Post #11

This week in calculus, we learned linearization, took a quiz, and had review packets for our test Wednesday.

The key word in linearization is approximate and that f(x) = f(c) + f'(c) (x-c).

For linearization you can be asked to approximate the tangent line to y= x^3 at x=2

Take the derivative : 3x^2

Plug in your x: 3(2)^2 = 12

Find y by plugging in for x: y= (2)^3 = 8

Plug into your equation: f(x) = 8+12(x-8)

You may be asked to use differentials to approximate something. The steps for doing these problems are:

1. Identify an equation

2. f(x)+f'(x)dx

3. Determine dx and x

4. Plug into your equation

EXAMPLE:

1. the square root of 65.4

Equation: the square root of x

f(x) + f'(x) dx = the square root of x + 1/ 2(the square root of x) dx

dx= .4

x= 65

Plug in: the square root of 65+ 1/ 2(the square root of 65) (.4)

= 65.155

We also learned how to solve for dy.

Example:

y= 5x^2-6

dy/dx= 10x

dy= 10x(dx)

I am still having problems with angle of elevation and I also have a question on number 22 on the packet I have. The question says: A man 6 feet tall walks at a rate of 4 feet per second away from a light that is 15 feet about the ground. When he is 10 feet from the base of the light, at what rate is the length of his shadow changing?

I know when you are looking for the rate of the tip of his shadow, you have to set up a proportion and solve for a, then take the derivative and plug in for dx/dt, but since this problem is asking for the rate of the length I think you have to work it differently. So if anyone knows how to work it please fill me in.