly provided….wait for it…the astronaut is NOT in orbit. What you have done is solved for the speed required to launch an apple up to an astronaut who sits at a distance of 2.5 x 107 meters away from the center of the earth. The key word here is “sits.” If the astronaut were “in orbit” AT THAT ALTITUDE he would be moving in a circle around the earth at a velocity of 4001 m/s (That “sideways” speed is what keeps the astronaut up there, in orbit, and prevents him from falling back to earth. Our “sitting” astronaut will immediately begin falling toward earth just like anything else dropped from rest at that altitude.) Which means that to put an apple” into orbit” you’d need to give it more energy than you calculated (about 17% more) to give it the same speed of 4001 m/s at that altitude that the astronaut had. Put another way, you are calculating the launch speed required to send an apple upward so that when it reaches a height of 2.5 x 107 meters it has no speed anymore (it begins to “nose over.”) So, what’s the easy fix here? Just DON’T use the phrase “get it up into orbit” otherwise you’ll have more explaining to do than you want. (Maybe just say “…to get it up to the astronaut.”) I do really like this example, however! YB: Good point, we will work on fixing this language. Too bad, because it’s an evocative phrase, but you’re right that technically speaking it’s not correct.

MB: You picked a hard problem. What about rotation of the Earth and motion of the astronaut???? The apple won’t get to the astronaut in your problem. And….use of the term ‘work’ here will be confusing. (It always is, even in more straightforward applications; the relationship of ‘work’ to forces is not intuitive to beginning students.) You just mean—how much energy must we give the apple? Use the symbol v[0] to indicate initial velocity. YB: This is similar to RH’s comments, except for the suggestion of discussing energy instead of work. We will ponder.

Pages 182-183: This is a tough physics problem… // …but the math boils down to this question.

Pages 184-185: We could calculate this integral the hard way… // It’s especially easy because…

Page 186: Calculus is obviously not sufficient to solve physics problems…

Quick links to Front matter, Back matter, and:
Part One: Ch 1: Introduction, Ch 2: Speed, Ch 3: Area, Ch 4: Fundamental theorem, Ch 5: Limits
Part Two: Ch 6: Derivatives, Ch 7: Toolkit, Ch 8: Extreme, Ch 9: Optimization, Ch 10: Economics
Part Three: Ch 11: Hard way, Ch 12: Easy way, Ch 13: Revisited, Ch 14: Physics, Ch 15: Conclusion

One response to “Cartoon Calculus: Ch 14 (Physics)”

  1. On page 186, you use the term “coup de grâce, “a deathblow or death shot administered to end the suffering of one mortally wounded.” Is that the sense you intend? Perhaps you might prefer “pièce de résistance“?

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