WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, and welcome back to AP Calculus.
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Today, we are going to talk about linear approximations and differentials.
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Let us jump right on in.
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Essentially, what we are going to do is we are going to have some function f(x).
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Instead of f(x), we are going to come up with something called a linearization of x
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which we are just going to call l(x).
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It is going to allow us to use this, instead of this, when we are not too far away from a particular point whose value we do know.
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Let us jump right on in.
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I’m going to start off by drawing a picture here.
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Let me go like this and something like that.
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Let us see we have got some curve, some function, and then, we have the tangent line.
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This is going to be our x0 and let us say over here is our x.
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If we go up, we got this point and this point.
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This point is going to be our x0, y0.
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This point over here, this is going to be our xy.
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It is going to be the y value at another point along the curve.
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This is our f(x), it is the actual f(x).
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Now this line here, this line, it is y - y0 = m × x - x0.
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m is the slope of the line, the derivative.
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y0 and x0 those are the points that it actually passes through.
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We know what this is.
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This height right here, this height is just y0.
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y is the same as that right there.
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This height, this is m × x - x0.
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x - this difference right here, this is x - x0.
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If I multiply, I have a bigger triangle here.
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If I'm here, if I move a distance x - x0, the height, this part is nothing more than the slope × the distance.
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I hope that makes sense.
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Remember from the slope is Δ y/Δ x =, the Δ y,
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in other words how high you move is nothing more than the slope × the Δ x.
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The slope × this is your Δ x, you have your slope.
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This height right here.
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Let me write this down now.
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The whole idea of the derivative and the tangent line which this is to a curve,
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is to be able to approximate the curve, approximate values along the curve.
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I will explain what I mean using the picture, after I finish writing this,
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is to be able to approximate the values along the curve by using the tangent line instead.
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As long as we do not move to far away from our x0.
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If we know some point on the curve, this tangent line gives me an approximation.
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I know if I do not move too far away from x0, either in this direction or in this direction,
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in this case, let us just worry about this direction.
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If I do not move too far away, the values of x along the curve are going to be all these values right here.
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But if I do not move too far away, then this value right here is a pretty good approximation to this value right here.
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It gets better and better, the closer I actually stay to x0.
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That is what this is all about.
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The whole idea of the derivative is to approximate a complicated curve near a certain point.
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If I want the value at a point near a point that I know, I do not have to use the function itself.
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I can just go ahead and use the derivative of the function.
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It gives me an approximation and that is what this is.
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Instead of finding this value, I find this value.
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This value comes from this.
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If I just move this y0 over to this side, I get this y value is actually equal to y0 + this height.
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That is what this equation is actually saying.
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If I rearrange this equation, rearranging the equation of the tangent line
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gives my y value is actually equal to my y0 value + m × x – x0.
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This is the value that I really like, but I’m going to approximate it by that.
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y0 is this height, m × x - 0 is this height.
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If I add those two heights which is what this says, I get this y value which is a pretty good approximation to that.
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That is the whole idea, this is the linearization.
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Now let me do this in terms of f(x).
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Let us do it again.
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I'm going to change the labels but it is still the same thing.
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I have got a graph, I have a tangent line.
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This is my x0, this is my x.
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This point right here, this is x₀, f(x)₀.
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Let me write this a little bit better.
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Instead of y, I’m going to call it f(x)₀.
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This point is x₀, f(x)₀.
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This point right here, this is nothing more than x f(x).
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Now this height, let me go ahead and go this height.
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This is nothing more than f(x0), right.
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If I go straight across that means this height is nothing more than f(x0).
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This height is f’ at x0 × x - x0 because the equation of this line is f(x) - f(x0) = f’ at x0 × x – x, y - y0 = m × x - x0.
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I just replaced y, y0 with f(x) and f(x0).
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Therefore, this point which is an approximation to this point is this height + this height gives me an approximation.
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Now we have f(x) rearranging this, moving this over to that side, = approximately f at x0 + f’ at x0 × x - x0.
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That will take me to that point right there.
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It is almost equal to that point right there.
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Again, this is greatly magnified.
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We call this the linear approximation.
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We call this the linear approximation of f(x) at x₀.
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It is symbolize with the l.
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The linear approximation is equal to f(x0) + f’ at x0 × x - x0.
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It is also called the linearization of f(x) at x0.
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It is also called the linearization of f(x) at x = 0.
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When you are given an x0, you are going to find the f(x0), you are going to find the f’.
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At x0, you are going to put those numbers here and here.
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You are going to form this thing.
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Now that you have a linearization, now you can replace f(x).
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You can replace, now l(x) replaces f(x).
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Instead of using f(x), you can use l(x) instead.
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That is what is going on here.
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When you create this linearization, this function, which is going to be some function of x,
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you can use that instead of f(x) directly.
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It is an approximation.
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Instead of dealing with the function directly, you are going to deal with the linear approximation to the function.
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You are going to deal with the equation of the tangent line, that is what is happening here.
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Let us do an example, I think it will make sense.
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Example 1, use linear approximations to find the natlog of 5.16.
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What it is that is actually happening here?
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I will draw it out real quickly.
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We know that the natlog function is something like that, it crosses at 1.
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We want to know what the natlog is at 5.16.
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Here is 5, let us just say that 5.16 is right there.
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This is 5.16.
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We want to know this value.
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We will use linear approximation not the actual function itself.
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5.16 is close to 5, we go to 5.
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We draw a tangent line at 5.
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I’m going to use this equation, the linearization.
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In other words, the equation of the tangent line to find that value.
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If I magnify this area, it is going to look something like this.
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The curve is going to be like that, then it is going to be like this.
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Here is where the actual point is, I’m going to find this one that is really close to it.
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Let us go ahead and do it.
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Let us recall our linearization is equal to f(x0) + f’ at x0 × x - x0.
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5.16 is close to 5, our x0 is 5.
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Our x that we are interested in is 5.16, we have that.
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Again, it needs an x0.
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Our x0, in this case is 5, just based on what the problem is asking.
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5.16 is closest to 5.
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The function itself is the natlog of x because the natlog function is what we are dealing with.
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The derivative f’(x) = 1/x.
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f(5) is equal to 1.609, f’ at 5 = 1/5.
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Therefore, the linear approximation, the substitution that I can make for f(x) for the natlog is f(x0).
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X0 was 5, f5 is 1.609 + f’ at x0.
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F’x0 is 5, f’ at 5 is 1/5 + 1/5 × x – x0.
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There you go, this is my actual function of x that I can substitute now for the log of x.
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This is the linearization function.
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Now I want to solve it.
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I want log of 5.16.
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Therefore, the l(5.16) is equal to 1.609 + 1/5, x is 5.16 – 5.
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When I solve this, I get 1.6414, that is my answer.
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I identify my function, log(x).
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I identify my x0, I have the linearization equation.
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I calculate the f at x0, I calculate f’ at x0.
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I put it in, now I have my linearization function.
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This is a substitute for the original function log(x).
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l(x) replaces f(x).
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l(x) approximates, that is why we can replace it, approximates f(x).
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This function is an approximation.
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When I'm close to 5 for ln(x).
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That is what is happening here.
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By the way, the actual value of the natlog of 5.16 is equal to 1.6409.
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Not bad, 6409, 6414, that is very good approximation.
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As long as we do not deviate too far from x0, either in this direction or this direction,
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the tangent line is a good approximation to the curve.
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Clearly, if that is your curve, as you get further and further away from the x value,
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it is going to deviate from the curve.
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Now you are here and here as opposed to here and here.
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The idea is to stay as close as possible.
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Choose your x0 wisely.
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Here are our f(x), once again, is the natlog of x.
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The linearization was 1.609 + 1/5 × x – 5.
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We asked for the natlog of 5.16 which is the l(5.16).
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In this case, the absolute value of the linearization, the actual f(5.16) – l(5.16),
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the actual value - the linear approximation.
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This is the actual - the linearization was 1.6409 - 1.6414, actually gave us 5 × 10⁻⁴.
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The error was on the order of 10⁻⁴.
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In general, what you can do is, if you know what error you want to be within, you can adjust your choice of x.
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Here we just happen to pick 5.16.
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We said it is not that far from 5, let us choose x0 = 5.
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If you want your error to be within a certain number, if you want to keep your error,
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let us say to a 0.1 or 0.01, 0.056, whatever it is that you want, you can actually adjust and
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it will tell you how far you can deviate from the x0 that you choose.
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In general, if you want to, I should say if you want,
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the error between f(x) and l(x) to be within a certain bound,
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then you have to solve the following.
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You have to solve that.
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You have to put in, in this case it would be ln(x) - 1.609 + 1/5 x – 5 under the absolute value sign.
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You have to solve the certain bound b.
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I should say a certain bound b.
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You have to solve this equation either analytically or graphically.
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As you will see in a minute, graphically is probably the best way to do this.
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Let us go ahead and do an example, I think it will make sense.
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Our example number 2, we are going to let f(x) equal the 3√5 - x.
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The question is for what values of x will the linearization at x₀ equal 1 be within 0.1?
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We are saying the x is equal to 1.
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How far can I move to the right of 1 and to the left of 1, to make sure that the error,
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the difference between the actual value of the function and the linear approximation of the function stays less than 0.1?
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Now I’m specifying the error that I want, now I need to know how far I can move away.
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Let us go ahead and do it.
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f(x) is equal to the 3√5 - x which is equal to 5 - x¹/3.
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l(x), we know that l(x) is f(x0) + f’ at x0 × x - x0.
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Let us find f(x0) first.
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Let me go ahead and do this in red.
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I have got f(x0).
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x0 is 1, f1 is equal to 5 - 1 is 4, 4¹/3.
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It is going to be the √4 which is going to equal 1.5874.
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I have taken care of that, now I need to find f’ at 1.
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f’ at x is equal to 1/3 × 5 – x⁻²/3 × -1.
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When I put in 1, I will write it this way, -1/3 × 5 – x⁻²/3.
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Therefore, f’ at x0 is f’ at 1.
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I put 1 into this and I end up with, -1/3 × 5 – 1⁻²/3 = -0.1323.
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I have taken care of that.
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I will go back here, I put this number and this number into here and here.
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Now I have my linearization, my l(x) is equal to 1.5874 - 0.1323 × x - x0 which is 1.
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This function can now replace this function, if I'm not too far from 1.
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If x is 1.1, 1.2, 1.3, whatever.
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If I’m not too far away, I can replace this function with a linearization.
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It is saying, how can I keep this error between the two?
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How can I keep the error less than 0.1?
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We have to solve f(x) – l(x) less than 0.1.
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We want the absolute value, we want the difference.
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We do not care about the positive or negative.
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That is why there is an absolute value sign.
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What we want is this, f(x) is 5 - x¹/3 - 1.5874 - 0.1323 × x – 1.
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We want that to be less than 0.1.
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This is the same as 1.5874 - 0.1323 × x - 1 less than greater than 5 – x¹/ 3 + 0.1, 5 - x¹/3 - 0.1.
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I hope that makes sense why that is the case.
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We have an absolute value sign, get rid of the absolute value sign.
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You put a -0.1 here.
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Then, I just move this function over that side, move this function over to that side.
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I think the best thing to do is graph this.
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You are going to graph the function, you are going to graph the function -0.1.
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You are going to graph the function + 0.1.
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You want this thing to be between those two graphs.
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This is a straight line, this is the equation of the tangent line to this function at the point 1.
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What you get is the following.
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When we graph this, we get this.
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The middle graph is the actual graph itself.
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Let me write that down.
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The middle graph that this is one, that one.
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The middle graph, that is the function itself, 5 – x¹/3.
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The lower graph, that one, that is 5 - x¹/3 - 0.1.
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The upper, exactly what you think, it is 5 - x¹/3 + 0.1, that is that one.
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The line, this, that is the linearization, that is the tangent line at the point 1.
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Notice it touches the graph at x = 1.
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The black line is the linearization.
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The black line is l(x).
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In order for the difference between l(x) to be between 0.1 above and 0.1 below,
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we need to make sure that this tangent line,
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we see where it actually touches either the upper or the lower graph and we read off the x values.
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As long as the linearization stays between the upper and lower which comes from what we just did in the previous page,
00:27:59.600 --> 00:28:03.800
remember we solve the absolute value, we rearranged it.
00:28:03.800 --> 00:28:08.300
We put 0.1, -0.1, we move the function over.
00:28:08.300 --> 00:28:12.600
That inequality has to be satisfied, that inequality is this graph.
00:28:12.600 --> 00:28:17.700
Instead of solving analytically, just do it graphically and you can just see
00:28:17.700 --> 00:28:22.200
where it touches the upper and lower, and read off the x values.
00:28:22.200 --> 00:28:23.100
Let us write that down.
00:28:23.100 --> 00:28:42.600
Wanting the absolute value of f(x) – l(x) to be less than 0.1 which is what this graph says means
00:28:42.600 --> 00:29:06.300
we want the tangent line which is l(x) between the upper and lower curves, the upper and lower graphs.
00:29:06.300 --> 00:29:15.600
Just read off the x values.
00:29:15.600 --> 00:29:22.200
When you read off the x values, you have -2.652.
00:29:22.200 --> 00:29:28.800
That means as long as x, here is 1, this is our x₀.
00:29:28.800 --> 00:29:39.000
As long as x goes that far and goes this far, as long as x is between there and there,
00:29:39.000 --> 00:29:52.300
my linear approximation will give me an approximation to the actual function itself, the 5 – x³, to an error of 0.1.
00:29:52.300 --> 00:29:53.300
That is what this means.
00:29:53.300 --> 00:30:00.500
As long as I stay within this and this, the tangent line itself does not go past 0.1.
00:30:00.500 --> 00:30:03.200
By specifying an error, I solve this thing.
00:30:03.200 --> 00:30:17.700
Graphing this thing, I see where the tangent line touches and I read off the x values, less than 3.398.
00:30:17.700 --> 00:30:18.900
This is there.
00:30:18.900 --> 00:30:28.200
As long as x is in this interval, the difference between f(x) and the actual linearization is less than 0.1.
00:30:28.200 --> 00:30:32.700
I hope that made sense, that is what you are doing.
00:30:32.700 --> 00:30:35.200
Now let us talk about differentials.
00:30:35.200 --> 00:30:41.800
Essentially the same thing except in calculus notation.
00:30:41.800 --> 00:30:46.000
It is really simple.
00:30:46.000 --> 00:30:48.700
Let y equal x³.
00:30:48.700 --> 00:30:54.100
We know that dy/dx is equal to 3x².
00:30:54.100 --> 00:30:58.600
Let us move this over, this is just a number, a small number.
00:30:58.600 --> 00:31:05.200
dy = 3x² dx, this is the differential.
00:31:05.200 --> 00:31:15.100
The differential of x³ at a particular x, at a particular x₀ is 3 × x₀² dx.
00:31:15.100 --> 00:31:42.400
This says, if I change my x value by a small amount, and the small amount is dx,
00:31:42.400 --> 00:31:52.300
then the y value changes by this small amount.
00:31:52.300 --> 00:32:06.700
The y value changes by this small amount, in terms of graphs.
00:32:06.700 --> 00:32:10.900
Same thing that we did, except now we are talking about really tiny motions.
00:32:10.900 --> 00:32:19.900
We have our function, we have our tangent line, this is our x0.
00:32:19.900 --> 00:32:33.100
Now this distance is dx, this distance is dy.
00:32:33.100 --> 00:32:34.900
We know what the distance dy is.
00:32:34.900 --> 00:32:41.000
It is just of the slope × dx.
00:32:41.000 --> 00:32:44.000
There you go, the slope at a given point.
00:32:44.000 --> 00:32:47.600
X0 is that, the slope is the derivative.
00:32:47.600 --> 00:32:50.100
This is just notation, that is all it is.
00:32:50.100 --> 00:32:52.500
It is the exact same thing.
00:32:52.500 --> 00:32:54.600
There you go.
00:32:54.600 --> 00:33:00.300
If you want to know how the y value changes when you make a small change in x,
00:33:00.300 --> 00:33:04.500
just move this x over and multiply it by the slope, the derivative at that point.
00:33:04.500 --> 00:33:11.700
From a given point, your change in y is going to be this much, if you change x by this much.
00:33:11.700 --> 00:33:15.300
That is all this says, the differential.
00:33:15.300 --> 00:33:27.000
If I said what is the differential of the function x³ at x0 = 2, I will in plug 2 to here.
00:33:27.000 --> 00:33:30.400
2 × 2 is 4, 3 × 4 is 12.
00:33:30.400 --> 00:33:34.900
We have got dy = 12 dx.
00:33:34.900 --> 00:33:45.200
If I move away from 2, a distance of 0.1, y is going to change by 12 × 0.1.
00:33:45.200 --> 00:33:46.800
That is what this is saying.
00:33:46.800 --> 00:33:50.200
If I change x by this much, how much is that going to change?
00:33:50.200 --> 00:33:52.000
It is the same thing here right.
00:33:52.000 --> 00:33:53.200
It is a rate of change.
00:33:53.200 --> 00:33:57.100
If I change x by a certain amount, how much is y going to change?
00:33:57.100 --> 00:34:00.700
Except now I have a dx here, all I have done is actually move it over.
00:34:00.700 --> 00:34:04.300
There is nothing different than what it is that we are doing.
00:34:04.300 --> 00:34:09.100
We are just looking at it slightly differently.
00:34:09.100 --> 00:34:14.800
Let us do an example, very simple.
00:34:14.800 --> 00:34:35.500
Example 3, what is the differential of f(x) = e ⁺x³ + sin x?
00:34:35.500 --> 00:34:38.300
Very simple, just take the derivative.
00:34:38.300 --> 00:35:03.000
This is just y = e ⁺x³ + sin(x), dy dx = e ⁺x³ + sin(x) × 3x² + cos(x).
00:35:03.000 --> 00:35:10.300
Therefore, the differential dy is equal to 3x² + cos(x).
00:35:10.300 --> 00:35:21.700
I just decided to move it over to the left, × e ⁺x³ + sin(x) dx.
00:35:21.700 --> 00:35:27.700
For a particular x value, at a particular x value, from that x value, from that x0,
00:35:27.700 --> 00:35:35.500
I should say from that particular x0 value, if I move away to the left or to the right by dx,
00:35:35.500 --> 00:35:40.600
the value of my function, the vertical movement is going to be that.
00:35:40.600 --> 00:35:42.400
That is all this is, the differential.
00:35:42.400 --> 00:35:48.200
If I have a differential movement in x, what is my differential change going to be in y, that is all this is.
00:35:48.200 --> 00:35:56.900
We actually call that the differential, when we have moved the dx over.
00:35:56.900 --> 00:36:02.000
Let us try another example here.
00:36:02.000 --> 00:36:20.000
Let us try example 4, use differentials to evaluate e⁻⁰.02.
00:36:20.000 --> 00:36:23.300
Linear approximation, differential, it is essentially the same thing.
00:36:23.300 --> 00:36:28.100
We are just different notation.
00:36:28.100 --> 00:36:40.400
-0.02 is close to 0, I’m going to take x0 = 0.
00:36:40.400 --> 00:36:42.500
That is going to be my base point.
00:36:42.500 --> 00:36:48.800
The function that I’m interest in is, clearly y = e ⁺x.
00:36:48.800 --> 00:36:54.200
Dy dx = e ⁺x.
00:36:54.200 --> 00:37:03.200
Therefore, dy = e ⁺x dx.
00:37:03.200 --> 00:37:13.500
At the point it is equal 0, we get the differential is equal to e⁰ × dx.
00:37:13.500 --> 00:37:21.000
e⁰ is 1, dy = just 1 dx.
00:37:21.000 --> 00:37:28.800
Our dx from 0, our 0 point, we are at -0.02, we are moving this way.
00:37:28.800 --> 00:37:32.200
Now we are at 0.02.
00:37:32.200 --> 00:37:44.800
Our dy is equal to 1 × -0.02.
00:37:44.800 --> 00:38:11.800
Therefore, e⁻⁰.02 is equal to e⁰ + dy + the differential from that point.
00:38:11.800 --> 00:38:18.700
I’m going to evaluate it at that point and if I change by dx, dy is going to change this much.
00:38:18.700 --> 00:38:36.100
From 0, if I go to 0.2, it is going to be e⁰ +, in this case I calculated that the differential is -0.02 =, e⁰ is 1 + -0.02 = 0.90.
00:38:36.100 --> 00:38:43.300
That is it, nothing particularly strange about this.
00:38:43.300 --> 00:38:46.600
I hope that makes sense.
00:38:46.600 --> 00:38:50.300
Let us finish off with a nice little problem here.
00:38:50.300 --> 00:38:58.100
The radius of a circular table is measured to be 30 inches with a maximum error and measurement of 0.4 inches.
00:38:58.100 --> 00:39:06.500
We have a table and measure the radius to be 30.
00:39:06.500 --> 00:39:15.200
At a possible error 0.4 inches which means it could be 29.6 or could be 30.4.
00:39:15.200 --> 00:39:22.100
It might actually be a slightly smaller or slightly bigger.
00:39:22.100 --> 00:39:31.400
What is the possible error in the calculated area of the table between the 30 + 0.4, the 30 - 0.4?
00:39:31.400 --> 00:39:33.000
What is that possible error?
00:39:33.000 --> 00:39:45.300
In other words, how much of a difference, what is the outside extra area or if it is smaller what is the inside extra area?
00:39:45.300 --> 00:39:54.300
What is the error in the actual area, if I have a possible error of 0.4 away from 30?
00:39:54.300 --> 00:39:57.300
Express the error as relative error as well.
00:39:57.300 --> 00:39:59.300
We will do both of those.
00:39:59.300 --> 00:40:04.700
The area, we know that the area of the circle is π r².
00:40:04.700 --> 00:40:19.100
The differential, if I change 30 to 30.4, or to 29.6, that 0.4 is my differential, that is my dr.
00:40:19.100 --> 00:40:24.500
Dr here equal 0.4 + or -.
00:40:24.500 --> 00:40:34.300
The change in the area, the differential of the area, da dr = 2π r.
00:40:34.300 --> 00:40:40.100
Therefore da = 2π r dr.
00:40:40.100 --> 00:40:41.900
I change my area by a certain amount.
00:40:41.900 --> 00:40:46.400
My area is going to change by a certain amount.
00:40:46.400 --> 00:40:50.300
This dr is the error in the radius.
00:40:50.300 --> 00:40:53.600
Da is going to be the error in the area.
00:40:53.600 --> 00:40:58.100
That is what they are asking, estimate the possible error in the calculated area.
00:40:58.100 --> 00:41:02.000
We are actually going to express it as a differential.
00:41:02.000 --> 00:41:09.200
Here r = 30, dr = 0.4.
00:41:09.200 --> 00:41:21.900
Therefore, da is equal it 2 × π × 30 × 0.4.
00:41:21.900 --> 00:41:34.500
da = 75.4 in², that is the error.
00:41:34.500 --> 00:41:41.400
The error is 75.4 in².
00:41:41.400 --> 00:41:51.300
If I made a measurement of 30, if they are telling me that my error is off by 0.4 inches + or -,
00:41:51.300 --> 00:42:02.700
that means the area that I calculated, the error in the area is going to be +75.4 or -75.4.
00:42:02.700 --> 00:42:07.200
That is how big of a difference my error in the area is going to be.
00:42:07.200 --> 00:42:10.500
It is just using differentials, that is all it is going on here.
00:42:10.500 --> 00:42:14.400
Relative error, let us go to blue.
00:42:14.400 --> 00:42:29.700
Relative error is the error divided by the calculated value.
00:42:29.700 --> 00:42:32.400
A couple of ways that we can do this.
00:42:32.400 --> 00:42:37.900
The error itself we calculated, that was 2π r dr.
00:42:37.900 --> 00:42:42.400
The actual value, the area is π r².
00:42:42.400 --> 00:42:45.400
The π cancels π, the r cancels r.
00:42:45.400 --> 00:42:51.100
You get 2 × dr/ r.
00:42:51.100 --> 00:43:06.800
Here the relative error is 2 × dr 0.4/ r which is 30, 0.027.
00:43:06.800 --> 00:43:12.500
If I express this is a percentage, I will move this over and that would be 2.7%.
00:43:12.500 --> 00:43:21.800
That means that if the error that I'm making measurement from 30 is 0.4,
00:43:21.800 --> 00:43:29.000
that means the difference in area is going to be 2.7% of the area calculated at 30.
00:43:29.000 --> 00:43:38.000
If it is 30.4, that means I'm going to be over by 2.7% of my actual area measured at 30.
00:43:38.000 --> 00:43:48.900
If I’m at 29.6, if the error is under the 30, that means I'm going to be short by 2.7% of my total area calculated at 30.
00:43:48.900 --> 00:43:49.900
That is what this means.
00:43:49.900 --> 00:43:59.500
Relative error is the error itself that you calculate divided by the actual area of the value that you measured.
00:43:59.500 --> 00:44:00.700
We could have done this directly.
00:44:00.700 --> 00:44:02.300
Here we actually did relative error.
00:44:02.300 --> 00:44:07.400
We use da/a, the error/ the actual value.
00:44:07.400 --> 00:44:11.100
We get it in terms of variables and then we plug the variables in.
00:44:11.100 --> 00:44:18.300
I could just have done it directly.
00:44:18.300 --> 00:44:29.100
We could also have just done it directly.
00:44:29.100 --> 00:44:39.200
In other words, the definition of relative error is equal to the actual error/ the actual value.
00:44:39.200 --> 00:44:46.700
Sorry, I should say the error/ the actual value, da/a.
00:44:46.700 --> 00:44:52.100
We found da, we found 75.4.
00:44:52.100 --> 00:44:59.900
Area is equal to π r², area = π × 30².
00:44:59.900 --> 00:45:07.100
It is equal to 2827.43.
00:45:07.100 --> 00:45:24.900
da/a, error/ actual value, error/ actual value = 75.4/ 2827.43 = 0.027.
00:45:24.900 --> 00:45:31.200
Again, this can give, if I want to speak in terms of percent, 2.7%.
00:45:31.200 --> 00:45:32.400
That is what a differential does.
00:45:32.400 --> 00:45:36.600
Anytime you have a given function, go ahead and differentiate, move the dx over.
00:45:36.600 --> 00:45:43.500
And that tells you, if I change x by a certain amount, how much does y change?
00:45:43.500 --> 00:45:45.900
That is it, it is really what we have been doing all along.
00:45:45.900 --> 00:45:48.900
Now we just want to think about it that way.
00:45:48.900 --> 00:46:00.900
Once again, if I have y = some f(x).
00:46:00.900 --> 00:46:07.800
I think it would be the best if I just use actual functions here.
00:46:07.800 --> 00:46:19.000
An example, if y = sin(x), dy dx, the rate of change of sin(x) is equal to cos(x).
00:46:19.000 --> 00:46:26.200
If I just want to concentrate on how much does y change when I change x, I move the dx over.
00:46:26.200 --> 00:46:34.300
If I change dx by 0.2 away from the point π,
00:46:34.300 --> 00:46:48.100
If I’m at the point π, and if I go 0.2 away from π to the left or 0.2 away from π to the right, y changes by cos(π) × 0.2.
00:46:48.100 --> 00:46:51.100
That is the differential, that is the whole idea of the differential.
00:46:51.100 --> 00:46:55.300
In this case, the differential actually gives me some error.
00:46:55.300 --> 00:47:00.400
If I have a measured value, and if I’m not quite sure about that measured value,
00:47:00.400 --> 00:47:07.300
let us say it is + this way and + this way, my measured value is my x0.
00:47:07.300 --> 00:47:11.900
The difference positive or negative is my dx.
00:47:11.900 --> 00:47:17.900
The y gives me the change in the overall function that I'm dealing with, in this case it was area.
00:47:17.900 --> 00:47:19.400
Area = π r².
00:47:19.400 --> 00:47:22.700
Here it is cos x dx.
00:47:22.700 --> 00:47:24.500
This is the dx, this is the dx.
00:47:24.500 --> 00:47:29.400
The differential itself gives me the total change of whatever it is that I’m calculating.
00:47:29.400 --> 00:47:31.800
I hope that makes sense.
00:47:31.800 --> 00:47:33.600
Thank you so much for joining us here at www.educator.com.
00:47:33.600 --> 00:47:34.000
We will see you next time, bye.