We continued to work on implicit derivatives, related rates and angle of elevation which i am not completely comfortable with. We were also introduced to linearization at the end of the week.
The key word in linearization is approximate
EQUATION: f(x) = f(c) + f'(c) (x-c).
related rates:
1. Identify all of the variables and equations.
2. Identify what you want to find.
3. Sketch and label.
4. Write an equations involving your variables.
5. Take the derivative with in terms of time.
6. Substitute the derivative in and solve.
implicit derivatives:
First Derivative
- 1. take the derivative of both sides
- 2. everytime you take the derivative of y note it with dy/dx or y^1
- 3. solve for dy/dx
Second Derivative
- first you find the first derivative and solve it for dy/dx by using the steps for the first derivative steps.
- you then take the second derivative of the solved equation. Plugging in d^2y/d^2x everytime you take the derivative of y again. and where you have dy/dx you plug in your solved equation for that.
- once you have everything plugged in and ready to go you then solve for d^2y/d^2x
EXAMPLE:
y=SIN(x)(solve for dy)
dy/dx = COS(x)
dy = [(dx)*(COS(x))]
what i don't understanddddddddddd. I think i would be able to grasp almost everything except i havent takin the time to really sit down and try to understand it, especially since everything was so busy last week, i was just exausted. but anyways, i don't really know what linearization is and the purpose of it
Think of linerization as having a porabala and pulling it to a stratight line. That's the purpose of it. Say you have the square root of 16.5 and you want to linerize it. All you really have to do is break up the decimal and plug it into an equation:
ReplyDeletef(x)+f'(x)dx
x=the suqre root of 16
dx=.5
Ok, so you plug it into your equation getting the swuare root of 16+the derivitive of the square root of 16(.5)
For linearization the keyword is APPROXIMATE. if you see that word, then you know that it is a linearization problem. the formula is f(x) + f'(c) (x-c)
ReplyDeleteEXAMPLE:
Approximate the tangent line to y=3x^2 @ x=2.
ok so we know that c=2
so, to find f'(C), we must take the derivative and plug in 2.
dy/dx=6x
dy/dx=6(2) = 12
now, to find y, we plug in x to the original.
y=3(2)^2...y=12
now plug into your linearization formula.
12 + 12 (x-2)
and there's your answer.
Linearization:
ReplyDeletef(x)=f(c)+f'(c)(x-c)
example: Approximate the tangent line to y=x^2 at x=1
you find all the different values: dy/dx=2x dy/dx=2 y=(1)^2=1
then you plug into the formula to get: f(x)=1+2(x-1)
example 2: use differentials to approximate: sq root(16.5)
steps:
1: identify an equation--- f(x)=sq root(x)
2:f(x)+f;(x)dx--- sqrt(x)+ (1/(2sqrt(x)))(dx)
3:determine dx-- .5
4:determine x--- 16
5:plug in--- sqrt(16)+(1/2sqrt(16))(.5)= 4.0625
error= .0005