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P2: Walk the Dog (Calculus: Chain Rule)
510 Plays5 years ago
Learn more about calculus, derivatives, and the chain rule with this Problem Episode about you walking your (perhaps fictional?) dog around a park.
This episode is distributed under a CC BY-SA license. For more information, visit CreativeCommons.org.
[Featuring: Sofía Baca, Gabriel Hesch]
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Transcript
Introduction to the Chain Rule in Calculus
00:00:00
Speaker
This episode is all about the chain rule and how it relates to a simple mathematical problem about walking a dog. And one of the topics we cover has to do with energy.
Exploring Brilliant.org's Calculus Courses
00:00:09
Speaker
To that end, our partner Brilliant.org has a series of courses called The Road to Calculus, which will give you a robust start in understanding calculus.
00:00:18
Speaker
I love how the course starts with pre-calculus and takes you all the way through subjects like trigonometry and integral calculus, and even simple motion problems similar to the ones here.
Listener Subscription Details for Free Courses
00:00:27
Speaker
To support your education in math and physics, go to www.brilliant.org slash breakingmath and sign up for free. The first 200 breaking math listeners can save 20% off the annual subscription, which we've been using. And now, back to the episode.
Hosts Introduction and Episode Theme
00:00:43
Speaker
This is Sophia. And this is Gabriel. And you're listening to problem episode two, walk the dog.
00:00:53
Speaker
And as you noticed, this week we're doing another problem episode. These are fun. These are something that we don't do too often, but we're going to do much more often. And we're going to start doing them after other episodes.
Recap: Ben Orlan's Calculus Book
00:01:05
Speaker
So our last episode was one on the Ben Orlan book called Change is the Only Constant, which is a book about calculus. So we're going to do a problem episode all about calculus.
00:01:16
Speaker
But first let's get some plugs out of the way.
Support on Patreon and Rewards
00:01:18
Speaker
You can support us on Patreon at patreon.com slash breakingmath. You can get the tensor poster for $22.46 monthly donation there. The tensor poster is a poster that teaches some of the fundamentals behind tensors, which are objects used in Einstein's general theory of relativity. They're basically like multi-dimensional matrices or vectors.
00:01:38
Speaker
This poster is a perfect gift for any fans of astrophysics or Einstein's mathematics or anything like that. They will love it. Or even Riemann. Because it has Riemannian geometry in the Riemann tensor. That's right. And if you don't feel like buying that for a monthly recurring donation... Oh, hold on. We also have $5 a month that you can get outlines of the show and ad-list versions of the show.
00:02:04
Speaker
And for $1, we'll give you a shout out. Nope. People have given us money before, but no one's wanted a shout out so far. So you could be the first. That's right. You could be the first one. That'd be quite, quite interesting. What kind of creative names would people send in? Creative names. See more bots. I don't know.
00:02:22
Speaker
Oh, that's right. That's right. I forgot to mention that the way that the shout outs work on the podcast is that we give you, um, like you say, like your name is like, uh, your name has seen more than your favorite math thing is like numbers. I don't know. Like we have a fan named Remon. We have a fan named, um,
00:02:46
Speaker
Can I remember the name I just said? Seymour. We have a friend names, whatever. You understand it by this time. Yeah. Anyway, you can also buy our poster outright on Facebook for $22.46 plus $12.50 shipping and handling at facebook.com, such breaking math podcast at twitter.com, slash breaking math pod, or I mean twitter.com at breaking math pod. And you can just check our website out, including some cool applets and stuff like that at breakingmathpodcast.com.
Sophia Presents a Calculus Problem
00:03:16
Speaker
So as we had mentioned earlier, this particular episode is a problems episode. Now I'm pretty excited, Sophia had told me that she had cooked up this episode herself and I was quite excited to see what she came up with. All that I knew is that it involved calculus. So Sophia, do you want to tell us a little bit about what you cooked up for everyone?
00:03:33
Speaker
What I cooked up was a proof that Ben Orland does not have a monopoly on calculus. He would have you believe so, no, but he's a great friend of the podcast. And it's basically a problem about walking your dog. So it's like, I mean, if we want to just drive right into it and then we'll do the sound.
00:03:54
Speaker
Oh, and also, I was going to add one more thing. We've actually had some requests from listeners of the show, including teachers, who have asked us specifically to work through certain problems. Now, I love mathematics and I love understanding things, so I think that that is a perfectly justifiable request, and this is our second attempt to do so.
00:04:16
Speaker
So, of course, the theme is calculus. We will explain to you all the problem, the setup. We will then explain the execution of how you solve it. Now, of course, we'll be talking, so it should be very listener-friendly and it should be very, very clear.
Analyzing the Dog-Walking Problem
00:04:36
Speaker
All right, so here's the problem. Suppose you're walking your dog around the park, and the park has a sidewalk on the outside part of it. Now suppose there's a corner in the sidewalk coming up that has a right angle in it and has two long straight segments of sidewalk leading up to the angle, like the corner of a square. So does that make sense so far? Following so far.
00:04:59
Speaker
All right, cool. So if you're pulled along by your dog, so your speed depends on your dog speed, and your dog chooses to move at a constant speed after turning the corner, and if your leash has a constant length, how do you determine how fast you are moving when the dog is a certain distance away from the corner? Okay, so you say, just to recap these, first of all, you've got a very, very well-behaved dog. That is, the dog is going at a constant rate and not accelerating, as my dog does. Yep, to make this problem workable.
00:05:28
Speaker
Now the other things to consider are you've got the lines, you've got a right angle turn. You also have a constant length of a leash, any length at all, but it's a constant, you know, it's a, it's a fixed length.
00:05:40
Speaker
I don't really just realize something. Um, for reasons we'll explain later, the dog does not have to be moving at a constant speed. It just has to be determining your speed. Oh, okay. Because, um, because what we're going to figure out a spoiler alert is how fast we are moving relative to the dog. So based on our position, um, so if we know how fast the dog is moving at that point, then we will know how fast we are moving. If we could figure out everything else. Okay. Very good.
00:06:09
Speaker
But yeah, cool. Does everything else make sense too? I think so. Yeah, so far so good. Do you want some of the problem? Sure, absolutely. So the ladder, as Sophie had explained, the ladder is the exact same thing as this problem with walking a dog around a corner. Now the right angle would be the path down the building and the right angle is at the ground where the ground goes horizontal.
00:06:36
Speaker
As a ladder is falling, you know, you've got the two legs that touch that wall and they are traveling down that wall You also have two legs on the ground that are traveling away from the wall Yep, right. Okay. And then as you had said, there is a change in speed
Comparing to the Ladder Problem
00:06:52
Speaker
Yeah. And so like, let's say you start lowering the top of the ladder down. Like let's say you put some wheels on the bottom of the ladder and you start lowering the ladder ladder down the wall with a rope. If you start at the very top, if you lower it down like a foot, the bottom is going to move pretty much about a foot, right? Yes. But if you, if the rope, if the ladder is almost horizontal with the ground and you move it a foot down, it's not going to move a foot horizontally. So that is the calculus in the problem.
00:07:21
Speaker
Now for this next part of the podcast, we are going to do the setup. That is, we're going to set up how we're going to solve this problem. We know that there is a triangle with a corner being the corner in the sidewalk and the two points connected by the hypotenuse being the position of you and your dog with the leech.
00:07:39
Speaker
Yeah. So like, um, let's say that this is like a triangle. I mean, let's say that like the dog is like three meters away from the corner, right? And you're four meters away from the corner. That would mean that you had a five meter long leash, right? Yes. Cause three squared plus four squared equals five squared. Yes.
00:07:55
Speaker
So, um, that is one thing that we have to notice, um, which we could, some, we're going to call that D person and D dog for distance of person and distance of dog, or actually we should just say distance because D will get that confused for the derivative D. Okay. Sure. Sure. So we, we, we use the term distance. And as you said earlier, this is, you can start with the Pythagorean theorem, which gives you the length of the leash as well as the distance from, from you to the corner and the dog to the corner.
00:08:25
Speaker
Yeah. So the distance from you to the corner squared plus a distance from the dog to the corner squared is equal to the length of the leash squared. And we're all over, of course, assuming that the leash stays completely taught. So basically the dog is pulling you along as much as you can go, basically. Yes. And in the, in the example of the ladder, when you've got a ladder placed across, you know, that's leaning up on a building, the length of the ladder is the hypotenuse. So that's obviously a constant length as well.
00:08:53
Speaker
Yeah. And we know that we have the velocity of the dog because it's moving at a constant speed, right? Correct. And so we know that's d distance of dog over dt because, which is the derivative of the dog with respect to time, because the dog is moving at a constant speed. So we can also, that's basically saying that the velocity of the dog, we could say that velocity of dog is equal to the derivative of the dog with respect to time.
00:09:20
Speaker
or the derivative of the distance between the dog to the corner with respect to time. We are solving for the derivative of the distance of the person or you to the corner with respect to time. So since we have a formula with the two unknowns being the distance between the person in the corner and the distance between the dog in the corner, we can find the derivative of our distance, the distance between the person in the corner with respect to the distance between the dog in the corner.
00:09:48
Speaker
This is useful because we have the derivative of the distance from the dog with respect to time already. And we can use the chain rule. Yeah, of course, since we already have the velocity. And the chain rule basically states, it allows you to take a derivative, take two derivatives and create a new derivative with it. It's kind of like multiplying fractions both derivatives.
00:10:14
Speaker
So it basically states that how fast some quantity A changes with respect to some quantity C is as fast as A changes with respect to some other variable or differentiator B times how fast B changes with respect to C.
00:10:31
Speaker
Now with respect to the ladder, if it's being lowered, and I do differentiate being lowered with a rope as opposed to being dropped in a free fall, in which case it would be accelerating, the ladder falling is not accelerating. It's being lowered at a constant speed, but let's just say we don't know the speed. The two variables would be how fast the top of the ladder is going toward the ground and how fast the legs of the ladder, where they touch the ground, how fast they're moving horizontally across the ground.
00:11:00
Speaker
Well, almost. The two variables in this case would be the distance between the floor and the top of the ladder, and the distance between the wall and the bottom of the ladder. Okay, fair enough. And then we could see, and it's an important distinction because if we know how fast the, if we know that, then we could say, then we can determine the distance between the wall and the bottom of the ladder
00:11:26
Speaker
in terms of the distance between the top of the ladder and the floor, which means that we could see how much the distance between the floor and the top changes when the distance between the floor and the bottom changes. So if it moves by like one meter in the bottom, or how much does the top move? And how about half a meter? How much does the top move? How about a quarter meter?
Calculating Relative Speed with Calculus
00:11:51
Speaker
And so on until we have the limit that represents a derivative.
00:11:55
Speaker
Now, before we go further, I think for an intuitive approach, I'm going to use a similar example. Suppose you had a bunch of stores with prices changing gradually from one area to another and all over time. If you were to walk a certain path, you could go to the stores and get the average price of, say, bread.
00:12:16
Speaker
you can notice that the faster you move, the faster the average price will change because you are moving from one area to another more quickly. This means that for you, the derivative of price with respect to time is equal to the derivative of price with respect to location times the derivative of location with respect to time.
00:12:39
Speaker
Yeah, so it is also like if you listen to the episode, it's like the needle scratch problem. Yeah, if you have two derivatives, you really can relate them like that. The intuition for this problem specifically is that if your dog goes twice as fast, right, then you're going to be going around the curve twice as fast at that point that you are at, right?
00:13:01
Speaker
Yes, so like if your dog goes twice as fast then the problem just speeds up by twice Correct. Therefore if you know how fast your dog is going and you know how much you and and you know like your position With respect to your dog's position
00:13:16
Speaker
Then you know how much your position changes with respect to your dog's position changing. Then you can figure out how fast you are moving at a certain point in time when your dog was moving at a certain speed at a certain place.
00:13:38
Speaker
Now we'll go ahead and move on to the actual solution of this problem with plugging in variables and cranking out the mathematics. To recap, we know that the distance between the person and the corner squared plus the distance between the dog and the corner squared, and again the dog is on the other side of the corner while the person is before the corner, equals the distance of the leash squared. That is the Pythagorean theorem.
00:14:02
Speaker
We also know that the distance of the person equals the square root of the length of the leash squared minus the distance between the dog and the corner squared. That's also the Pythagorean theorem just rearranged.
00:14:18
Speaker
Yeah, and so now since we have D person or distance of the person with respect to the corner, from now on, we're just going to not say with respect to the corner, we'll just say the distance of the person. Since we know the distance of the person with respect to the distance of the dog, we can figure out D distance of person over D distance of dog.
00:14:42
Speaker
Which turns out, and we're not going to go over this part because this is just a bunch of plugging in. You could do this on Wolfram Alpha. This episode is mostly about concepts. It's negative distance of dog over square root of leash length squared minus distance of dog squared.
00:15:04
Speaker
And if you'll notice, this is actually negative distance of dog over distance of a person. Yep. Now we apply the chain rule and we see that the derivative of the distance of the person with respects to time is equal to the derivative of the distance of the person with respect to the derivative of the distance of the dog times the derivative of the distance of the dog with respect to time.
00:15:29
Speaker
And of course, because derivative of the distance of something with respect to time is equal to the velocity of that thing, you can also say that this is the velocity of the person is equal to the distance of the person with respect to the distance of the dog times the velocity of the dog.
00:15:48
Speaker
So as you'll notice, when the dog just starts out, you don't move at all because it's V a V the velocity of dog times distance of dog over distance of person, right? It's all hashed out. Yes. That's that's how it works out. So let's say your dog is moving three meters per second. And it's a one in it's three meters away from the corner and you're four meters away from the corner. Then it is
00:16:19
Speaker
is moving at 3 quarters of the speed. You're moving at 3 quarters of the speed of your dog. And you can just keep plugging in numbers like that. But it's seriously just that easy. The negative velocity of your dog times the ratio of the dog's distance to the corner to your distance to the corner is how fast you're moving.
Real-Life Example with Changing Rates
00:16:43
Speaker
Now to illustrate this problem, you could actually do a side-by-side comparison using a plot of the rate of the dog and the rate of the person or of yourself. And you'd notice that they are not the same at any given point. Well, they might be at one point, but...
00:16:59
Speaker
Yeah, if you have in front of you a laptop and a phone, get your phone and lean it up against your laptop screen and put your laptop screen up at a right angle. Now notice that when you slide the bottom of your phone a little bit, when it's resting right against the screen,
00:17:16
Speaker
The top doesn't move almost anything. Now put it at like a 45 degree angle resting between your keyboard and the screen. Now you'll move the bottom the same amount and the top will move a lot more. And that's the intuition behind this problem. And if you think about a thing sliding down a wall, it goes slow, slow, slow fast. Yeah.
00:17:35
Speaker
What really surprised me was when I first heard this problem without actually examining it further, I assumed that the two rates of change would be the exact same, but they most certainly are not. Yeah, it's interesting, right? And I mean, some of the assumptions that we made, I think, don't even apply for
00:17:51
Speaker
for other shapes. For example, if there was a weird squiggly shape and we were being led by our dog, there wouldn't even be a guarantee of the leash staying the same length, for example, because then the dog moving as fast as they could wouldn't necessarily keep it taut, like if the path looped back, for example.
00:18:17
Speaker
So I hope you all enjoyed this problem episode of Breaking Math and be sure to, well I was going to tell you where you can listen to the podcast, but you're listening to it. Yep, exactly. Be sure to send us an email if there's a problem that you would like us to explore in great detail on this show. There's so many out there that we would love listener input.
00:18:38
Speaker
That's a wrap. Oh wait, email for listener input. Breakingmathpodcast.gmail.com. That's right. And that's it. With that, you have your problem solved and you can figure out the velocity of you and your dog.