Welcome to Calculus, I'm professor Greist.

We're about to begin lecture 12 on linearization.

Derivatives are useful in so many contexts, moving far beyond computing

slopes. In this lesson, we'll cover one of the

primal application to linearization. This stems from our understanding of

Taylor series. Linearization is the first step in a

Taylor series, using the first derivative to approximate.

Thinking of derivatives in terms of linear variation can be frustrating, if

you're trying to visualize it. It's maybe not so easily visualized as is

slope, however in some samples you cant early see the variation.

Consider the area A, the square of side length x.

What happens if we increase x by small amount?

The area becomes x squared plus 2xh plus h squared.

We can see those terms by increasing the side length a little bit and decomposing

the extra area into a pair or rectangles the area x times h and a small square,

whose area is h squared. So, ignoring the terms in big O of h

squared, we see that the linear variation is 2x times h.

For a triangle, we would get something similar, let's say a right triangle with

length x and height x, then the area, one half x squared, has the following

variation. A of x plus h is one half x squared plus

x times h plus one half H squared. Visualizing the additional area, we can

break it up into a parallelogram. Whose width is h, and whose height is x,

giving a linear variation term of x times h.

The leftover material is in big O of h squared.

Lastly, for a circular disc of radius x. We have that the area is pi times x

squared. What does the variation look like in this

case? I ahve x plus h is pi x squared plus 2 pi

x h. Plus pi h squared.

that means if we increase The area a little bit then the, the first order term

has area two pi X times H. But then you've got this leftover higher

order term. How does that work?

That's a little harder to see. It's maybe a bit easier if you

approximate the circular disk by a regular polygon with many sides.

And extend those sides out to a height of h.

Then you see that there are a number of small triangles.

Left over, whose areas will add up, to that of a disc of radius h in the limit

as the number of sides is going to infinity.

These first order of variations are the basis for doing linear approximation.

For example, let's say you want to estimate the length.

Of a bannister for a staircase. If you know something about the

dimensions, let's say it's 13 units across and 9 units high, then that length

would be square root of 9 squared plus 13 squared or the square root of 250.

Let's say you want to come up with an estimate for that.

I don't know what that exact number is, and let's say I don't have a calculator.

Well, I could consider the function f of x equals square root of x.

And I do know what the square root of 256 is.

So, that's, 16 squared. To get from there to 250, I can use a

value of H, a perturbation of negative six.

So that from a Taylor Expansion, f of 250 is going to be f of 256 plus, well, the

derivative evaluated at 256 times the difference times h.

Negative six. Although the terms in the Taylor

expansion are of higher order, well, let's see.

The square root of 256, that's 16. Minus six over two times 16.

And that gives us 15 and 13 16ths. As as answer, and that's not so bad of an

approximation. That's about 15.8.

The true answer is the same up to the first three digits.

So, in this case, linear approximation works pretty well.