Calculus: Instantaneous Rate

Calculus: Instantaneous Rate

09/30/2009

~ Ferddy

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An Introduction to Derivatives
Using the Derivative
Instantaneous Rate Page [1 of 3]
Okay, folks. Now, it’s time for, basically, a marathon session of doing a whole bunch of these calculus problems. Really getting a sense of how they work and understand them and hopefully have some fun. Here’s the first one. Let’s take a look.
A crab is crawling along the edge of your desk. Its location, in feet, is given at a time t—times and t’s in seconds—is given by p of t equals t squared plus t. So that means that if you give me the time in seconds, then I could tell you the location of the crab p in feet. Now you’re thinking this is sort of a lame example—I mean, you know—when’s the last time a crab is crawling along the edge of your desk? Hey, but you never know. I mean—oh my God! Look! A crab is crawling at the edge of my desk! Week, week, week, week, week. Look, it’s even heading toward me. Look at that. Wow! Well, I’m going to stop this thing.
Now, the question, though is where is the crab located after two seconds? And there’s a follow-up question here that I want us to think about, and that is how fast is the crab moving at that instant? So, you can imagine that the crab is moving. Let’s see if I can resuscitate the crab. By the way, resuscitating the crab is a lot harder than you may think. Now, I want to count out loud to two, and look how fast it’s traveling after two seconds. See if you can see the instantaneous speed.
Thousand one, thousand two. Now, at that instant, when I said “two,” how fast was the crab traveling? That’s the second question here. Okay. Well, how are we going to actually tackle that problem? Well, let’s tackle the first question, though, first.
First question is where is the crab after two seconds? Now, how would you find that? Would you need to find the derivative? Well, we’ve been talking about the derivative. Maybe we need to find the derivative. Well, let’s actually think about what’s being asked here. I’m asked, “Where is the object?” or “Where is the crab located after two seconds?” Not how fast it’s going, not a rate, not a speed. I’m looking for a location. I’m looking for a position, and we actually are given the position formula.
So, in fact, to answer the first question, what I need to do is I need to find out, what is the position when the time equals two seconds? So, actually, the first question is a pretty is a pretty easy one. All I’ve got to do is take a look at p and plug in two for t—or t for two. (Laughs heartily.) I made a little joke. And the staff grimaces, by the way, in case you’re wondering the reaction of that one.
Okay. Anyway, let’s give this a shot. I’ll take this question and just move it over to there, so you can see it. But what we’ve got going on here is that, again, we have the p function, and I want to find p evaluated at two, because that’s exactly where the crab is located after two seconds. So, what do I do? I just plug in a two wherever I see a t. So, I see a two—two squared plus two. And that’s just four plus two is six, and that’s in feet. So six feet.
So, where is the crab located after two seconds? It’s six feet away from where it started. It’s going to climb across the desk. It started here, and so, two seconds later, it was—wow—six feet. So this thing was actually really booking. Shoom, and it’s six feet along.
Okay, great. So that answers the first question, which wasn’t that hard at all. But now let’s sort of see if we can warm up to the second question, which was how fast is it moving at that instant. How fast was the crab moving at the very instant of that second. Well now, what we want to do is we want to find instantaneous velocity. And how do we do that? Well, I remind you that the derivative of a function at a point represents the instantaneous rate of change or the instantaneous velocity of the function at that point. So, if you want to find velocity and you’re given position, like we are over there, then all I’ve got to do is take the derivative and that will allow me to actually figure out what the velocity is. So, now I need to go off and find the derivative. Okay.
Well, how do we find the derivative? We’ll come back to this question in a second. We first need a derivative. So, I’ve reduced this question to the question of finding a derivative. Well, what I’ve got to do is I’ve got to begin with the definition of a derivative. The derivative equals that, we’ve seen that a lot. We’re going to see that a lot more.
So, what do I do now? Well, now all I’ve got to do is insert in all the data—in this case with our function p of t. So, the language is a little bit different here. In fact, maybe you would like to see me write this out for the particular p function that we call it. So, here it would look like with the p. It would be p prime of t equals the limit as delta t approaches zero of p of t plus delta t—minus p of t all divided by delta t. In other words, it’s the exact same formula here. Wherever I see an x, I replace it by a t. Wherever I see an f, I replaced it by a p. Notice that there really is nothing going on here except these things are just placeholders telling me where to insert the data. So, if you like to look at this, given the way the function looks over there, that’s fine. Otherwise, we could just forget about this and just insert the stuff correctly.
Okay, let’s move on now and actually try to find this limit. So, it’s the limit as delta t approaches zero. And now, I have to insert. Now, I’ve got to insert the data. So, I’ve got find out p of t plus delta t. And you can see over there what p is. P of t equals t squared plus t. So, what do I need to do? Wherever I see a t there, I’m going to plug in t plus delta t. Notice that I see t there twice. So, I’m actually going to have plug in t plus delta t in two separate places. So, what’s that first thing look like here? It would look like the quantity t plus delta t all squared plus t plus delta t.
I want you take a look at this for a second, and I want you to verify that that right there is in fact the same as this. That little piece is the same as that piece. Look at the function and notice all I did was take every occurrence of t and just erased it, and in its place, put in t plus delta t. So you can see the function something squared plus the something. That’s exactly what this tells me to do.
Okay. And now, I’m supposed to subtract off p of t. Well, p of t is pretty easy. It was given to us. It’s just t squared plus t. That’s pretty easy. And then, I divide the whole thing through by delta t. I want you take a look at that for a second and sign off on that. I want you to sign off and say, “Yep, I like it,” or “No, I don’t like it.” What do you think? Is it correct or not? Did I make a mistake or not?
Well, let’s see. This certainly is equal to this. So, that’s okay. And I’m supposed to subtract off p of t. Well, that’s a p of t. Oh wait, but I’m supposed to subtract all of it off. I only subtracted off that first person. It’s that same wacko, classic mistake that everyone makes, and I keep making it. I’ve got to put parentheses around that thing. Because that negative sign has to hit that t as well. And in fact, that negative sign hitting that t is really important, because you know the coincidence that we’ve seen happen again and again? If we don’t distribute, then that coincidence is not going to happen in this problem.
So, in fact, this is actually an important little lesson that you don’t actually see in textbooks. But when you’re working through finding the derivative, by definition, many times you work through these things; you don’t see a tremendous coincidence occur, and you don’t see things drop out. I bet that you made some sort of algebraic mistake. And just go back and see if you can find the mistake. It’s an important hint.
Anyway, coming back to here, I think we now corrected our mistake. I put the parentheses there, and now, all I have to do is simplify this. Well, of course, actually, the one thing you might try to do is take the limit. Let delta t go to zero and see what we get. If delta t goes to zero, then that term goes away. So, I’m left with t squared plus—and then that term goes away—t. So I’m left with t squared plus t minus t squared plus t. So these two things cancel. I get zero on the top, zero on the bottom. Voila, indeterminate form. No big surprise. We need to do something.
And what I would do, if I were me, and I am, is I would just expand this out. There’s too many parentheses here. You know what? If you have too many parentheses, it spoils the soup. So, let’s square this and distribute the negative sign and see what we’ve got here. This would be t squared plus two t delta t plus—that was just this piece here—delta t squared. That’s the squaring of this. Then I’ve got all this stuff to write down. I don’t have to write the parentheses, since I’m just adding those, but if you want to keep the parentheses in, that’s okay. And sadly, it’s time for the crab to move.
And now, I’ve got to distribute that negative sign, so I see a negative t squared—and this is really important—minus t, all divided by delta t. Now, you know, a lot of people, you know, find this really scary to look at it and stuff. I love it. I love it when they get really, really long like this. I really love it! Because it’s fun to watch everything fall out. Let’s see if we can watch everything fall out. In fact, everything that doesn’t have a delta t in it should drop out.
There’s a t squared. I hope there’s a minus t squared somewhere. There is. So, this cancels with this. That has a delta t. That has a delta t. That’s just a lone t. But, happily, it mates up with this lone minus t to give me zero. See how important it was to distribute that negative t? Otherwise, we’d be in big trouble.
And hence, with everything that remains has a common factor of delta t. So, I can actually suck that common factor out—factor it out—and what are we going to be left with? Well, what we will be left with is the following. The limit as delta t approaches zero of delta t times two t plus delta t plus—and if I factor out that delta t, I’m left with a one—all over delta t.
Look what I did. I factored out the delta t here. I factored out one delta t here. And I factored out the delta t here. It leaves a one here. Sometimes people go, “Oh, if I take out the delta t, I have zero.” No, you won’t have zero, because that delta t is actually delta t multiplied by an invisible one. If you don’t believe me, just distribute, and notice when you multiply that by the one, you get back the delta t. So, we don’t want a zero there. We actually want a one.
And now, once again, we’ve isolated the zero over zero. The delta t divided by delta t. Delta t factor on the top; delta t factor on the bottom. I can just cancel those away, as long as you promise me that delta t is not zero. Is it zero? Absolutely not. I’m only approaching zero, never equal zero. So, we’re completely safe. So, this is left with the limit as delta t approaches zero of two t plus delta t plus one.
Now let’s take the limit. Let’s let delta t approach zero, so that delta t will get smaller and smaller and smaller. So this term actually drops out to zero. I’m going to sneeze. Achoo! Excuse me! Thank you! Phew! That felt good. Okay, so that delta t goes to zero, and as that goes to zero, I’m left with—well, that remains—two t plus one. This term goes to zero._____ but everything else remains the same.
So, way back here, we see now that the derivative of the position—the position of the crab—is given by that. So, the derivative equals this. Just found that out. A lot of algebra there. Isn’t that neat? All this algebra. But again, remember it’s not too bad. Just look at a distance divided by a time. Took a limit. Did a little bit of algebra and cleaned it up. Not that bad. Not that scary.
Now we got to answer the question. How fast was the crab moving? In fact, I’ll just put the question over. The question I’ll put over there: how fast is the crab moving at the very second when time is two. So two seconds into it, how fast is it going? How do I find that out? Well, this is a machine—remember—look at that. The derivative represents the instantaneous rate of change. So, all I’ve got to do is look at the derivative, right? And all I have to do is plug in the time t equals two. And that will tell me the instantaneous velocity.
So, what I need to do is just look at p prime and plug in the two seconds. And what do I get? I see two times two is four plus one is five. So, five—and what are the units? Feet per second. So, that crab is going at a rate of five feet per second at the very instant it passes its second second of travel. And where is it located at that point? We’ve already seen where it’s located. We saw it’s located at six feet. So, at that second—at that very, very moment—it’s six feet away from where it starts. How fast was it going at that instant? The answer: five feet per second.
Pretty neat. Not so bad. So, this gives you a sense of the kind of method we use to find the derivative. Now, I’ll let you think about one, and we’ll come back and do a whole bunch more. See you in a bit.