Derivatives can be used to calculate limits via l'Hôpital's rule. Given a real-world equation involving two changing quantities, differentiating yields "related rates."
Why would I ever want to take derivatives?
Now, where are we in this course? Well, a long, long time ago, we were looking at limits, right? And we first studied the, the concept of the limit, and we learned how to compute with limits. And then went into application of limits, which was limits of difference quotients, which was the derivative. And we studied some of the concepts of the derivative. What is the derivative telling us about a function? And then we've been studying a turn of techniques for computing derivatives. And we learned about the product rule, the quotient rule, the chain rule. Derivatives of some special functions like e to the x, log x, sines and cosines.
Where does that leave us? We've studied the concepts we've studied some of the computations. Now we're going to study applications. Why would anyone care about derivatives in the first place? Right, what's the real world pay off? There's a couple applications I want to look at now.
One of those applications is just back to limits. If you want to calculate limits. Turns out, sometimes you can use derivatives to calculate limits. That's the story of lopital's rule.
All the time in the real world, you've got equations between different quantities. And maybe you know how one of those quantities is changing, well, if you start with that equation and you differentiate it, that gives you some relationship between the rates, right? The derivative of that equation will tell you how one's things rate of change is related to something else's rate of change. And by translating sort of story problems, sort of real world scenarios. Into equations, and then differentiating those equations. We can learn, about the relationships between these different rates of change. That's the story of related rates. We're going to do a turn of the examples of those, right now
How can derivatives help us to compute limits?
There are plenty of limits that are really quite hard to evaluate. Here's an example. Take the limit as x approaches 4 of the square root of x minus 2, divided by x minus 4. Admittedly, that limit's not really that difficult to calculate. Alright? The limit is 1 4th. But we can also think about this, maybe a little bit more intuitively. Look at what's going on. The limit of the numerator is 0, right? As x approaches 4, the square root of a number close to 4 minus 2 is close to 0. And the limit of the denominator is also 0. And because the limit of the denominator is 0, I can't just use my limit of the quotients as the quotient of limits deal to evaluate this limit, right? I'm genuinely in a difficult situation.. And the problem is that the numerator being close to zero, is trying to make this ratio small. But when the denominator is close to zero, it's trying to make this ratio really big. And neither the numerator nor the denominator win the game. Right? Those 2 forces balance off, giving the answer of 1 4th. But why? Well, we can see this geometrically. So here I've graphed in green the function squared of x minus 2. That's the numerator. And in orange I've graphed x minus 4. Y equals x minus 4. And you can see that when x is equal to 4, these curves cross the x axis because when I plug in 4 here, I get zero. When I plug in 4 here, I get zero. But also note that the green curve is getting closer to zero more quickly than the orange curve. Let's replace those two functions with the approximations that we get by using the derivative. Now, let me replace this green curve with the tangent line to the green curve at the point where the green curve crosses the x-axis. Well, here I go. Here's a picture just of that tangent line. Now I could compute that tangent line by doing a little bit of calculus. Alright. This function is the square root of x minus 2. If I differentiate that I get 1 over 2 square root sub x, that's the derivative of the square root function. And the derivative of minus 2 is just 0. Now I can evaluate that derivative at 4, and I get the derivative is 1 4th, and that's the slope of the tangent line. Here to the green curve. Now I know that my tangent line should cross through the point 4, 0. So here's equation that tangent line in point slope form and I can rewrite this a little bit more nicely as y equals a quarter of x minus 1. So what happens to the limit problem when we replace those expressions by their tangent lines? So instead of thinking about this geometrically, I can just look back at this original limit problem, and replace these pieces by their tangent lines. And we calculated the tangent line to the graph of the square root of x minus 2. To be one quarter times x minus 4, and the denominator is already a straight line. So, if I replace the numerator and denominator by the straight line approximations that are good for inputs near 4, this original limit problem is transformed into this limit problem. But this limit problem is really easy to do, right? I've got an x minus 4 in the numerator and an x minus 4 in the denominator. So this limit is 1 quarter. It's important to emphasize that we really haven't done anything new here. We could have calculated this limit even without thinking about derivatives and tangent lines, but this perspective opens up a new way of approaching some limit problems. So here's how I can package this up. Into a situation that's maybe useful more generally. Suppose I got two functions, I got a function f and a function g. And let's suppose the limit of f as x approaches a is 0 and the limit of g of x as x approaches a is also 0. And just for kicks, let's suppose that f and g are also differentiable, twice differentiable, you know, I want these to be really nice functions. Now let's suppose I want to try to calculate the limit of f of x over g of x, as x approaches a. This might be a difficult limit to calculate. But what we just saw is that we could try to understand this a little bit better by replacing f and g by their tangent line approximations. Another way to say that Is I'm just going to replace f of x and g of x by their approximations that I get from considering their derivative, right? So, the limit of f of x over g of x should be close to this because f of x is close to this, right? It's the value of f at a plus the derivative of f at a times how much the input changes when I go from a to x. That's approximately f of x. And this denominator should be about g of x. It's g of a plus the derivative of g at a times x minus a, that's about g of x. So if you believe that the numerator and denominator here are approximately f of x and g of x, you might then believe that these limits are also equal. But this is a really great situation to be in because f of a say is equal to 0 if f is continuous there, maybe since it's differentiable. And g of a is also equal to 0 since g is continuous. So that these f of a and g of a terms go away. And then I've got an x minus a and an x minus a term there, so those also cancel, and all I'm left with is just f prime of a divided by g prime of a. So it seems like if you try to calculate the limit of this ratio, what's really relevant is understanding something about the derivatives of your numerator and denominator separately. At that point a. The precise statement of this is usually called L'Hopital's rule. So here's a statement of L'Hopital's rule. Suppose I got two functions f and g and they're differentiable for inputs near a, and the limit of f of x and the limit of g of x as x approaches a are both 0. So these functions output is very small when their input is close to a. And the limit of the derivative of f divided by the derivative of g as x approaches a, just exists. And the derivative of g is not 0, when I evaluate that derivative for inputs near a. If all of these conditions are true, then I get the fantastic conclusion that the limit of f of x over g of x. Is the limit of the derivative of f over the derivative of g? And the hope is that this limit is easier to compute then the original limit. Let's try to use L'Hopital's rule in a much trickier context. So here's an example, let's compute the limit as x approaches zero of cosine x minus one minus x quote over two all over x 2 the 4th power. Now we can do it using L'Hopital's rule. So the limit of the numerator is zero and the limit of the denominator is zero. And these are really nice functions, so they're nice enough for us to apply L'Hopital. So I should calculate the limit of the derivative of the numerator, divided by the derivative of the denominator and that might help me compute this original limit. So here we go. The derivative of the numerator is minus sign x, that's the derivative cosine. Minus zero, that's the derivative of 1. Minus x, that's the derivative of x squared over 2. Divided by, what's the derivative of x to the 4th? It's 4x cubed. Now here, I'm in a situation where the limit of the numerator is 0, because the limit of sin x as x approaches 0 is 0 and the limit of x as x approaches 0 is 0. So, the limit of the numerator is 0 and the limit of the denominator is also 0. So here's the funny thing. I could try to apply L'Hopital again to understand this limit. So here we go. I'll use L'Hopital again. I'll differentiate the numerator. The derivative of minus sin is minus cosine. And the derivative of x is 1. So this is minus cosine of x minus 1. Divided by the derivative of the denominator which is 4 times 3x squared, or 12x squared. Now what kind of situation am I in here? Well here, the numerator has limit 0 because cosine's limit is 1 and it's attracting 1. And the limit of the denominator is also 0, so I'm again in a situation where the numerator and denominator are heading toward 0. So to understand this, I could again use L'Hopital to look at how quickly these terms are dying. So I apply L'Hopital again, and L'Hopital tells me to look at the derivative of the numerator. The derivative of minus cosine of x, is sine x. Minus 1, well that's just 0, divided by the limit of the denominator. Well, what's the derivative of the denominator? It's 24x. So now I'm again in a situation where the numerator and the denominator have limit 0. So I could again apply L'Hopital, and it would tell me to look at the derivative of the numerator over the derivative of the denominator. The derivative of the numerator is cosine x and the derivative of the denominator is 24. But now look, the numerator has limit 1, and the denominator has limit 24, it's just a constant. So this is a limit of the quotient and the quotient of limit, so this limit is, Is 1 24th. Now, L'Hopital then tells me that this limit being equal to 1 24th makes this limit equal to 1 24th. And because this limit exists, it then tells me that this limit is also equal to 1 24th. And because this limit then exists it then tells me that this limit is equal to 1 24th. And then because this limit exists it tells me that the original limit is 1 24th.
How can l'Hôpital help with limits not of the form 0/0?
We already know how to calculate limits that are a very small number divided by a very small number a zero over zero form. So we've seen Patel's rule for the situation when limit the numerator limit of the denominator of both zero, but it turns out that rule is also valid when the limit of the numerator and the limit of the denominator are both infinity. I mean you still got all the other conditions. You gotta check that the limit of the ratio of the derivative exists, the derivative of the denominator is not zero. For values of x near a. But given these conditions, you then get the same fantastic conclusion. That the limit of f over g is the limit of the derivative of f over the derivative of g. Let's take a look at an example. Well, here's an example. This is an example, you really don't need to use a lot, but at least it demonstrates what the technique is. The limit on the numerator is infinity, the limit ofthe denominator is also infinity, though that's maybe a little bit harder to see. And consequently We could use lopatol/g, and it would tell us to compute instead the derivative the numerator. Which is 4 x plus 0, divided by the derivative the denominator, which is 6 x minus 1. Now can I calculate the limit of this as x approaches infinity. Yes, of course I could just do this directly too but I could again use lowbatal of I really wanted to. The limit of the numerator is infinity, the limit of denominator is infinity. So lobatall tells me to instead consider the ratio of derivatives, the derivative of the numerator is 4, the derivative of the denominator is 6. And now I'm in a situation where I'm just taking the limit of a constant over a constant. That means that limit is 4 6th, which then tells me how to compute the original limit, but, and this isn't the best way to do this problem, but at least it demonstrates that you could use Lopital in a situation where you got infinity over infinity. Infinity. But not every limit is something going to zero over something going to zero, or something going to infinity over something going to infinity. Sometimes you might have a limit where it's a product, the first term is heading towards zero and the second term is heading towards infinity. For example, here I'm asking what the limit as x approaches infinity of sin of one over x times x. This first term, sine of 1 / x, that's getting close to 0 when x is very large. And the second term, just the x, that is getting very big. So this first term is having a tendency to made this quantity smaller but the second term is having a tendency to make the quantity bigger. You know, who wins?. Well with lobitol/g the only thing that we can really deal with is sort of a 0 over 0 situation or an infinity over infinity situation and that's neither of these. But we can transform this problem into one of these situations. So instead of calculating this limit I'll calculate the equivalent problem, the limit as x approaches infinity. Of sine 1 over x in the numerator divided by 1 over x. So instead of multiplying by something which is going towards infinity, I'm going to divide by its reciprocal. The reciprocal of something going to infinity, is going to 0. So now I've got a situation where the numerator is going to 0, and the denominator is going to 0, and this problem, equivalent to the original problem. Is now amenable to L'Hospital's rule. So by L'Hospital's rule I would want to differentiate the numerator, differentiate the denominator and then look at that limit to try to understand this original limit. Well, what's the derivative of, Of sine of 1 over x, it's cosine of 1 over x, because the derivative of sine is cosine, times the derivative of the inside, which is minus 1 over x squared. And I'm going to divide by the derivitative of the denominator, which is minus 1 over x squared. So, now how do I evaluate this limit? Well, the good news is that I've got a minus 1 over x squared in the numerator and a minus 1 over x squared in the denominator. So this limit is the same as the limit as x approaches infinity of just cosine of 1 over x. But now, 1 over x is getting very close to 0 and what's cosine of a number close to 0? It's 1. So this original limit, is 1. You might have something near 1, being raised to a very high power. Here's an example. The limit as x approaches infinity, of 1 plus 1 over x, raised to the xth power. So for very large values of x, the base here, 1 plus 1 over x, that's close to 1, but the exponent is very large. If you take a number close to 1 and raise it to a very. Very high power it's actually unclear what you're going to get. Depending as to how quickly this is moving towards 1 and how quickly this thing is growing. You get wildly different answers. Now this is not 0 over 0 or infinity over infinity. So I've got to transform this problem into something. Thing that L'Hopital can handle. The trick here is to use exponential functions. So I'm going to rewrite this as e to the log of the limit as x approaches infinity of 1 plus 1 over x to the xth power. E to the log of something does nothing, ritght? These are inverse functions. But, log of a limit is the limit of the log. So this is. E to the limit as x approaches infinity of the log of 1 plus 1 over x to the xth power. Now log as something to a power is that power times the log, so this is e. To the limit as X approaches infinity of X times the log of one plus one over X. Now what kind of situation am I in here? X is very large but log of a number close to one is close to zero. This is big number times number close to zero. That's the infinity times zero indeterminate form, so how am I going to handle this? We just saw a minute ago that I'm going to handle this by. Putting the infinity in the denominator with the reciprocals to make this 0 over 0, the sort of thing that can handle. So this is e to the limit as x approaches infinity of lg 1 plus 1 over x divided by 1 over x. This is the same as this, but now I've got something approaching 0 divided by something approaching 0. This is the sort of situation that can help me with. This is e to the limit, by lopital. The derivative of the numerator divided by the derivative of the denominator. The derivative of log is one over the inside function, times the derivative of the inside, which is minus one over x squared. Divided by what's the derivative of 1 over x, well it's the same thing here minus 1 over x squared and this is the limit as x approaches infinity. Now I've got a minus 1 over x squared which cancels with the minus 1 over x squared in the denominator. And I've got 1 over 1 plus 1 over x as x approaches infinity. This is getting very close to one. So this is e to the 1st power, which is e. And that means that this original limit, the limit of 1 plus 1 over x to the x power as x approaches infinity, is equal to e. Or you might have a very large number, being raised to a very small power. For example, let's say I want to come up with the limit as x approaches infinity of x to the 1 over xth power. So the base here is very large, which have a tendency to make this number very big, but the exponent is getting close to 0, which would have a tendency to pull this back down closer to 1. So what is this? Well we can try to transform this into the sort of limit problem that can handle, and I can again do that with exponential functions. So I can rewrite this as e to the log of the limit of x to the 1 of xth power. And this is the limit as x approaches infinity. Now, this is the log of a limit. Which is the limit of the log of x to the 1 over xth power. And this is the limit as x approaches infinity. But the log of something to a power is that power. So 1 over x times. The log of the base as x approaches infinity. Now I've got 1 over x, which is the number close to zero. Times log of x, which is a very large number. This is zero times infinity, so to speak. So I should try to transform this indeterminate form into something that labetalol can handle. Well, we'll write this a e to the limit of x approaching infinity, of say log x over x. This is infinity over infinity, so to speak. That's the sort of thing that L'Hopital's okay with, so instead of taking this limit. I could look at the ratio of the derivatives. The derivative of log x is 1 over x. The derivative of x is 1. So I should look at the limit of 1 over x over 1 as x approaches infinity. Well, that limit is 0 and e to the 0 is 1, so the limit of x to the 1 over x as x approaches infinity is equal to 1. Or you might have a limit that looks like something going to infinity minus something going to infinity. So let's try to compute the limit as x approaches infinity of the square root of x squared plus x minus x. So this is a very large number minus a very large. The infinity minus infinity situation. So we should try to factor this or rewrite this to get it into a zero over zero or infinity over infinity. The sort of thing I could apply lopitol/g to. 2. So I could try to pull out an x from this, because x is going to infinity. I know x is a large positive number here. So I could rewrite this as x times, so what if I pull out an x from here? That's the square root of 1 plus 1 over x. Minus, when I pull out an extra one here I get minus 1. So really, this limit, for large values of x, so as x approaches infinity, is the same as x times this quantity, the square root of 1 plus 1 over x minus 1. This is a large number. What do I know about this number? Well this is the square root of 1 plus the reciprocal of a large number. This is close to 1 minus 1. This second term is close to zero. This is infinity times 0 in determinate form. So I could rewrite this using our standard trick as 1 plus 1 over x minus 1. So this is now the numerator. This thing's going to 0 divided by 1 over x. The reciprocal of this. But I'm dividing by it, and that's the same as multiplying. So now I've got 0 divided by a number close to 0. I, it's 0 over 0 in determinate form. So lopital tells me I can analyze this by looking at the ratio of the, The derivatives, so I should look at the limit as x approaches infinity of, what's the derivative of this? Well, the derivative of the square root is one over two square root of the inside function, one plus one over x, times the derivative of the inside function. So the derivative of 1 over x. And I don't have to worry about the minus 1, because the derivitive of that's 0. And I divide by the derivitave of the denominator, so I'll just write derivative 1 over x. So now I've got derivative 1 over x in the numerator, derivative 1 over x in the denominator. This limit then, should be same as just the limit of this. So I should be looking at the limit as x approaches infinity of 1 over 2 square root. Of one plus one over x. Well, what's one plus on over x as x approaches infinity, that's just one. So this is one over two square root of a number close to one, this is one half. And, indeed, this original limit really is one half. I mean, honestly, we didn't need l'hopital to calculate that, but we could use l'hopital if we wanted to. To evaluate this limit. Okay, okay. Let's summarize all the possibilities. So here's this summary of everything you might see in the. If you see zero over zero, or infinity over infinity in a limit problem. You can just apply L'Hopital, in that case. But here, I've lifted off some of the other things that you might see. And these in equations, there's no nonsense equations, right? But I hope they kind of tell you what you should do. So if you see, say 0 times infinity in a limit, and by that I mean, it's a product of things, one of which is limit 0, the other of which has limit infinity. Well, you should transform this into one of these cases so you can apply L'Hopital. So you could do that by moving the 0 into the denominator, and taking its reciprocal This is really infinity over infinity. Or, you could move the infinity in the denominator, and take its reciprocal. And now you've got 0 over 0. If you see 1 to a very large power, or something close to 1 to a large power, right? You can use either the log to transform this. And e to the log of this is e to the large number times log of number close to 1. Log of a number close to 1 is close to zero. This is an infinity times 0 indeterminate form. But you know how to handle those. Because you can convert them back into these cases. Which you can then use. If you see infinity to the 0. Well, you could use the same e to the log trick. And then the exponent here is 0 times log of a big number. Or a number close to 0 times log of a big number. But this is the 0 times infinity indeterminate form, which is right here. We should transform you to these cases, which you then use L'Hopital on. If you see 0 to the 0, by which I mean, a number close to 0 raised to a power close to 0, you can again use the e to the log trick And this is e to a number close to 0, times log and a number close to 0. What's log of a very small number? That's very negative. So, this exponent is, again, the 0 times infinity in determinate form, which you can then convert into this and apply L'Hopital To it. The last case is this infinity minus infinity case, and there, one thing you could try to do is put it over some common denominator, so I'm thinking of the common denominator here as being 1 over the product of these two terms, and here I've got 1 over the second term minus 1 over the first term. But the point here is that you can rewrite this difference of very large quantities. As a something getting close to 0 divided by something getting close to 0. Which is the sort of thing that you could then apply L'Hopital.
Why shouldn't I fall in love with l'Hôpital?
People often have this idea that l'Hopital's rule is really great. I'm going to tell you why you shouldn't fall in love with l'Hopital's rule. For one thing, l'Hopital's rule often leads to circular reasoning. Let's go to the blackboard and take a look.
For example, here's a problem that you might be tempted to use l'Hopital on. You know, in a limit as x approaches zero of sine x over x. Now, what do you have to check? Well, let's take a look at the limit of the numerator by itself. The limit of sine x as x approaches zero, sine is continuous so that limit is just sine of zero, which is zero. Numerator's limit is zero. Limit of our denominator? the limit of x as x approaches zero is zero. So, yeah. This is a zero over zero indeterminate form. So, we have tried using l'Hopital. l'Hopital tells us that this limit is equal to the limit as x approaches zero of the derivative of the numerator, which is cosine x over the derivative of the denominator, which is one. Now, this limit is quite a bit easier to do, right? This is really just the limit of cosin x as x approaches zero. Cosine is continuous so this is just the value of cosine at zero, which is one. And yeah, I mean, the limit of sine x over x as x it approaches zero, it is, in fact, one. This looks good, alright? The, the question though is how did you know that the derivative of sine X was cosine x? You needed to know that in order to apply l'Hopital. How did you know the derivative of sine x was cosine x? Well, let's think back.
Right. So, how did I know that the derivative of sine x was cosine x? Well, taking a derivative is the same thing as a limit of a difference quotient. So, to calculate, say, the derivative of sine at zero, I take a limit like this. The limit if h goes to zero of sine 0 + h - sine 0 / h. Sine of zero is just zero. So, calculating this limit, which is calculating the derivative of sine at 0, is really the same as calculating the limit as h goes to zero, sine 0 + h, that's just sign h - 0 / h.
Now, what just happened? I wanted to calculate the derivative of sine at zero, and to calculate the derivative of sine at zero, is really the same as calculating the limit at h goes to zero of sine h over h. It's the definition of derivative. And I was going to use this to evaluate the limit of sine x over x using l'Hopital, right? l'Hopital doesn't just reduce a limit problem down to another limit problem. It reduces a limit problem down to a limit problem and two derivative problems, right? To, to do l'Hopital, you have to be able to differentiate the numerator and denominator. So, if you're going to use l'Hopital to evaluate sine x over x as x goes to zero, you're going to have to differentiate sine x. But then, to differentiate sin x at zero, which is what you're trying to do, to, to use l'Hopital, you have to be able to evaluate this limit, the limit of sine h over h. You have to be able to do the limit you're trying to do already. It's just circular reasoning. So, what just happened? I was trying to calculate the limit of sine x over x as x approaches zero. I used l'Hopital. l'Hopita told me to differentiate sine x. But then, to differentiate sine x, I would've needed to be able to evaluate the limit of sine h over h as h goes to zero, the exact problem I'm trying to solve. It's an example of circular reasoning. There's other things that can go wrong with l'Hopital. One of the conditions of l'Hopital is that the limit of the ratio of the derivatives has to exist. let's see an example where that fails to happen.
Here's an example. Let's take a look at the limit of x plus sine x over x as x approaches infinity. Try to use l'Hopital. So, I want to look at the limit of the numerator by itself. The limit of x plus sine x as x approaches infinity is infinity. The limit of the denominator by itself, the limit of x as x approaches infinity, also infinity. This is an infinity over infinity indeterminate form. It seems good. Now, l'Hopital's tells me that to calculate this, I should look at the limit as x approaches infinity. The derivative of the numerator, which is one plus cosine x over the derivative of the denominator, which is one. Now, what's the limit of one plus cosine x as x approaches infinity? Well, the limit of one is just one. But what's the limit of cosine x as x approaches infinity, right? Cosine just oscillates between -1 and 1. This limit doesn't exist. This limit doesn't exist. Now, does that mean that this limit doesn't exist? I, I, I wrote equals here, but remember, I'm using l'Hopital's rule, alright. And what does l'Hopital's rule actually say? It says, that this limit is equal to the limit of the derivative over the derivative, provided this limit exists. And this limit doesn't exist in this case. So, l'Hopital is silent as to the value of this limit. Let's see if we can actually compute that limit some other way. Alright. So, let's try to do this limit some other way. We're going to do the limit of x plus sine over x as x goes to infinity without using l'Hopital. This is a limit of fraction, we could split up the fraction as a limit as x goes to infinity of x over x plus sine x over x. Now, that's a limit of a sum, which is the sum of the limits provided the limits exist. So, this is the limit as x goes to infinity of x over x plus the limit as x goes to infinity of sine x over x. Now, what's the limit x over x as x goes to infinity? Just one. What's the limit of sine x over x as x goes to infinity? Well, sine x is always between -1 and 1. And x here is going to infinity. I can make x as big as I like. So, question is can I make this as close to zero as I like? Yeah. By making x large enough, a number between -1 and 1 / x can be made as close to zero as I like. So, this limit is zero. So, the sum of the limits is one and that's the limit of the sum. That's the limit of x plus sine x over x. It's one as x goes to infinity. And this is true in spite of the fact that l'Hopital failed us, right? When we did l'Hopital's in this problem, I was told to differentiate the numerator, divided it by the derivative of the denominator as x goes to infinity, and that derivative didn't exist. It was oscillating between zero and two. And in that case, l'Hopital is just silent. l'Hopital requires that limit of the ratio of the derivatives to exist, and if it doesn't exist, l'Hopital doesn't say anything. This limit does exist and then to see that requires some, some algebraic computation. So, we manage to evaluate that limit. The limit was one, but no thanks to l'Hopital. The limit of the ratio of the derivatives didn't exist, it was oscillating. We evaluated that limit by algebraic manipulation. And don't get me wrong. l'Hopital is awfully powerful. There's plenty of situations where you want to use l'Hopital. But often, you can get away with just doing some algebraic manipulation. So, I encourage you, when you're doing those limit problems, don't forget that you can just algebraically manipulate things. There might be an easier way than bringing out l'Hopital. Only use l'Hopital if you absolutely have to use l'Hopital. In other words, don't fall in love with l'Hopital.
How long until the gray goo destroys Earth?
>> Suppose that we have some sort of substance, I'll call it gray goo, that converts anything it touches into more of itself. The rate of growth of this gray goo is proportional to its current size. Well let's say that the constant of proportionality is just 1. So f of t will be the amount of gray goo, and the derivative of f of t is the rate of change in the amount of gray goo. So if the proportionality constant is 1, whatever they're saying is that the rate of change in the amount of gray goo is just the amount of gray goo. So if there's more gray goo, the rate of change in the quantity of gray goo is also higher. What are the units of the derivative? Well let's have time measured in seconds and the amount of gray goo measured in grams. That means the rate of change in the amount of gray goo will be measured in grams per second. Let's suppose that when the experiment begins at time 0, we have 1 gram of gray goo. So in symbols, that says that f of 0 is equal to 1 gram. So after 0 seconds have elapsed, so at the beginning of the experiment, I've just got 1 gram of gray goo. We can write down an equation for f of t. Right. I know a function whose value at 0 is 1 and whose derivative is itself, whose rate of change is proportional to itself with proportionality constant 1. I know that function. That function is just e to the t. How much gray goo is there after 10 seconds have elapsed? So that means I want to calculate f of 10 seconds. I don't know how much material there is after 10 seconds. It's e to the 10. That's approximately 22,000 grams. And 22,000 grams is 22 kilograms. So after 10 seconds, there's about 22 kilograms of gray goo. Okay, here is the big question. How long will it take until the entire Earth is converted into gray goo? So the mass of the Earth is about 6 times 10 to the 27th grams. So I'm looking for a time so that f of t is 6 times 10 to the 27th. In other words, I'm trying to find a value of t so that e to the t is 6 times 10 to the 27th. So I'm looking for t so that e to the t is 6 times 10 to the 27th. Well I can take log of both sides. I'm looking for t equals log of 6 times 10 to the 27th. This is natural log. And this is log of a product which is the sum of the log. So it's log of 6 plus log of 10 to the 27th. Now this is log of something to a power. So that's log of 6 plus 27 times log of 10, and I can approximate these. Log of 6 is about 1.8, and log of 10 is about 2.3. So I've got 1.8 plus 27 times 2.3. That's 63.9, and this is in seconds. So you've got 63.9 seconds until the amount of gray goo is about 6 times 10 to the 27th grams. Once you unleash 1 gram of this material, you've got, what, 64 seconds. In other words, you have no chance to survive. Make your time.
What does a car sound like as it drives past?
>> Calculus is all about rates. And one main example of rates is velocity. Remember velocity is the derivative with respect to time of your position. Velocity measures how quickly your position is changing. Here's a particularly nice example that we can look at. A car driving past you. So I've drawn a diagram of the situation. I've placed you at the origin, so your coordinates are 0, 0. And I've placed the car here at this red dot. Its x-coordinate is v t. I'm thinking of v as the velocity of the car, and t as the current time. And I placed this y coordinate at B so the car is going to drive along the line y equals b at a constant velocity v. Let's compute how far the car is from you. I've drawn a line segment between you and the car. I don't know how long that line segment is. Well, imagining invisible right triangle here, I can use the Pythagorean theorem to measure the length of the hypothenus if I know the lengths of the two legs. This horizontal leg has length vt, the vertical leg has length b. The length of this hypotenuse, the length of this line segment is the square root of the sum of the squares of the lengths of the legs of the right triangle. How fast is the distance between you and the car changing? So I don't know how fast this, the distance between you and the car is changing. So take a derivative with respect to t. So this is the derivative of the square root of something. And the derivative of the square root function is 1 over 2 square root. So I'm going to be evaluating the derivative of the square root at the inside, which is vt squared plus b squared times the derivative of the inside function. So the derivative of bt squared plus b squared. Now I'll just copy down the same thing, 1 over 2 square root, v t squared plus b squared times. What's the derivative of this? Well, it's the derivative of vt squared. So that's 2v squared t, plus the derivative of b, which is 0. So now I could write this all together. I could, for instance, cancel this 2 and this 2, and get that the derivative of the distance between you and the car is v squared t over the square root of v t squared plus b squared. Let's take a look at this with a graph. So here's a graph of that function. Remember, that function is the derivative of the distance between you and the car, so it's negative over here, because the car's distance to you is going down, and it's positive over here, because the car's distance to you is increasing. You can look at it here as a little tiny model. Here's the car. Here's you right? The car maybe starts over here and then it drives past and how does the distance between you and the car change? Well here the distance is getting less, less, less, less, less. This is as close as the car gets to you. More, more, further, further, further away and that's exactly what you see from this graph, right? The derivative's negative here because the car is getting closer to you. It's zero here when the car's as close as it ever gets to you and the derivative is positive over here because the car is getting further away from you. We can hear this in one of the explorations on the website. Here I've got a car, that's driving past. And here in blue is you, the observer. Let's look at what happens when I turn on the sound. It's higher frequency and lower frequency. Higher, lower. You're hearing the derivative.
How fast does the shadow move?
Here is a classic related rates problem. So here's the question. You're walking away from a light source, at some speed, and you don't know how fast is your shadow's length changing. There's a standard method to solving a related rates problem, four steps. So, step 1 is draw a picture, alright? You're told some sort of story, you should convert it into some diagram. Step 2, find an equation. Given that picture, you should label everything in the picture so you can write down some equation. Step 3 is to differentiate that equation which would probably involve the chain rule. And step 4 is to solve that equation for whatever you're interested in finding out. I mean, hopefully you're trying to find some rate of something so, you're going to try to find one of these derivatives in there. Let's draw a picture to represent this person moving away from the light source. Here's my picture, I've got a light source right here and I've got this person right here. And I imagine that person walking away from the light and I've drawn in their shadow. And I've drawn this beam of light just glancing across this person's head. Now we'll turn this picture into an equation using similar triangles. You might suppose that the original problem told us that the lamppost was 3 meters tall, and the person was, say, two meters tall. And I can label the other relevant lengths on this diagram. I'll call x the distance from the bottom of the lamppost to the person's feet, and the distance from the feet to the tip of the shadow, I'll call that s for shadow. So, now I've got a nicely labeled diagram, and I can replace that nicely labeled diagram with a slightly more extract diagram, where the person is now a vertical line, the lamppost is now a vertical line, and I've got these triangles. The up shot here is that I've got similar triangles. This little triangle is similiar to this big triangle. Well, because the angles are all the same, right? Both of these triangles have a right angle here, they have the same angle here, consequently these angles are the same. They're two triangles that have the same angles, they're similar triangles. And as a result, I get this equation out of the diagram. This distance to this distance, x plus s to 3, must be the same as the corresponding ratio in to the other triangle, which is this distance to this distance s over 2. With an equation in hand we can now differentiate. Thinking of my position and the shadow's length as a function of t, I could rewrite this equation as say, x of t plus s of t divided by 3 equals s of t divided by 2, right? My position is a function of time and my show's length is a function of time, and I can differentiate both sides of this equation. The derivative of this is 1 3rd, the derivative of x plus the derivative of s, and the derivative of the other side is one half the derivative of s. And I could solve for the derivative of s. Let me first expand this out. I've got 1/3 derivative of x plus 1/3 derivative of s, is 1/2 derivative of s. I'll subtract 1/3 s prime from both sides. Got 1 3rd x prime is 1 6th s prime. 'Because a half minus a 3rd is a 6th. And then I multiply both sides by 6. So I find out that s prime. This is the rate of change in the shadow's length is twice. X prime this is the speed for which the person is moving. There's a bit of a surprise to this answer. Take a look at this equation, It's telling you that the speed that your shadows length is changing is just twice the speed that your walking. But it doesnt matter where your standing. The value of x doesn't appear on this side only x prime appears on this side. So that's kind of interesting right? The speed with which your shadow is growing only has to do with how fast you're walking. In fact, the speed at which your shadow is growing is exactly twice the speed at which you are walking. Assuming of course, you are two meters tall and the lamp post is three meters tall. I think the important thing to take away from this example is that similar triangles are hugely helpful. If you see similar triangles, use them. There's also one thing that I think is very funny about this particular solution to the problem. Here's the challenge for you to think about. If your velocity is 90% the speed of light, so you're almost going the speed of light. Then this formula would be telling you that your shadow's length would be moving faster than the speed of light, is that really possible?
How fast does the ladder slide down the building?
. A classic related wraith problem involves a ladder leaning up against the side of a building. Well, here's the setup. I've got a wall and the ground, and I've got a ladder resting on one side on the ground, the other side, against the wall. Ladder is 5 meters long and the bottom of the ladder is 3 meters from the base of a wall. I'm going to start pulling the bottom of the ladder away at a speed of one meter per second. Just as I do this, how fast does the top of the ladder start moving down the side of the wall? That's the question. I'm moving the bottom of the ladder. The ladder's sliding down the wall. How fast is this side moving down this wall? So, remember we've got this four step process. The first step is to draw picture. Of course, I think this is already a pretty good picture. The second step is to write down an equation. So, I'm label everything in my diagram and then figure out what the equation is. So, now I want to label this, instead of 3, I'll call this distance from the wall to the base of the ladder, x. And I'll call this distance from the top of the ladder to the bottom of the wall y. And this speed, the speed with which I'm pulling the ladder away, that's really asking how quickly this distance is changing. So that's dx dt. So now I've labeled everything on my picture. So the equation here is just x squared plus y squared equals 5 squared because this ladder, even as it slides down the side of a wall, it still makes a right triangle with this leg, this leg and this is the hypotenuse. And the length of the ladder doesn't change. It's always 5 meters long. So, by Pythagorean theorem, tells me that x squared plus y squared is 5 squared. The third step is to differentiate. I'm going to differentiate the equation, regarding the variables as functions that depend upon t, time. So let's take this equation and differentiate it. So I'm going to be differentiating with respect to time. So d dt of x squared plus y squared is d dt of the length of the ladder. Now the length of the ladder is not changing, right? The derivative of this constant is just zero. How do I differentiate x squared plus y squared? Well, that's the derivative of a sum. So, it's the sum of the derivatives. It's d dt x squared and d dt y squared. Now, I'm thinking of x and y as functions of t. So, I'm going to differentiate these with a chain rule. The derivative of something squared is twice the inside times the derivative of the inside plus, and now I gotta differentiate y thinking of it as a function of t, and that's twice y times dy dt. That's zero. Now I'll evaluate. So, now I just want to evaluate, all right? I know the values of x, dx, dt, and y, right? At this particular moment x is 3 meters, and dx dt is 1 meter per second, and y is 4 meters. And what I'm trying to figure out is dy dt. So, once I plug in what I know, I get this equation. And then I can easily solve for dy dt. And when I solve for dy dt, I get that dy dt is a negative 3/4 meters per second. Does the sign, the sign of that answer make sense? So, yeah. Does it make sense that this is negative 3/4 of a meter per second? Well, take a look at our picture, right? I'm pulling the bottom of the ladder this way, right? So I'm moving the bottom of the ladder this way. And as I do that, the top of the ladder moves down the side of the building. Right? So, as this distance, which is x, is increasing, this distance, which is y, is decreasing. And so, yeah, it totally makes sense that dy dt is a negative.
How quickly does a bowl fill with green water?
We can graph the height of the water. On the x-axis, I've plotted time in seconds. And on the y-axis, I've plotted the water's height in pixels. And here's the curve that I get. Each of these little crosses is one of the measurements that I made at some given time, I measured the height of the water and I get this curve. What do you notice about this graph? So, one thing I notice right off the bat, is that the water level increases fairly rapidly at first, but at the end of this process, the water level is still increasing but it's increasing more slowly. Of course, this makes sense since this bowl is narrower at the bottom than it is at the top, and the water is coming in at a constant rate. A bit more formally, alright, dv/dt, the change in volume of water over time, that's constant. I'm pouring in water at a constant rate, but dh/dt starts off very large, the water's height is increasing very rapidly at first. And by the end of this process, dh/dt is small. So, I'd like a function that relates water height to volume. Well, if I take a look at the side of the bowl, the bowl starts off fairly slanted, and then sort of flattens out, and that's what I'm graphing here. I'm graphing the radius of the bowl. On the x-axis, I'm plotting the height, that's how far I am from the bottom of the bowl. And as I move up the bowl, the radius of the bowl is increasing and that's what I'm plotting on the y-axis. And if you look at the bowl, right, the bottom of the bowl starts off very slanted and then it sort of flattens out. And that's exactly what I'm representing here with, with this broken line, right? The bowl starts off fairly slanted and then flattens out. Once I know what the radius of the bowl is at a particular height, I can then think about the volume of water when then water level is at some given height. So, once I know the radius of the bowl at a particular height, from the bottom of the bowl, I can then make a graph that shows the volume of water in the bowl, given a particular height of water. So, here on the x-axis, I'm plotting the height of the water. And on the y-axis, I'm plotting the volume of water in the bowl. And you can see that when the water is deeper, there's more water in the bowl. It's an increasing function. Now, let's think a little bit about the slope of the tangent line to this graph. The tangent line initially has not very large slope, compared to after a while, the tangent line has a much higher slope. And what that really means is that initially, for a given change in volume, there's quite a change in the height of the water. But after a while, the same change in volume leads to a less dramatic change in the height of the water. That really makes sense from our original experience, we poured all that water into the bowl. We think back to a graph of the water level in the bowl. Here, I've got time. Here, I've got the height of the water. The water is flowing in at a constant changing volume per unit time. And initially, aIright, the slope of this tangent line is quite high, because the water height is increasing very rapidly, but after a while, because there's the same change in volume during this entire process, the change in the water height is less dramatic, right? It's the same amount of water coming in, but the water's moving up the sides of the bowl less quickly.
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