Calculus: More on Instantaneous Rate

Calculus: More on Instantaneous Rate

10/01/2009

~ Ferddy

YAY! This video actually showed what the professor wanted us viewers to see on the bottom of our screen! Great job!

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An Introduction to Derivatives
Using the Derivative
More on Instantaneous Rate Page [1 of 4]
Now, I want us to think about an example, which involves a red sports car. Now, I love sports cars, and here is the red sports car that I want to show you. Look at this thing. This is a beauty, by the way. I mean, you can see the doors actually open on this. Isn't that beautiful? And it goes pretty fast. Look at that. Isn't that great? Oh, in fact, watch this. Watch this. Can you see that? It actually drives. Wooo.
Okay, anyway, there's the car, and now let me tell you the question that I wanted to pick up here today. See, the question is the following. First of all, we're told a little bit of information. We're told that the red sports car is traveling, and its position--in terms of miles--is at time t. So, t is going to represent time in hours--is given by this function. This is a function, remember, that if you insert time, it's going to spit out location. It's going to spit out position. Okay?
So, let me actually show you --looking at this function--what the action of the car is. Okay, I'm actually going to show you how this looks. So, here's what the car actually did. The car started here--in fact--in fact, just for thinking purposes, let me show you exactly what happened. The car actually started, like, in this position. But the car started off first by going backwards. Turns out the car actually went backwards, and then went forward. Oh, geez. So, the motion, again, of this thing--this position tells me that--this position function tells me what happened was the car first went backwards, and then zoomed off. That's what this says. So, I'm going to visually show you exactly what that looks like backwards and then forwards.
Well, that's the travels of the red car. And now, here are a whole bunch of questions that we want to answer. And, in fact, I want to tell you all the questions up in front, and then go through, and work them out one at a time. You might say, "Well, gosh. Why doesn't he just give us one question, do it, the next question, and do it and so forth. Because in the real academic world, if you're taking a quiz or if you're doing a homework problem, they just list all the questions right there. And it's our responsibility to actually focus a little bit and concentrate and make sure we actually answer all the questions that we're supposed to answer.
In fact, this is such a sad thing. When I grade exams with my students, it always kills me when they do really, really great stuff, but they never went back to read and see if they actually either addressed the actual question I asked, or if they did all the parts. A lot of times, I ask questions with multiple parts. They answer, like, the first one, the second one, and they never answer the rest, and I got to take off for it.
So, concentrate. It's important to make sure, when we start a job, at the end of the day, you finish the job. But here are all the things we want to actually figure out. The first thing we want to answer is when is the car 30 miles from where it started. Let me show you, it starts here--let's say. This is not, of course, drawn to scale, because the car is too small for me to fit in. And so then, the car--you know--backs up and then goes forward. And the question is -- when is it 30 miles from where it started? That's the first question.
The next question that I want us to address is what is the velocity at the very moment the car is 30 miles away. So, as the car is traveling--right--if you watch this. Let's suppose this is 30 miles away. Now, watch what happens. At that moment that it crosses my finger, it's moving. How fast was it moving at that moment? Okay? So, that's the second question we want to address.
And then, the last question--there's three parts to this--when does the car stop moving? Now actually, here--I guess--I've given you a major, major hint. A hint that you wouldn't actually see, maybe, on a quiz, or you certainly wouldn't see in a book. Because a book, of course, doesn't have video--live video--like this does. But, I did show you the motion of the car. You might remember that the car's motion backed up and then went forward.
Now, notice that to go from backing up to going forward, it is required you must stop--just for a second--but you must stop. And so, you can see it backs up, it stops, and then goes forward. And so, basically, it's that stopping point that we're going to be interested in there. So, that actually will help us there. That's actually a little bit of a hint. And so what, we're just doing these for fun. A little hint. That's always friendly, isn't it?
Okay. All right, now, go back and start working on these questions one at a time. All right, so I'm going to, in fact, take them all them all away. I'm going to put them all over there, so you can see them. At least I'm actually going to just put the first one over there. I'll even write the question over there, so you can read it if you want to see it again. Now, we have a clean slate.
So, what do we need to do? Well, the first question asks us, at what time is the car--well, in fact, let me just take a look at the question. I actually forgot the question. No, I can't read that________, that's right. When is the car 30 miles from where it started?
Now, how would we do this? How would we tackle this problem? When is the car at a certain position? What I need is a relationship that links what two things? Well, that links time and position. Maybe you're thinking, "Gee, I need take the derivative." But we need to think about that. Derivative gives us velocity. Derivative gives us speed--right--rate of change. I'm not asking for how things are changing. I'm just asking for a time. I'm asking for when is the car 30 miles away.
So, what I need is position and time, and of course, we have that. That's the position function that we were given. So, we start off with the position function. So, p of t equals t squared minus seven t. And so, when is the car 30 miles from where it started? Well, that's asking for what? Should I plug in 30 for t? Well, to plug in 30 for t, that would be saying that I'm saying that the time is 30. I don't want that. I want the distance to be 30, and I wish to find the t. So, what I need to do is actually figure out for what t will this thing equal 30. So, what I'd like to do is, actually, set this equal to 30 and solve. Because if I can then solve this, that would tell me what time would make the position 30, which is exactly what this question is asking.
By the way, you know I think a real hard part of solving these kind of questions--any kind of questions--is to first make sure you understand what the question is asking. You know, a lot of times, I see students that go off, they look a problem. It looks really, really complicated, and they start working and doing all this stuff, and they get stuck. And then, they come to me. They say, "Professor Berger, I'm stuck." And they have all this work in front of them. It looks really interesting and impressive. And I say to them, "Well, tell me. What's the question?" And they don't know. You can't answer a question if you don't understand the question first. And it sounds really silly, and it sounds sort of basic, but it's really fundamental. Don't start attacking a problem blindly. Make sure you understand what the question's asking. And it turns out, by the way--in math and in life--that if you really understand the question completely, the answer will drop out. That's a good way of knowing if you really understand what's being asked, if you can answer the question pretty easily.
Anyway, here you can see once we understand what these words mean, then it's clear how to solve it. We just set up this kind of equation. Now we need to solve that. Well, how would you solve this kind of equation? Well, you could notice you could factor out a t. You might be tempted to do that. But, if I have a quadratic like this, and I want to factor, I want to have factors that actually all equal zero. So, the thing to always remember if you've got a quadratic equation like this is to pull everything over to one side and have the other side equal zero, and then try to factor.
So, let's bring the 30 over. It becomes a minus 30. So, I see t squared minus seven t minus 30 equals zero. And now, our task is to factor this. So, you know how I do this. I brazenly start with that. Almost done. Okay now, how do I do this? Well okay, so this t will probably go in both of these spots. This sign tells me that they're going to have opposite signs here. So, I'm going to put a plus and a minus here.
And now, what's the mission now? I've got to put two numbers in here whose product gives me 30, and yet when I combine them this way, I get negative seven. So, you might think, for example, of 30 and one, but that won't actually give me the negative seven. You might think to yourself of five and six, but that actually will give me either plus or minus one. Maybe this can't even be factored. Until you remember that 30 is also three times ten. Now, three times ten, actually--I think if we put the numbers in the right spot--give us a three t minus ten t, which is, indeed, minus seven t. Great! So, we factored it. Terrific.
And remember, if I have two numbers, when you take their product, they equal zero, it must mean that either this number's zero or that number's zero. Let me just make a little comment here, for you algebra fans-- or our nemesis' in the audience--and that is that this is why it has to equal zero. You see, if you tried to factor this, you just can't say, "Well, if two products of two things equal 30, then one is something, and the other's something else."
See, it's not that way, because 30 could be factored into one times 30, five times six, could be factored as three times ten, and a lot of ways. But, there's only one way to have a product of two things equaling zero. Either one is zero, or the other is zero. So, if you've got a big equation like this to solve, make sure you have it equal to zero.
Okay. Well, now on with this. Either t plus three equals zero, which would actually mean that t is the number negative three. Or t minus ten equals zero, which would mean that t would have to be the number ten. So, we seem to be getting two answers. I claim that one of them is a silly answer. Well, this is time. I'm assuming, by the way--I guess maybe I should have said this, but I think it's understood--that I'm starting at the beginning of the day here. So, t is zero. And so, this negative answer doesn't make a lot of sense in terms of the context of the physical problem. Here's the car. I'm traveling.
So, what does it mean to be negative time? Well, I don't know what that means, and so therefore, I will think of that as a not good answer here. And so, I'm left with t equals ten as the only answer. T equals ten, and what are the units here? The units here are hours. So, this is ten hours. But wait, is it hours? Yeah, it's hours. And so, the answer is ten hours.
By the way, that's an answer. And what was the question again? Well, take a look over there, and remember the question was - exactly when is the car 30 miles away from where it started. And now, we see that the answer is exactly ten hours after it starts its travels, it is exactly 30 miles away. So, that actually answers the very first question. The very first question, which is over here. I almost lost it. When is the car 30 miles from where it started?
Great! Are we done? No. Two more questions to go. So, let's then take a look at the next question. The next question is, "What is the velocity at the very moment the car is 30 miles away?" So now, we know that the car is 30 miles away after ten hours. And now, we want to find out the velocity -- velocity. Now, is the position function going to help us here? Well, it will, but not directly. We're not going to plug in 30, or set things equal to 30 here, because what we need here is what? What we need is a velocity. And the moment you think of velocity, what I hope you begin to think about is one thing - velocity, you should be thinking derivative. When you see a velocity question, if you see a rate question, if you see a slope question, the first thing that should be triggered is derivative.
And I remind you here that, in fact, we've already said many times that the derivative of a function represents the instantaneous rate of change, or the instantaneous velocity of that function. So, we have a function here. It's the position function. I want to find out the velocity. I need to take the derivative of it. Ah ha! So, finally, I come to a question that requires finding a derivative. So, let's find the derivative of this function.
Okay, now we're going to go off and do that. Do that. Now, let me remind you the definition of a derivative. This is the definition of a derivative. Now, again, this is the card maybe--and let's really emphasize this--this is the card which says f prime. That's just the notation for derivative. And it equals that limit. That limit, again remember, is just the distance over time--just the rate. Now, it's all in terms of f's and x's here. And you'll notice that, of course, the problem is actually posed in terms of t's and p's. But, we've understood, from the previous examples, that we'll just interpret this as a p of t, and we're just going to substitute in.
So, let's find the derivative now. In fact, I'll just take this card and I'll put it over there, too, so you can see it. But I want the space here to write. So now, we're going to write down the definition of p prime of t. It equals the limit as delta t approaches zero of what? Well, of p of t plus delta t minus p of t, all over delta t. Now, let's plug in p of t plus delta t and see what that is. So, wherever I see a t in that p function, what I'm going to do is replace every single occurrence of t by the quantity t plus delta t.
Okay. And I'm going to tell you right now, I'm going to write all this out, and I'm going to make a lot of mistakes. So, now I'm actually giving you a little bit of warning. I'm going to make tons of mistakes right now, just in case you thought that all those mistakes I've been making were just unintentional. They really weren't. I'm going to make tons of mistakes. See how many of them you can catch on the fly.
Okay. So, here we go. I see here t plus delta t squared minus seven times t plus delta t. That's just this thing. If you plug in and replace every occurrence of t by t plus delta t, you get this. You see this thing squared minus seven times that. And then, I subtract off just p of t. So, I just write down p of t--that's pretty easy--all over delta t.
How many mistakes did I make? I made a lot of mistakes and these are great mistakes and mistakes that a lot of people make. I hope you can identify at least some of them. Let me show you all of them. First mistake was that it's not just this thing squared, it's this entire thing that has to be squared. It's this entire blop that has to be squared. And then, I have to multiply this entire thing by the negative seven, not just that first term. This, as written, says minus seven t plus delta t. If you want to tell me that everything has to be multiplied by minus seven, which is what the function tells me, it's minus seven times t. But now, that t is going to be replaced by all this stuff. I've got to multiply all that by this, and there are no more mistakes. Are there no more mistakes? Actually, there is one more mistake. I have to subtract off all of this stuff. You always need parentheses around here.
Okay. Well now, what do you do? Well, I'm now going to untangle this mess by getting rid of all the green parentheses by carefully expanding this. And we've done this a lot. I'm going to do this really sort of fast. Now, if I square this, I get t squared plus two t delta t plus delta t squared, and I'm going to distribute that minus seven. And see, minus seven t minus seven delta t. And I'm going to distribute that minus sign. That minus sign has to hit everybody, so it's a minus t squared, and then a minus and a minus is a plus seven t all over delta t.
And, again, this is my favorite part. It looks really, really complicated here. But actually everything drops out that we don't like. There's a t squared and, hopefully, a minus t squared. They go away. There's a minus seven t and a plus seven t. They go away. And notice, everything that's left has at least one factor of delta t, so I can factor it out. And so, this equals the limit as delta to goes to zero of delta t times two t plus delta t minus seven, all over delta t. There's the zero over zero. I cancel them away, and what does that equal? That equals the limit as delta t goes to zero of two t plus delta t--delta t--minus seven. And if I let delta t now approach zero, this term--this delta t term--drops out. And so, this equals two t minus seven. And so the derivative is two t minus seven. Great!
Well, we did a lot of work there--all that work to find a derivative. And what does the derivative mean? Well, derivative means--two t minus seven--that gives me the velocity. And what was the question? The question was, "What is the velocity at the very moment the car is 30 miles away?" Well, do I plug in 30 for t? No. Because what I need to do is plug in a time. I need to know the time that the car was at when it crossed the 30-mile mark. Happily, that was the previous question and we saw that the answer was ten.
So, I plug in ten here, because ten is the time when the car is 30 miles away, and I see ten times two, which is 20 minus seven, which equals around 13. So 13 miles per hour. 13 miles per hour. And that's--so, that gives me the velocity when the car's 30 miles away. Notice that, in fact, I had to first the time. I had to first find the time and then plug in the time.
And the very last question--let's do this really, really fast--"When does the car stop?" What does it mean for the car to stop? What would its velocity be? What's the velocity when something stops? Zero. So, to find out when the car stops, I want to set the velocity equal to zero. So, I set this equal to zero, because it has to stop. Zero velocity.
And if I solve this, I see p equals seven over two. So, seven over two, which is roughly three and half hours. So, after three and half hours, the car stops. So, the action, basically, is quite long. It goes backwards for a long time. It goes backwards for three and half hours. That's a long time to go backwards, isn't it? And then, it sort of stops and then goes forward, and then, ten hours after it started, it got to the 30-mile mark. It was going 13 miles and hour.
It's sort of amazing that car this pretty and this fast is only going 13 miles and hour. But it was. Hope you enjoyed it. See you soon.