Preview
Buy lesson
Buy lesson
(only $2.97) 
You Might Also Like

Calculus: Using Integrating Factors 
Calculus: Differentiability 
Calculus: 1stOrder Linear Differential Equations 
Calculus: Solving Separable Differential Equations 
Calculus: Power Series  Integration 
Calculus: Power Series  Differentiation 
Calculus: Power Series DifferentiationIntegration 
Calculus: Fibonacci Numbers 
Calculus: Applying Implicit Differentiation 
Calculus: The Number Pi 
College Algebra: Solving for x in Log Equations 
College Algebra: Finding Log Function Values 
College Algebra: Exponential to Log Functions 
College Algebra: Using Exponent Properties 
College Algebra: Finding the Inverse of a Function 
College Algebra: Graphing Polynomial Functions 
College Algebra: Polynomial Zeros & Multiplicities 
College Algebra: PiecewiseDefined Functions 
College Algebra: Decoding the Circle Formula 
College Algebra: Rationalizing Denominators

Calculus: The Number Pi 
Calculus: Applying Implicit Differentiation 
Calculus: Fibonacci Numbers 
Calculus: Power Series DifferentiationIntegration 
Calculus: Power Series  Differentiation 
Calculus: Power Series  Integration 
Calculus: Solving Separable Differential Equations 
Calculus: 1stOrder Linear Differential Equations 
Calculus: Differentiability 
Calculus: Using Integrating Factors
About this Lesson
 Type: Video Tutorial
 Length: 18:21
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 198 MB
 Posted: 06/26/2009
This lesson is part of the following series:
Calculus (279 lessons, $198.00)
Calculus: Applications of Differentiation (15 lessons, $31.68)
Calculus: Related Rates (5 lessons, $8.91)
Taught by Professor Edward Burger, this lesson comes from a comprehensive Calculus course. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/calculus. The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'Hopital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus II topics.
Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.
He has also taught at UTAustin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".
Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, "Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas" and of the textbook "The Heart of Mathematics: An Invitation to Effective Thinking". He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The "Journal of Number Theory" and "American Mathematical Monthly". His areas of specialty include number theory, Diophantine approximation, padic analysis, the geometry of numbers, and the theory of continued fractions.
Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.
About this Author
 Thinkwell
 2174 lessons
 Joined:
11/14/2008
Founded in 1997, Thinkwell has succeeded in creating "nextgeneration" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technologybased textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.
Thinkwell lessons feature a starstudded cast of outstanding university professors: Edward Burger (PreAlgebra through...
More..Recent Reviews
This lesson has not been reviewed.
Please purchase the lesson to review.
This lesson has not been reviewed.
Please purchase the lesson to review.
Optimization
Related Rates
The Baseball Problem Page [1 of 1]
Let's go from the ladders to a baseball game. That's more fun anyway. It's better to play and have fun than look at ladders falling and hitting people and suing them. So let's take a walk to the baseball diamond here. So here's a baseball diamond, which I remind you is actually a perfect square where each side is 90 feet long. And let's suppose that today there's a game and there's some opposing team playing the Detroit Derivatives.
And so what happened was, of course, the batter's up here at bat and the pitch comes in. The batter swings and hits it way out in left field. And the batter begins to run to first base. And the batter runs to first base at a rate of 24 feet per second. So that person's really racing off to that first base.
Let me show you exactly what that looks like here. I have some sneakers. Let's see if we can just do it. So here we go, and now the batter's about to go and there the batter's going off, you see, running, heading toward first base. Now one thing you can notice, as that batter is heading toward first base and gets there and  trying to steal, trying to steal, trying to steal. Don't steal. Now as that batter is running from home plate out to first base, you'll notice that we could measure, if we wanted to, the distance between second base  the second base person  and the runner. And you'll notice, in fact, that as the batter is running, that distance is changing. In fact, let me try to do this again for you. I hope we're going to be as successful as the last one. That was pretty good. But now I'm going to try to actually show you that distance, which of course is just a straight line distance, and see if you can see that in fact that distance is really changing. So right now, there's the distance, and let's see what happens as the batter actually takes off.
So there's the distance. It's actually shrinking. You see there? It's shrinking, changing, it's shrinking, and then the batter finally lands. And the question that I want us to think about right now is, how fast was the distance between second base and the batter changing at the very instant the batter was located midway between home and first? So let me say that again. The batter is taking off and heading toward first base at a rate of 24 feet per second. While that batter is running, the distance between second base and the batter is changing. It's changing from this length all the way to this length. In fact, can you see, is the length shrinking or actually expanding? Well that length you can see is actually shrinking. So it's shrinking. And the question is, how fast is that distance shrinking the very instant the batter is halfway between first and home? That's the question.
Okay, well let's try to resolve this question, and I think I might actually have to bring these props back, so I'm just going to move them close by here. And so how do we start this? Well, the first thing I think we need to do is draw a picture. So let's draw a picture of exactly what we just saw here. So we have this baseball diamond, which is in the shape of a square. I'll try to draw you a perfect square here. And so you can see the bases are here. Here's home. And remember that each side of this square is 90 feet long. So each side is 90 feet all around. You get this square. Now the batter starts here and starts to run in this direction. And then at some point  let's just pick an arbitrary point. Suppose that the batter is right here, we can actually measure the distance between second base and the batter. We can measure that distance. And the question is, how fast is that distance changing? So now we're ready, I think, with this picture to use our method for solving all of life's problems.
So the first thing we have to do is understand what is asked of us. That actually, I think, in many of these problems is sometimes the hardest part. So first of all, let's give a name to this length right here. So we can call that length, let's say, s for second base. So s represents the length from wherever the batter is at any instant to second base.
So what are we being asked here? Well, we're asked to find out how that length is changing at the very instant the batter is halfway between home and first. So here's what we want. What we want is to find how this distance, s, is changing. Now I could write that out, "how the distance between the batter and second base is changing," and you can see it all there. I'm not going to write that out again. But instead, let's try to use mathematical symbols and the notion of calculus. If s is changing, it's changing with respect to time. So I want to find out how this distance is changing with respect to time. So that's a derivative. And what would the derivative be? Well, it would be find . That's the rate of change of s with respect to time, which is the change in this distance, and that's what I'm after.
So I want to find , and when do I want to find it? Well, when the batter is halfway between home and first. So I could write that out, like we did there, "The batter is halfway between..." but let me give it some symbols. Let me call the distance the batter is from first base  let me give that a name. Why don't I call that f, for first. So f represents the distance the batter is from first base at any particular time. And so when f equals what? Well, at that moment, when the batter is halfway between this 90footlong run, f must be equal to half of 90, which is 45, so when f equals 45 feet. So that is what we want to find: , how s is changing with respect to time, at the very instant when, in fact, the batter's in the center, which means that the distance, f, is 45 feet.
Well, that's what we want. Now, the next step is to figure out what we already know. Well, we already know something, so what do we know? Well, I know how fast, in fact, this batter is running. How fast is the batter running? Well, I know the batter is running  and we're told that  the batter is running at a rate of 24 feet per second. So that means that this distance is changing at a rate of 24 feet per second. So we know that the rate of change of this length, f with respect to time, is 24. So I could write that as = 24 feet per second.
Now actually, if we really want to be careful, we should maybe think about this in saying, well, if he's running that way and heading toward first base. He's running out there with a rate of 24 feet per second. So that's and that's what we now know, and that's all we know, I think, is that.
Well, now what's the last step? The last step is to find a connection between what we know  how f is changing  and what we want  how s is changing. So really, what I would like is a connection that links f and s together. But what is that connection? Well, maybe you noticed it already. That connection turns out to be the fact that if you look at this part of the field, we have a perfect beautiful right triangle. Now this side may seem as serious until you remember it's a perfect square, so in fact, this is actually 90 feet. So this is 90 feet. This is f and this is s, so we can find a relationship between s, f and 90. In particular, we now note this is the connection, that f^2 + 90^2  those are the two legs squared  that would have to equal s^2. So that's a connection that links the people that are in the story together.
And this is great, because now I have a relationship between f and s. If I now take the derivative of this with respect to time, then what would I see? I would see how s is changing. I would see how f is changing. I would see how f is changing. I know how f is changing. I want to find out how s is changing, and so therefore, I can actually use this relationship to do that. So that's the connection or relationship between the quantities that we're after.
Great, so now we just have to differentiate. So let's now differentiate with respect to time. As always, we're differentiating with respect to time, because that's the independent variable. That's the variable that we want to differentiate with respect to. And in problems about rates, it's always time. So let's take the derivative of this. Since these are not t's  there are other variables that depend upon t implicitly  we must use implicit differentiation. So let's do that. If I take the derivative of x^2 with respect to t, I have to use a little chain rule here. I've got blop squared. The derivative of that is 2 blop  I peeled off the 2  times the derivative of f with respect to t. Well, that has a fancy name. It's called , the derivative of f with respect to t, plus the derivative of 90^2.
Well 90^2 is a constant, so it's derivative is always zero. And what about s^2? Well the derivative of s^2  that's blop squared. The derivative of blop squared is 2 blop, multiplied by the derivative. I peel off the 2. I just peeled off the 2  the derivative of s with respect to t. We have a name for that. We call it .
And that's terrific, because that is now a relationship  a connection  between the very speed  the very rate  that I want and a rate that I already know, so this is perfect. So we can actually insert all the information and actually solve this for the speed we want. So let's see how we do that.
First of all,  now what is ? , we already know, is equal to 24. So this is 24 and we already know what f is going to be at that particular moment. At that particular moment, we'd say that f is 45 feet. So I could put in here 45 x 2, which is 90, times 24, and that equals 2 times s. But what's s? Well, we don't know what s is, though, do we? How are we going to find s at that particular moment in time? Well, we know that f at that moment would be 45. We know this is 90, so we can again use the Pythagorean theorem to figure out what that hypotenuse is.
Let me go off on a little sideline here and compute that for you. So if we have a picture that looks like this, we know this is 90, we know this is 45  it's not drawn to scale  and this s is a question mark. So the Pythagorean theorem says this squared plus that squared equals that squared, so we can solve for s. So therefore, s would equal the positive square root  since it's a length, after all  of 90^2 + 45^2. Well, you can actually work that through or, if you're like me and lazy, I think of 90 as 2 x 45. So here's a little trick I'm going to do here. I'm going to write 90 as 2 x 45. But then I'm squaring it, so that's actually going to be 2^2 x 45^2. With me? 90 is just 2 x 45. When I square it, I get 2^2 x 45^2.
Well, if I have 4 45^2's, and I add an extra 45^2, I have five of them. So this equals . And that's . But is just 45. So actually, I'm going to save myself a little bit of arithmetic. I see it's 45 x . That's that length. You might have done just sort of by working through on a calculator, which is fine. But I just want to point out that if you're a little bit sneaky, you can actually write the answer in nice reduced form without doing any calculation at all.
So 45 is the length of s at the very moment when f = 45. So let's actually write that down. So this is going to be 45 x . So 2 x 45 is 90, times x . There we go. The 2 times the 45 gave me the 90.
And now I can solve for . First of all, notice the 90s drop out. I can cancel away the 90s. And if I divide, I see that = , and that's feet per second and that's the answer. Is the distance between second base and the batter, at that point, changing faster or slower than the batter is actually running toward first base? Well, the batter is running toward first base with a constant rate of 24 feet per second. And here we see that the change  this distance here is . This is actually a smaller number. That means that this distance is actually changing at a slower rate. And that's because there's some offset here, because this point is far away. It's not in the same direction. This is direct going inch is our straight line whereas this thing is offset. So in fact, the rate actually a little bit slower.
Now I want to make just two points here. The first point is, there's actually maybe a slightly different way of writing this problem out which I personally prefer and I want to tell you about that really, really fast. And right over here, we know that the derivative, , is 24 feet per second. That's certainly correct. But if you think about the picture here and think about how I'm measuring it, that f there, you'll notice that as that player is moving through there, that distance is actually shrinking. So if I want to put a sign to this, I might actually want to make this a negative number, because in fact, the distance is shrinking. The thing is closing in. So everything I've done here is completely correct, because this is the way I set the problem up. However, if we thought of the problem a little differently, we might have wanted to think about this fact that it's shrinking as a negative, and I could put in a negative sign here. Now that would only affect things right here when I put in a negative sign here. Everything else is fine. And then I have a negative sign here. This would, though, show that the distance itself is shrinking, which, of course, is what's happening. And now we see that a little more clearly since I've made a better choice of direction, this direction being a negative direction. If you didn't do that, by the way, it's completely correct, because you just force, so you just said that direction will be positive. And so therefore, even though it's shrinking, now in our point of view, shrinking is positive. But if you try to keep shrinking as a negative object, then you can actually write things this way.
And then the last comment I want to make is to tell you about a very classic mistake that people make. And here's the mistake that people make. They come to this relationship and they say the following. They say, "Well, wait a minute. I know that f = 45, so I'll plug in 45 right now into here before I take the derivative, because f is equal to 45." So if we plug in 45 for f, then we get 45^2 + 90^2 = s^2. So if I take the derivative now, that whole thing is just one big constant, so it's derivative is zero and then I have the 2s. So if I plugged in 45 for f right here, then I have all that equals zero, which means = 0.
So what went wrong here? Well, this is classic calculus mistake. What went wrong is, this student inserted the 45 a little too soon. Then you're saying, "Well, wait a minute. Is there a time lag? How do I know when it's too soon?" Well, the answer is, the moment you put in the 45 for f, what you're now telling the problem is that, okay, this runner is no longer moving. In fact, that runner is stationary at 45. There's no movement there. f, this distance, is not changing. It's 45. Well, if the runner were to stop right there, then it's definitely the case that this distance is not changing, so would equal zero. So you found the correct answer for the wrong problem. So what you have to remember is you don't want to insert the particular moment that you want to consider until after you do all the calculus and after you put in all the rates. Otherwise, you lose that.
In particular, when I drew this picture, you'll notice that I did not put the player right in the center, which is the point of action here. And you may have thought that that was either stupid or I made a mistake. But in reality, I never put whatever the thing is at that spot. Because I want to think about this picture as not a static picture, but as a dynamic picture that's changing. And so, in fact, this person is going to slide in and pass that instant in time. And by drawing it there, I might actually get confused and think it's fixed and then, all of a sudden, run into that problem. So please be careful that you don't plug in the data until after you take the derivative.
Anyway, our answer is, the fact that this distance from the little sneakers to second base turns out to be . Okay, we'll try another one up next and we'll see how we make out. Bye for now.
Get it Now and Start Learning
Embed this video on your site
Copy and paste the following snippet:
Link to this page
Copy and paste the following snippet: