 Hi everyone, I have for you here two examples of working with differentials. As we go through we're going to talk about the difference, the distinction between dy and delta y. We have here a cubic polynomial x cubed plus x squared minus 2x plus 1 and we're going to consider that function at the x value of 2 and a delta x of .05. Think of the .05 as almost that delta neighborhood, it's similar to that, that we talked about when we did the formal definition of a limit. The first thing to recognize is that this delta x of .05 that is the same as what we're going to use for dx throughout the problem, so that's one thing to keep in mind. The first thing we're asked to do here is to calculate the differential dy and evaluate it for the x value of 2. Remember how it is that we calculate our differential, dy is going to be equal to f prime of x times dx, we know dx that's .05, we just need to calculate the derivative and we are going to do this for the x value of 2. Our derivative of course is given by 3x squared plus 2x minus 2 and we're going to need to multiply that by dx value, we're going to evaluate this x value at 2. If we substitute 2 in place of our x's you should get 14 and dx of course was given to us to be .05. That gives us an answer in the end of .7. Now think about what this tells us. This is the approximate difference based upon the tangent line approximation. Let me go to a graph of this and I'll show you that. We have here a graph of the cubic polynomial and you can see indicated the actual point of tangency 2, 9. Now the way I set this up is this x value right there that should be 2.05. Remember our delta x is .05, so if you trace that up you can see right in here how there is so little difference between the location of the actual curve and the location of the tangent line, they are so, so close together. So the .7 represents the approximate difference between the function value and what we would obtain by using the tangent line approximation instead. Let's now compute delta y for the same equation. Delta y really represents the real difference between two function values on the original curve. dy is given to us by the function value of x plus delta x minus f of x and notice how each of these is referring to the original function, that part's really important. Our x value is 2, delta x is .05, so we have the function value at 2.05, we're going to get that by substituting it to the original function remember, minus f of 2. I would suggest putting into your calculator the original function, remember you can use the function notation so that you do not have to truncate and round off your decimals. So when you calculate the function value at 2.05 minus the function value at 2, the first should come out as approximately 9.718 just so you can make sure you're getting the right answer and you get a difference of .718. Now what that represents is the real difference between the two function values on the original curve at x equals 2 versus x equals 2.05. Again let me show you that on the graph. What this is saying is if you compare the function value on the curve at 2 that's going to be 9, that's what you should have gotten for that, compare that to the function value on the curve at 2.05, that's going to be maybe around here. The difference between those y values is .718. For delta values very, very small it should be true that delta y is very close in value to dy. So here we got a delta y of .718, so remember that represents the difference in these two y values on the original curve. What dy represents is the approximate difference based upon the tangent line approximation. Delta y refers only to the original function values on the original given curve, dy is what makes use of the tangent line approximation. Let's take a look at an application problem in which we're going to approximate error. We are told the radius of a ball bearing is measured to be .7 inches. If the measurement is correct to within .01 inch, estimate the maximum error in using this to calculate the volume of the ball bearing. We are told that the radius is .7, so let's just refer to that as r. We are told the measurement is correct to within .01 inches, so that's going to be dr. We can also refer to it as delta r. What we're asked to do is calculate the error in the volume. We can refer to that as dv, where we could refer to it as delta v, and that's what we're trying to determine. We need to know how to calculate volume of a sphere. That formula is four thirds pi r cubed, and all we're going to do is differentiate. So the derivative of v would be dv. Using the power rule on the right side of the equation, we arrive at 4 pi r squared dr, and now it's just a matter of substituting. And if you multiply that all out, we can go ahead and multiply in pi if you care to. You obtain approximately .062. So what this represents is the maximum error in calculating the volume. This would be in inches cubed, where you could do cubic inches. Either representation of the units of measure is fine, but it's cubic units because we are talking about a volume measurement.