2 Let’s start with the function at x = 4 because we Mysterious radiation has caused all calculators to stop working. It just so happens that at this time, you need to find out the square root of 5. The best you are going to be able to do now is approximate it. But how?Let’s start with the functionat x = 4 because weknow what isWhen we make a graph of this process, you will get a better sense of why we are doing this.First of all, what is the tangent line at x = 4?Now graph both the function and its tangent line…
3 Use the tangent line to the function at x = 4to approximateNotice that for numbers close to 4, the tangent line is very close to the curve itself…So lets try plugging 5 into the tangent line equation and see what we get…
4 Use the tangent line to the function at x = 4to approximateNotice how close the point on the line is to the curve…(5, 2.25)So let’s plug 5 into the tangent line to get…(4, 2)As it turns out…So in this case, our approximation will be a very good one as long as we use a number close to 4.
5 For any function f (x), the tangent is a close approximation of the function for some small distance from the tangent point.Notice that at points very close to a, the tangent line and the curve are almost the same.
6 For any function f (x), the tangent is a close approximation of the function for some small distance from the tangent point. (a+h, y(a+h))(a, f(a))(a+h, f(a+h))Where y = mt(x – a) + f(a)
7 For any function f (x), the tangent is a close approximation of the function for some small distance from the tangent point. (a+h, y(a+h))(a, f(a))(a+h, f(a+h))We call the equation of this tangent the linearization of the function at a and it is denoted by the function notation L(x).
8 Problems in the book will also refer to “the linearization of f at a” y – y0 = m(x – x0)We take the point-slope form…Make a few notational substitutions…and…is the standard linear approximation of f at a.Remember: The linearization is just the equation of the tangent line. The use of the term L(x) is to make it known that you are using the tangent line to make a linear approximation of the function in question. But we are still just talking about the tangent line.
9 Try a few simple linearizations near x = 0: The graph would look like this:
10 Notice how close the graphs are to each other near x = 0
11 Try a few simple linearizations near zero: Now try the other three.
12 We can also approximate using another approach called DifferentialsChemistry students should recognize y and x. Remember that taking a derivative assumes that both x and y go to 0. When we did thisFor this method, we are going to allow dx to be a small number but not 0.We’ll start with our function
13 DifferentialsdxdyyxNote that as long as dx and dy are going to 0, then we are getting more precise. But as soon as we start making dx > 0, then we are going in the other direction so whatever we get by this method is an approximation of the change in y
14 For this method, we are going to allow dx to be a small number but not 0. We’ll start with our functionBecause dy and dx originally had limits of 0, we aren’t supposed to think of as a fraction. But in this case we are going to let dx > 0 so we can treat like a fractionIn which dy will be a small change in y which we will determine by multiplying the derivative by the small change in x (dx).
15 Let’s remember what we are trying to accomplish here. We are going to use starting at x = 4 because we know without a calculator that y = 2Since dy = approximate change in y,let’s set dx = 1Why choose 1 in this case?Because like linearization, we are going to use to approximate from x = 4And when we add 0.25 (the change in y) to 2 (our original y), we get the same approximation of that we had with linearization.
16 Use linearization and differentials to approximate using the functionSinceWe’ll use the tangent line at x = 8 for the linearization
17 Use linearization and differentials to approximate using the functionSinceNow let’s use differentials:Again,simply because it is the difference between 8and 9Do a slide about percentage of error and then one more differentials problem as an application like the area problem. Maybe take one from the HW problems. Then delete the area problem slidesSame as with linearization
18 A square with sides x has an area If a 2 X 2 square has it’s sides increase by 0.1, use differentials to approximate how much its area will increase.
19 What will the actual difference in area be? Actual change in areaSo our approximation is off byOur percentage error here is:
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