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Inflection points (graphical) AP Calculus AB Khan Academy.mp3 | So it's getting a little bit, it's getting a little bit flatter. So our slope is at a very high level, but it's decreasing. It's decreasing, decreasing, decreasing. Slope is decreasing, decreasing even more. It's even more. And then it's actually going to zero. Our slope is zero. |
Inflection points (graphical) AP Calculus AB Khan Academy.mp3 | Slope is decreasing, decreasing even more. It's even more. And then it's actually going to zero. Our slope is zero. And then it becomes negative. So our slope is still decreasing. And then it's becoming more and more and more negative. |
Inflection points (graphical) AP Calculus AB Khan Academy.mp3 | Our slope is zero. And then it becomes negative. So our slope is still decreasing. And then it's becoming more and more and more negative. And then right around, and then right around here, it looks like it starts becoming less negative, or it starts increasing. So our slope is increasing, increasing. It's really just becoming less and less negative. |
Inflection points (graphical) AP Calculus AB Khan Academy.mp3 | And then it's becoming more and more and more negative. And then right around, and then right around here, it looks like it starts becoming less negative, or it starts increasing. So our slope is increasing, increasing. It's really just becoming less and less negative. And then it's going close to zero, approaching zero. It looks like our slope is zero right over here. But then it looks like right over there, our slope begins decreasing again. |
Inflection points (graphical) AP Calculus AB Khan Academy.mp3 | It's really just becoming less and less negative. And then it's going close to zero, approaching zero. It looks like our slope is zero right over here. But then it looks like right over there, our slope begins decreasing again. So it looks like our slope is decreasing again. So it looks like our slope is decreasing. It's becoming more and more and more and more negative. |
Inflection points (graphical) AP Calculus AB Khan Academy.mp3 | But then it looks like right over there, our slope begins decreasing again. So it looks like our slope is decreasing again. So it looks like our slope is decreasing. It's becoming more and more and more and more negative. And so it looks like something interesting happened right over there. We had a transition point. And then right around here, it looks like it starts, the slope starts increasing again. |
Inflection points (graphical) AP Calculus AB Khan Academy.mp3 | It's becoming more and more and more and more negative. And so it looks like something interesting happened right over there. We had a transition point. And then right around here, it looks like it starts, the slope starts increasing again. So it looks like the slope starts increasing. It's negative, but it's becoming less and less and less negative. And then it becomes zero. |
Inflection points (graphical) AP Calculus AB Khan Academy.mp3 | And then right around here, it looks like it starts, the slope starts increasing again. So it looks like the slope starts increasing. It's negative, but it's becoming less and less and less negative. And then it becomes zero. And then it becomes positive. And then more and more and more and more positive. So inflection points are where we go from slope increasing to slope decreasing. |
Inflection points (graphical) AP Calculus AB Khan Academy.mp3 | And then it becomes zero. And then it becomes positive. And then more and more and more and more positive. So inflection points are where we go from slope increasing to slope decreasing. So concave upwards to concave downwards. And so slope increasing was here to slope decreasing. So this was an inflection point. |
Inflection points (graphical) AP Calculus AB Khan Academy.mp3 | So inflection points are where we go from slope increasing to slope decreasing. So concave upwards to concave downwards. And so slope increasing was here to slope decreasing. So this was an inflection point. And also from slope decreasing to slope increasing. So that's slope decreasing to slope increasing. And this is also slope decreasing to slope increasing. |
Undefined limits by direct substitution Limits and continuity AP Calculus AB Khan Academy.mp3 | Let's see if we can figure out the limit of x over natural log of x as x approaches one. And like always, pause this video and see if you can figure it out on your own. Well, we know from our limit properties this is going to be the same thing as the limit as x approaches one of x over, over the limit, the limit as x approaches one of the natural log of x. Now, this top limit, the one I have in magenta, this is pretty straightforward. This, if we had the graph of y equals x, that would be continuous everywhere. It's defined for all real numbers and it's continuous at all real numbers. And so it's continuous, the limit as x approaches one of x is just going to be this evaluated at x equals one. |
Undefined limits by direct substitution Limits and continuity AP Calculus AB Khan Academy.mp3 | Now, this top limit, the one I have in magenta, this is pretty straightforward. This, if we had the graph of y equals x, that would be continuous everywhere. It's defined for all real numbers and it's continuous at all real numbers. And so it's continuous, the limit as x approaches one of x is just going to be this evaluated at x equals one. So this is just going to be one. We just put a one in for this x. So the numerator here, we just evaluate to a one. |
Undefined limits by direct substitution Limits and continuity AP Calculus AB Khan Academy.mp3 | And so it's continuous, the limit as x approaches one of x is just going to be this evaluated at x equals one. So this is just going to be one. We just put a one in for this x. So the numerator here, we just evaluate to a one. And then the denominator. Natural log of x is not defined for all x's and therefore it isn't continuous everywhere, but it is continuous at x equals one. And since it is continuous at x equals one, then the limit here is just going to be the natural log evaluated at x equals one. |
Undefined limits by direct substitution Limits and continuity AP Calculus AB Khan Academy.mp3 | So the numerator here, we just evaluate to a one. And then the denominator. Natural log of x is not defined for all x's and therefore it isn't continuous everywhere, but it is continuous at x equals one. And since it is continuous at x equals one, then the limit here is just going to be the natural log evaluated at x equals one. So this is just going to be the natural log, the natural log of one, which of course is zero. E to the zero power is one. So this is all going to be equal to, this is going to be equal to, we just evaluate it, one over, one over zero. |
Undefined limits by direct substitution Limits and continuity AP Calculus AB Khan Academy.mp3 | And since it is continuous at x equals one, then the limit here is just going to be the natural log evaluated at x equals one. So this is just going to be the natural log, the natural log of one, which of course is zero. E to the zero power is one. So this is all going to be equal to, this is going to be equal to, we just evaluate it, one over, one over zero. And now we face a bit of a conundrum. One over zero is not defined. If it was zero over zero, we wouldn't necessarily be done yet. |
Undefined limits by direct substitution Limits and continuity AP Calculus AB Khan Academy.mp3 | So this is all going to be equal to, this is going to be equal to, we just evaluate it, one over, one over zero. And now we face a bit of a conundrum. One over zero is not defined. If it was zero over zero, we wouldn't necessarily be done yet. That's indeterminate form. As we will learn in the future, there are tools we can apply when we're trying to find limits and we evaluate it like this and we get zero over zero. But one over zero, this is undefined, which tells us that this limit does not exist. |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | So we have the graph of y equals f of x right over here, and we want to figure out three different limits. And like always, pause this video and see if you can figure it out on your own before we do it together. Alright, now first let's think about what's the limit of f of x as x approaches six. So as x, let me do this in a color you can see, as x approaches six from both sides, well as we approach six from the left-hand side, from values less than six, it looks like our f of x is approaching one, and as we approach x equals six from the right-hand side, it looks like our f of x is once again approaching one. And in order for this limit to exist, we need to be approaching the same value from both the left and the right-hand side. And so here, at least graphically, so you never are sure with the graph, but this is a pretty good estimate, it looks like we are approaching one. Right over there, let me do it in a darker color. |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | So as x, let me do this in a color you can see, as x approaches six from both sides, well as we approach six from the left-hand side, from values less than six, it looks like our f of x is approaching one, and as we approach x equals six from the right-hand side, it looks like our f of x is once again approaching one. And in order for this limit to exist, we need to be approaching the same value from both the left and the right-hand side. And so here, at least graphically, so you never are sure with the graph, but this is a pretty good estimate, it looks like we are approaching one. Right over there, let me do it in a darker color. Now let's do this next one. The limit of f of x as x approaches four. So as we approach four from the left-hand side, what is going on? |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | Right over there, let me do it in a darker color. Now let's do this next one. The limit of f of x as x approaches four. So as we approach four from the left-hand side, what is going on? Well as we approach four from the left-hand side, it looks like our function, the value of our function, it looks like it is approaching three. Remember, you can have a limit exist at an x value where the function itself is not defined. The function, if you said what is f of four, it's not defined. |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | So as we approach four from the left-hand side, what is going on? Well as we approach four from the left-hand side, it looks like our function, the value of our function, it looks like it is approaching three. Remember, you can have a limit exist at an x value where the function itself is not defined. The function, if you said what is f of four, it's not defined. But it looks like when we approach it from the left, when we approach x equals four from the left, it looks like f is approaching three. And when we approach four from the right, once again, it looks like our function is approaching three. So here, I would say, at least from what we can tell from the graph, it looks like the limit of f of x as x approaches four is three, even though the function itself is not defined there. |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | The function, if you said what is f of four, it's not defined. But it looks like when we approach it from the left, when we approach x equals four from the left, it looks like f is approaching three. And when we approach four from the right, once again, it looks like our function is approaching three. So here, I would say, at least from what we can tell from the graph, it looks like the limit of f of x as x approaches four is three, even though the function itself is not defined there. Now let's think about the limit as x approaches two. So this is interesting. The function is defined there. |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | So here, I would say, at least from what we can tell from the graph, it looks like the limit of f of x as x approaches four is three, even though the function itself is not defined there. Now let's think about the limit as x approaches two. So this is interesting. The function is defined there. F of two is two. But see, when we approach from the left-hand side, it looks like our function is approaching the value of two. But when we approach from the right-hand side, when we approach x equals two from the right-hand side, our function is getting closer and closer to five. |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | The function is defined there. F of two is two. But see, when we approach from the left-hand side, it looks like our function is approaching the value of two. But when we approach from the right-hand side, when we approach x equals two from the right-hand side, our function is getting closer and closer to five. It's not quite getting to five, but as we go from 2.1, 2.01, 2.001, it looks like our function, the value of our function is getting closer and closer to five. And since we are approaching two different values from the left-hand side and the right-hand side as x approaches two from the left-hand side and the right-hand side, we would say that this limit does not exist. So does not exist, which is interesting. |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | But when we approach from the right-hand side, when we approach x equals two from the right-hand side, our function is getting closer and closer to five. It's not quite getting to five, but as we go from 2.1, 2.01, 2.001, it looks like our function, the value of our function is getting closer and closer to five. And since we are approaching two different values from the left-hand side and the right-hand side as x approaches two from the left-hand side and the right-hand side, we would say that this limit does not exist. So does not exist, which is interesting. In this first case, the function is defined at six, and the limit is equal to the value of the function at x equals six. Here, the function was not defined at x equals four, but the limit does exist. Here, the function is defined at x equals two, but the limit does not exist as we approach x equals two. |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | So does not exist, which is interesting. In this first case, the function is defined at six, and the limit is equal to the value of the function at x equals six. Here, the function was not defined at x equals four, but the limit does exist. Here, the function is defined at x equals two, but the limit does not exist as we approach x equals two. Let's do another function, just to get more cases of looking at graphical limits. So here, we have the graph of y is equal to g of x. And once again, pause this video and have a go at it. |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | Here, the function is defined at x equals two, but the limit does not exist as we approach x equals two. Let's do another function, just to get more cases of looking at graphical limits. So here, we have the graph of y is equal to g of x. And once again, pause this video and have a go at it. See if you can figure out these limits graphically. So first, we have the limit as x approaches five of g of x. So as we approach five from the left-hand side, it looks like we are approaching this value. |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | And once again, pause this video and have a go at it. See if you can figure out these limits graphically. So first, we have the limit as x approaches five of g of x. So as we approach five from the left-hand side, it looks like we are approaching this value. So let me see if I can draw a straight line that takes us, so it looks like we're approaching this value. And as we approach five from the right-hand side, it also looks like we are approaching that same value. And so this value, just eyeballing it off of here, it looks like it's about.4. |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | So as we approach five from the left-hand side, it looks like we are approaching this value. So let me see if I can draw a straight line that takes us, so it looks like we're approaching this value. And as we approach five from the right-hand side, it also looks like we are approaching that same value. And so this value, just eyeballing it off of here, it looks like it's about.4. So I'll say this limit definitely exists, just when we're looking at a graph, it's not that precise. So I would say it's approximately 0.4. It might be 0.41. |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | And so this value, just eyeballing it off of here, it looks like it's about.4. So I'll say this limit definitely exists, just when we're looking at a graph, it's not that precise. So I would say it's approximately 0.4. It might be 0.41. It might be 0.41456789. We don't know exactly just looking at this graph. But it looks like a value roughly around there. |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | It might be 0.41. It might be 0.41456789. We don't know exactly just looking at this graph. But it looks like a value roughly around there. Now let's think about the limit of g of x as x approaches seven. So let's do the same exercise. What happens as we approach from the left from values less than seven? |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | But it looks like a value roughly around there. Now let's think about the limit of g of x as x approaches seven. So let's do the same exercise. What happens as we approach from the left from values less than seven? 6.9, 6.99, 6.999. Well, it looks like the value of our function is approaching two. It doesn't matter that the actual function is defined. |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | What happens as we approach from the left from values less than seven? 6.9, 6.99, 6.999. Well, it looks like the value of our function is approaching two. It doesn't matter that the actual function is defined. The g of seven is five. But as we approach from the left, as x goes 6.9, 6.99, and so on, it looks like our value of our function is approaching two. And as we approach x equals seven from the right-hand side, it seems like the same thing is happening. |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | It doesn't matter that the actual function is defined. The g of seven is five. But as we approach from the left, as x goes 6.9, 6.99, and so on, it looks like our value of our function is approaching two. And as we approach x equals seven from the right-hand side, it seems like the same thing is happening. It seems like we are approaching two. And so I would say that this is going to be equal to two. And so once again, the function is defined there, and the limit exists there, but the g of seven is different than the value of the limit of g of x as x approaches seven. |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | And as we approach x equals seven from the right-hand side, it seems like the same thing is happening. It seems like we are approaching two. And so I would say that this is going to be equal to two. And so once again, the function is defined there, and the limit exists there, but the g of seven is different than the value of the limit of g of x as x approaches seven. Now let's do one more. What's the limit as x approaches one? Well, we'll do the same thing. |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | And so once again, the function is defined there, and the limit exists there, but the g of seven is different than the value of the limit of g of x as x approaches seven. Now let's do one more. What's the limit as x approaches one? Well, we'll do the same thing. From the left-hand side, it looks like we're going unbounded. As x goes.9, 0.99, 0.999, 0.99999, it looks like we're just going unbounded towards infinity. And as we approach from the right-hand side, it looks like the same thing, same thing is happening. |
Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3 | Well, we'll do the same thing. From the left-hand side, it looks like we're going unbounded. As x goes.9, 0.99, 0.999, 0.99999, it looks like we're just going unbounded towards infinity. And as we approach from the right-hand side, it looks like the same thing, same thing is happening. We're going unbounded to infinity. So formally, sometimes informally people might say, oh, it's approaching infinity or something like that, but if we wanna be formal about what a limit means in this context, because it is unbounded, we would say that it does not exist. It does not exist. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | He was a contemporary of Isaac Newton. These two gentlemen together were really the founding fathers of calculus, and they did some of their, most of their major work in the late 1600s. And this right over here is Usain Bolt, Jamaican sprinter, who's continuing to do some of his best work in 2012. And as of early 2012, he's the fastest human alive, and he's probably the fastest human that has ever lived. And you might not, you might have not made the association with these three gentlemen. You might not think that they have a lot in common, but they were all obsessed with the same fundamental question. And this is the same fundamental question that differential calculus addresses. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | And as of early 2012, he's the fastest human alive, and he's probably the fastest human that has ever lived. And you might not, you might have not made the association with these three gentlemen. You might not think that they have a lot in common, but they were all obsessed with the same fundamental question. And this is the same fundamental question that differential calculus addresses. And the question is, what is the instantaneous rate of change of something? And in the case of Usain Bolt, how fast is he going right now? Not just what his average speed was for the last second, or his average speed over the next 10 seconds. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | And this is the same fundamental question that differential calculus addresses. And the question is, what is the instantaneous rate of change of something? And in the case of Usain Bolt, how fast is he going right now? Not just what his average speed was for the last second, or his average speed over the next 10 seconds. How fast is he going right now? And so this is what differential calculus is all about, instantaneous rates of change. Differential calculus. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | Not just what his average speed was for the last second, or his average speed over the next 10 seconds. How fast is he going right now? And so this is what differential calculus is all about, instantaneous rates of change. Differential calculus. Calculus. It's all, and Newton's actual original term for differential calculus was the method of fluxions, which actually sounds a little bit fancier. But it's all about what's happening in this instant. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | Differential calculus. Calculus. It's all, and Newton's actual original term for differential calculus was the method of fluxions, which actually sounds a little bit fancier. But it's all about what's happening in this instant. In this instant. And to think about why that is not a super easy problem to address with traditional algebra, let's draw a little graph here. So on this axis, on this axis, I'll have distance. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | But it's all about what's happening in this instant. In this instant. And to think about why that is not a super easy problem to address with traditional algebra, let's draw a little graph here. So on this axis, on this axis, I'll have distance. So and I'll say y is equal to distance. I could have said d is equal to distance, but we'll see, especially later on in calculus, d is reserved for something else. We'll say y is equal to distance. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | So on this axis, on this axis, I'll have distance. So and I'll say y is equal to distance. I could have said d is equal to distance, but we'll see, especially later on in calculus, d is reserved for something else. We'll say y is equal to distance. And in this axis, we'll say time. And I could say t is equal to time, but I'll just say x is equal to time. X is equal to time. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | We'll say y is equal to distance. And in this axis, we'll say time. And I could say t is equal to time, but I'll just say x is equal to time. X is equal to time. And so if we were to plot Usain Bolt's distance as a function of time, well at time zero, he hasn't gone anywhere. He is right over there. And we know that this gentleman is capable of traveling 100 meters in 9.58 seconds. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | X is equal to time. And so if we were to plot Usain Bolt's distance as a function of time, well at time zero, he hasn't gone anywhere. He is right over there. And we know that this gentleman is capable of traveling 100 meters in 9.58 seconds. So after 9.58 seconds, we'll assume that this is in seconds right over here, he's capable of going 100 meters. And so using this information, we can actually figure out his average speed. Let me write it this way. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | And we know that this gentleman is capable of traveling 100 meters in 9.58 seconds. So after 9.58 seconds, we'll assume that this is in seconds right over here, he's capable of going 100 meters. And so using this information, we can actually figure out his average speed. Let me write it this way. His average speed is just going to be his change in distance over his change in time. And using the variables over here, we're saying y is distance, so this is the same thing as change in y over change in x from this point to that point. And this might look somewhat familiar to you from basic algebra. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | Let me write it this way. His average speed is just going to be his change in distance over his change in time. And using the variables over here, we're saying y is distance, so this is the same thing as change in y over change in x from this point to that point. And this might look somewhat familiar to you from basic algebra. This is the slope between these two points. If I have a line that connects these two points, this is the slope of that line. The change in distance is this right over here. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | And this might look somewhat familiar to you from basic algebra. This is the slope between these two points. If I have a line that connects these two points, this is the slope of that line. The change in distance is this right over here. Change in y is equal to 100 meters. And our change in time is this right over here. So our change in time is equal to 9.58 seconds. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | The change in distance is this right over here. Change in y is equal to 100 meters. And our change in time is this right over here. So our change in time is equal to 9.58 seconds. We started at 0, we go to 9.58 seconds. Another way to think about it, the rise over the run, you might have heard in your algebra class, is going to be 100 meters over 9.58 seconds. So this is 100 meters over 9.58 seconds. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | So our change in time is equal to 9.58 seconds. We started at 0, we go to 9.58 seconds. Another way to think about it, the rise over the run, you might have heard in your algebra class, is going to be 100 meters over 9.58 seconds. So this is 100 meters over 9.58 seconds. And the slope is essentially just the rate of change, or you could view it as the average rate of change, between these two points. And you'll see, if you even just follow the units, it gives you units of speed here. It would be velocity if we also specified the direction. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | So this is 100 meters over 9.58 seconds. And the slope is essentially just the rate of change, or you could view it as the average rate of change, between these two points. And you'll see, if you even just follow the units, it gives you units of speed here. It would be velocity if we also specified the direction. And we can figure out what that is. Let me get a calculator out. Let me get the calculator on the screen. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | It would be velocity if we also specified the direction. And we can figure out what that is. Let me get a calculator out. Let me get the calculator on the screen. So we're going 100 meters in 9.58 seconds. So it's 10.4. I'll just write 10.4. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | Let me get the calculator on the screen. So we're going 100 meters in 9.58 seconds. So it's 10.4. I'll just write 10.4. I'll round to 10.4. So it's approximately 10.4. And then the units are meters per second. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | I'll just write 10.4. I'll round to 10.4. So it's approximately 10.4. And then the units are meters per second. And that is his average speed. And what we're going to see in a second is how average speed is different than instantaneous speed, how it's different than the speed he might be going at any given moment. And just to have a concept of how fast this is, let me get the calculator back. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | And then the units are meters per second. And that is his average speed. And what we're going to see in a second is how average speed is different than instantaneous speed, how it's different than the speed he might be going at any given moment. And just to have a concept of how fast this is, let me get the calculator back. This is in meters per second. If you want to know how many meters he's going in an hour, well, there's 3,600 seconds in an hour. So he'll be able to go this many meters 3,600 times. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | And just to have a concept of how fast this is, let me get the calculator back. This is in meters per second. If you want to know how many meters he's going in an hour, well, there's 3,600 seconds in an hour. So he'll be able to go this many meters 3,600 times. So that's how many meters he can, if you were able to somehow keep up that speed in an hour. This is how fast he's going meters per hour. And then if you were to say how many miles per hour, there's roughly 1,600. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | So he'll be able to go this many meters 3,600 times. So that's how many meters he can, if you were able to somehow keep up that speed in an hour. This is how fast he's going meters per hour. And then if you were to say how many miles per hour, there's roughly 1,600. And I don't know the exact number, but roughly 1,600 meters per mile. So let's divide it by 1,600. And so you see that this is roughly a little over 23, about 23 and 1 half miles per hour. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | And then if you were to say how many miles per hour, there's roughly 1,600. And I don't know the exact number, but roughly 1,600 meters per mile. So let's divide it by 1,600. And so you see that this is roughly a little over 23, about 23 and 1 half miles per hour. So this is approximately, I'll write it this way, this is approximately 23.5 miles per hour. And relative to a car, not so fast. But relative to me, extremely fast. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | And so you see that this is roughly a little over 23, about 23 and 1 half miles per hour. So this is approximately, I'll write it this way, this is approximately 23.5 miles per hour. And relative to a car, not so fast. But relative to me, extremely fast. Now, to see how this is different than instantaneous velocity, let's think about a potential plot of his distance relative to time. He's not going to just go this speed immediately. He's not just going to go, as soon as the gun fires, he's not just going to go 23 and 1 half miles per hour all the way. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | But relative to me, extremely fast. Now, to see how this is different than instantaneous velocity, let's think about a potential plot of his distance relative to time. He's not going to just go this speed immediately. He's not just going to go, as soon as the gun fires, he's not just going to go 23 and 1 half miles per hour all the way. He's going to have to accelerate. So at first, he's going to start off going a little bit slower. So his slope is going to be a little bit lower than the average slope. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | He's not just going to go, as soon as the gun fires, he's not just going to go 23 and 1 half miles per hour all the way. He's going to have to accelerate. So at first, he's going to start off going a little bit slower. So his slope is going to be a little bit lower than the average slope. He's going to go a little bit slower. Then he's going to start accelerating. And so his speed, and you'll see the slope here, is getting steeper and steeper and steeper. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | So his slope is going to be a little bit lower than the average slope. He's going to go a little bit slower. Then he's going to start accelerating. And so his speed, and you'll see the slope here, is getting steeper and steeper and steeper. And then maybe near the end, he starts tiring off a little bit. And so his distance plotted against time might be a curve that looks something like this. And what we calculated here is just the average slope across this change in time. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | And so his speed, and you'll see the slope here, is getting steeper and steeper and steeper. And then maybe near the end, he starts tiring off a little bit. And so his distance plotted against time might be a curve that looks something like this. And what we calculated here is just the average slope across this change in time. But we could see at any given moment, the slope is actually different. In the beginning, he has a slower rate of change of distance. Then over here, then he accelerates. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | And what we calculated here is just the average slope across this change in time. But we could see at any given moment, the slope is actually different. In the beginning, he has a slower rate of change of distance. Then over here, then he accelerates. Over here, it seems like his rate of change of distance, which would be roughly, or you could view it as the slope of the tangent line at that point, it looks higher than his average. And then he starts to slow down again. And when you all average it out, it gets to 23 and 1 1 half miles per hour. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | Then over here, then he accelerates. Over here, it seems like his rate of change of distance, which would be roughly, or you could view it as the slope of the tangent line at that point, it looks higher than his average. And then he starts to slow down again. And when you all average it out, it gets to 23 and 1 1 half miles per hour. And I looked it up. Usain Bolt's instantaneous velocity, his peak instantaneous velocity, is actually closer to 30 miles per hour. So the slope over here might be 23 whatever miles per hour. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | And when you all average it out, it gets to 23 and 1 1 half miles per hour. And I looked it up. Usain Bolt's instantaneous velocity, his peak instantaneous velocity, is actually closer to 30 miles per hour. So the slope over here might be 23 whatever miles per hour. But the instantaneous, his fastest point in this 9.58 seconds, is closer to 30 miles per hour. But you see, it's not a trivial thing to do. You could say, OK, let me try to approximate the slope right over here. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | So the slope over here might be 23 whatever miles per hour. But the instantaneous, his fastest point in this 9.58 seconds, is closer to 30 miles per hour. But you see, it's not a trivial thing to do. You could say, OK, let me try to approximate the slope right over here. And you could do that by saying, OK, well, what is the change in y over the change of x right around this? So you could say, well, let me take some change of x and figure out what the change of y is around it, or as we go past that. So you get that. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | You could say, OK, let me try to approximate the slope right over here. And you could do that by saying, OK, well, what is the change in y over the change of x right around this? So you could say, well, let me take some change of x and figure out what the change of y is around it, or as we go past that. So you get that. But that would just be an approximation, because you see that the slope of this curve is constantly changing. So what you want to do is see what happens as your change of x gets smaller and smaller and smaller. As your change of x gets smaller and smaller and smaller, you're going to get a better and better approximation. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | So you get that. But that would just be an approximation, because you see that the slope of this curve is constantly changing. So what you want to do is see what happens as your change of x gets smaller and smaller and smaller. As your change of x gets smaller and smaller and smaller, you're going to get a better and better approximation. Your change of y is going to get smaller and smaller and smaller. So what you want to do, and we're going to go into depth into all of this and study it more rigorously, is you want to take the limit as delta x approaches 0 of your change in y over your change in x. And when you do that, you're going to approach that instantaneous rate of change. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | As your change of x gets smaller and smaller and smaller, you're going to get a better and better approximation. Your change of y is going to get smaller and smaller and smaller. So what you want to do, and we're going to go into depth into all of this and study it more rigorously, is you want to take the limit as delta x approaches 0 of your change in y over your change in x. And when you do that, you're going to approach that instantaneous rate of change. You could view it as the instantaneous slope at that point in the curve, or the slope of the tangent line at that point in the curve. Or if we use calculus terminology, we would view that as the derivative. So the instantaneous slope is the derivative. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | And when you do that, you're going to approach that instantaneous rate of change. You could view it as the instantaneous slope at that point in the curve, or the slope of the tangent line at that point in the curve. Or if we use calculus terminology, we would view that as the derivative. So the instantaneous slope is the derivative. And the notation we use for the derivative is dy over dx. And that's why I reserved the letter y. And you say, well, how does this relate to the word differential? |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | So the instantaneous slope is the derivative. And the notation we use for the derivative is dy over dx. And that's why I reserved the letter y. And you say, well, how does this relate to the word differential? Well, the word differential is relating this dy is a differential. dx is a differential. And one way to conceptualize it, this is an infinitely small change in y over an infinitely small change in x. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | And you say, well, how does this relate to the word differential? Well, the word differential is relating this dy is a differential. dx is a differential. And one way to conceptualize it, this is an infinitely small change in y over an infinitely small change in x. And by getting super, super small changes in y over change in x, you're able to get your instantaneous slope, or in the case of this example, the instantaneous speed of Usain Bolt right at that moment. And notice, you can't just put a 0 here. If you just put change in x is 0, you're going to get something that's undefined. |
Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3 | And one way to conceptualize it, this is an infinitely small change in y over an infinitely small change in x. And by getting super, super small changes in y over change in x, you're able to get your instantaneous slope, or in the case of this example, the instantaneous speed of Usain Bolt right at that moment. And notice, you can't just put a 0 here. If you just put change in x is 0, you're going to get something that's undefined. You can't divide by 0. So we take the limit as it approaches 0. And we'll define that more rigorously in the next few videos. |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | So the way I'd like to think about it, let's think about what the volume, if I were to just take the bottom function, if I were to just take f of x, if I were to just take f of x, and if I were to rotate that function around y equals x, if I were to rotate this thing around, sorry, around y is equal to one, what would the volume of that be? And then I'm going to subtract from that the volume if I were to take the top function, if I were to take g of x and rotate it around. So let's first of all think about what volume I would get, and this is really the disk method, and I go into it in much more detail earlier in the calculus playlist. But let's think about the volume if f of x is rotated around that axis. And to do that, let's imagine each sliver of that volume. So this is, let me draw a little thing right over here. And you can imagine once this little sliver is rotated, it forms, it forms, or you can imagine this length is the radius of a disk. |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | But let's think about the volume if f of x is rotated around that axis. And to do that, let's imagine each sliver of that volume. So this is, let me draw a little thing right over here. And you can imagine once this little sliver is rotated, it forms, it forms, or you can imagine this length is the radius of a disk. And so, and just to imagine that, let me draw the entire disk. So if this is rotated around, if this is rotated around, it will become a disk. It will become a disk. |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | And you can imagine once this little sliver is rotated, it forms, it forms, or you can imagine this length is the radius of a disk. And so, and just to imagine that, let me draw the entire disk. So if this is rotated around, if this is rotated around, it will become a disk. It will become a disk. It will become a disk that looks something like that. And I'll just call the depth of the disk, so the disk, if you imagine a coin, this is kind of the side of the coin, the depth of the coin, the depth of the coin right over there. I know I can draw that better than that. |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | It will become a disk. It will become a disk that looks something like that. And I'll just call the depth of the disk, so the disk, if you imagine a coin, this is kind of the side of the coin, the depth of the coin, the depth of the coin right over there. I know I can draw that better than that. Let me, so the depth of the coin is just like that. It's a fixed depth, and I'm going to call that dx, so it's just this distance right over here. It is dx. |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | I know I can draw that better than that. Let me, so the depth of the coin is just like that. It's a fixed depth, and I'm going to call that dx, so it's just this distance right over here. It is dx. And what is going to be the area of that coin? Well, the area of the surface of this coin, so let me do it like this. I want to do a different color. |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | It is dx. And what is going to be the area of that coin? Well, the area of the surface of this coin, so let me do it like this. I want to do a different color. So I want to do it in blue. The area of this coin is just pi times the radius of that coin squared. And what is the radius of the coin? |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | I want to do a different color. So I want to do it in blue. The area of this coin is just pi times the radius of that coin squared. And what is the radius of the coin? Well, the radius of the coin is this height, is this height right over here. And what is that height? Well, it is one minus f of x. |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | And what is the radius of the coin? Well, the radius of the coin is this height, is this height right over here. And what is that height? Well, it is one minus f of x. So that is equal to the radius. So the area, the area of kind of the face of this coin is going to be pi, the area of the face of this coin is going to be pi times the radius squared, which is equal to pi times one minus f of x, one minus f of x squared. That's this blue area right over here. |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | Well, it is one minus f of x. So that is equal to the radius. So the area, the area of kind of the face of this coin is going to be pi, the area of the face of this coin is going to be pi times the radius squared, which is equal to pi times one minus f of x, one minus f of x squared. That's this blue area right over here. And then if I want to find the volume of this coin, I would multiply it by the depth of the coin. So times dx. And if I wanted to find the volume of this entire solid, this entire rotated solid, I would want to find the sum of all of these volumes. |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | That's this blue area right over here. And then if I want to find the volume of this coin, I would multiply it by the depth of the coin. So times dx. And if I wanted to find the volume of this entire solid, this entire rotated solid, I would want to find the sum of all of these volumes. So this is just the disk right over here, but I could have another disk, a similar disk that I do right over here. I could have another disk right over here. And I want to take the sum of all of those disks. |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | And if I wanted to find the volume of this entire solid, this entire rotated solid, I would want to find the sum of all of these volumes. So this is just the disk right over here, but I could have another disk, a similar disk that I do right over here. I could have another disk right over here. And I want to take the sum of all of those disks. So I want to take, so the volume is going to be the sum over all of those disks. So x goes from zero, which is this bounding point, to x is equal to 1 1⁄2, times pi times one minus f of x squared. This is the area of the face of each of those disks. |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | And I want to take the sum of all of those disks. So I want to take, so the volume is going to be the sum over all of those disks. So x goes from zero, which is this bounding point, to x is equal to 1 1⁄2, times pi times one minus f of x squared. This is the area of the face of each of those disks. And then I multiply it times the depth of each of those disks. Now this is the volume of each of those disks, and I'm taking the sum of all of them. So this is the volume, this is the volume if I were to just rotate f of x around y is equal to one. |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | This is the area of the face of each of those disks. And then I multiply it times the depth of each of those disks. Now this is the volume of each of those disks, and I'm taking the sum of all of them. So this is the volume, this is the volume if I were to just rotate f of x around y is equal to one. And actually I should just write dx here. And so this right over here, this expression, I just did that, so they really are equal. This is obviously just the volume of each of those disks. |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | So this is the volume, this is the volume if I were to just rotate f of x around y is equal to one. And actually I should just write dx here. And so this right over here, this expression, I just did that, so they really are equal. This is obviously just the volume of each of those disks. So this is the volume if I were to take the f of x around y equals one. Let's figure out, so let me call this volume of f of x. And by the same logic, the same exact logic, we can figure out the volume if we take g of x, if we rotate g of x around, if we rotate g of x, if we construct disks like this and rotate them around y is equal to one. |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | This is obviously just the volume of each of those disks. So this is the volume if I were to take the f of x around y equals one. Let's figure out, so let me call this volume of f of x. And by the same logic, the same exact logic, we can figure out the volume if we take g of x, if we rotate g of x around, if we rotate g of x, if we construct disks like this and rotate them around y is equal to one. And so the volume, if I take g of x around y equals one, would be zero to 1 1⁄2 times pi times one minus g of x, because one minus g of x is each of these radiuses right over here, each of these radiuses. That squared dx. And so the volume of what they're asking us, the volume of the solid generated when r is rotated, well, r is kind of the space in between f of x and g of x. |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | And by the same logic, the same exact logic, we can figure out the volume if we take g of x, if we rotate g of x around, if we rotate g of x, if we construct disks like this and rotate them around y is equal to one. And so the volume, if I take g of x around y equals one, would be zero to 1 1⁄2 times pi times one minus g of x, because one minus g of x is each of these radiuses right over here, each of these radiuses. That squared dx. And so the volume of what they're asking us, the volume of the solid generated when r is rotated, well, r is kind of the space in between f of x and g of x. So it's going to be, the volume is going to be the difference between these volumes. It's going to be this volume, this is kind of the outer volume, and we're gonna take out its hollow core. We're gonna hollow it out by subtracting out this volume. |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | And so the volume of what they're asking us, the volume of the solid generated when r is rotated, well, r is kind of the space in between f of x and g of x. So it's going to be, the volume is going to be the difference between these volumes. It's going to be this volume, this is kind of the outer volume, and we're gonna take out its hollow core. We're gonna hollow it out by subtracting out this volume. So the volume of that region is going to be the integral, I'll do this in a new color, integral from zero to 1 1⁄2 of pi times one minus f of x squared dx minus the integral from zero to 1 1⁄2 of pi times one minus g of x squared dx. And this is a completely valid answer, but you might want to simplify it. We have the same bounds of integration. |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | We're gonna hollow it out by subtracting out this volume. So the volume of that region is going to be the integral, I'll do this in a new color, integral from zero to 1 1⁄2 of pi times one minus f of x squared dx minus the integral from zero to 1 1⁄2 of pi times one minus g of x squared dx. And this is a completely valid answer, but you might want to simplify it. We have the same bounds of integration. We have the same variable of integration. And actually we have this pi over here, so we could factor that out. And so this is the same thing as pi times the integral from zero to 1 1⁄2 of one minus f of x, one minus f of x squared minus one minus g of x, one minus g of x squared, and then all of that dx. |
2011 Calculus AB free response #3 (c) AP Calculus AB Khan Academy.mp3 | We have the same bounds of integration. We have the same variable of integration. And actually we have this pi over here, so we could factor that out. And so this is the same thing as pi times the integral from zero to 1 1⁄2 of one minus f of x, one minus f of x squared minus one minus g of x, one minus g of x squared, and then all of that dx. And then, and actually you probably would want to do this, you probably would want to do this if you're taking the AP exam, not just leave it in terms of f of x and g of x. You would actually want to write the expressions for what f of x, f of x, and g of x are. So really the best answer would probably be pi times the integral from zero to 1 1⁄2 times one minus, well f of x is eight x to the third power, eight x to the third power squared, minus one minus g of x, g of x is sine of pi x, that squared, that squared, times dx. |
Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3 | Let f be the function given by f of x is equal to the square root of x plus 4 minus 3 over x minus 5, if x does not equal 5, and it's equal to c if x equals 5. And say if f is continuous at x equals 5, what is the value of c? So if we know that f is continuous at x equals 5, that means that the limit, the limit as x approaches 5 of f of x is equal to f of 5. This is the definition of continuity. And they tell us that f of 5, when x equals 5, the value of the function is equal to c. So this must be equal to c. So what we really need to do is figure out what the limit of f of x as x approaches 5 actually is. Now if we just try to substitute 5 into the expression right up here, in the numerator you have 5 plus 4 is 9, the square root of that is positive 3, the principal root is positive 3, 3 minus 3 is 0, so you get a 0 in the numerator, and then you get 5 minus 5 in the denominator, so you get 0 in the denominator. So you get this indeterminate form of 0 over 0. |
Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3 | This is the definition of continuity. And they tell us that f of 5, when x equals 5, the value of the function is equal to c. So this must be equal to c. So what we really need to do is figure out what the limit of f of x as x approaches 5 actually is. Now if we just try to substitute 5 into the expression right up here, in the numerator you have 5 plus 4 is 9, the square root of that is positive 3, the principal root is positive 3, 3 minus 3 is 0, so you get a 0 in the numerator, and then you get 5 minus 5 in the denominator, so you get 0 in the denominator. So you get this indeterminate form of 0 over 0. And in the future we will see that we do have a tool that allows us, or gives us an option to attempt to find limits when we get this indeterminate form. It's called L'Hopital's Rule. But we can actually tackle this with a little bit of fancy algebra. |
Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3 | So you get this indeterminate form of 0 over 0. And in the future we will see that we do have a tool that allows us, or gives us an option to attempt to find limits when we get this indeterminate form. It's called L'Hopital's Rule. But we can actually tackle this with a little bit of fancy algebra. And to do that I'm going to try to get this radical out of the numerator. So let's rewrite it. So we have the square root of x plus 4 minus 3 over x minus 5. |
Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3 | But we can actually tackle this with a little bit of fancy algebra. And to do that I'm going to try to get this radical out of the numerator. So let's rewrite it. So we have the square root of x plus 4 minus 3 over x minus 5. And any time you see a radical plus or minus something else, to get rid of the radical what you can do is multiply by the radical doing the, or if you have a radical minus 3 you multiply by the radical plus 3. So in this situation you just multiply the numerator by square root of x plus 4 plus 3 over the square root of x plus 4 plus 3. We obviously have to multiply the numerator and the denominator by the same thing so that we actually don't change the value of the expression. |
Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3 | So we have the square root of x plus 4 minus 3 over x minus 5. And any time you see a radical plus or minus something else, to get rid of the radical what you can do is multiply by the radical doing the, or if you have a radical minus 3 you multiply by the radical plus 3. So in this situation you just multiply the numerator by square root of x plus 4 plus 3 over the square root of x plus 4 plus 3. We obviously have to multiply the numerator and the denominator by the same thing so that we actually don't change the value of the expression. If this right over here had a plus 3 then we would do a minus 3 here. This is a technique that we learn in algebra or sometimes in pre-calculus class to rationalize usually denominators but to rationalize numerators or denominators. It's also a very similar technique that we use oftentimes to get rid of complex numbers usually in denominators. |
Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3 | We obviously have to multiply the numerator and the denominator by the same thing so that we actually don't change the value of the expression. If this right over here had a plus 3 then we would do a minus 3 here. This is a technique that we learn in algebra or sometimes in pre-calculus class to rationalize usually denominators but to rationalize numerators or denominators. It's also a very similar technique that we use oftentimes to get rid of complex numbers usually in denominators. But if you multiply this out, and I encourage you to do it, you notice this has the pattern that you learned in algebra class. This is the difference of squares. Something minus something times something plus something. |
Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3 | It's also a very similar technique that we use oftentimes to get rid of complex numbers usually in denominators. But if you multiply this out, and I encourage you to do it, you notice this has the pattern that you learned in algebra class. This is the difference of squares. Something minus something times something plus something. So the first term is going to be the first something squared. So square root of x plus 4 squared is x plus 4. The second term is going to be the second something or you're going to subtract the second something squared. |
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