This week we pretty much reviewed for a test.
One thing that i feel pretty sure of myself with is related rates
1) first determine what is given in the problem. All given variables and even what your looking for.
2)set up your equation. Most of the time this equation will be simple such as patheagorean theroem or area or something like that.
3) you then take the derivative and put dy/dx,dx/dr, etc. the correct term for each variable.
4) plug in your given and then you have set up a simple equation.
5) once your equatin is set up and all given is plugged in you then solve for you unknown.
Because this week was hecktic and busy i guess i just missed linearization or something but i have no clue what to do with it. I need help
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Linearization:
ReplyDeletethe keyword is APPROXIMATE
you use the formula ...
f(x) = f(c) + f'(c)(x-c)
EXAMPLE:
Approximate the tangent line to y=(x)^2 at x=1
we know that c=1 and we'll take the derivative and plug in x to find the slope...
dy/dx = 2x...dy/dx = (2)(1)
dy/dx = 2
now we'll take the original and plug in x
y=(1)^2...y=1
so using the formula from linearization:
f(x) = f(c) + f'(c)(x-c)...we'll plug in what we have
f(x) = 1+2(x-1)
THAT'S HOW YOU DO LINEARIZATION..hope it helps
ALSO..with linerarization we learned that a differential is when something is solved for dy or dx. For example:
y= sin[x] so we'll solve for dy
dy/dx = cos(x)
dy = [dx]cos(x)
THAT'S IT...it's like first derivative and then multiplying by dx because you wanta get dy by itself!
Using differentials, we learned how to approximate.
For example:
APPROXIMATE the square root of 99.4
f[x]=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 = 99
plug in:... the square root of 99 + [(1)/(2*the square root of 99)]*[.4]
giving you...9.96997
when we plug in the square root of 99.4 in our calculater we are only 0.00002 off so that's our error!
Linerization is actually really easy if you look at it. For linerization, you also need to know how to find differientials (solving for either dx or dy). The equation you use for linerization is f(x)=f(c)=f'(c)dx
ReplyDeleteEX: Use differentials to approximate the square root of 25.6
First you have to identify an equation. The equation for this would be f(x)=the square root of x. You now plug your numbers into your equation giving you the square root of 16 + 1/2square root of 16 (.5)
for linerization find differentiables solving for dx or dy. the equation is f(x)=f(c)=f(c)dx. hope this helps
ReplyDeleteLinearization isn't that hard if you just sit down and look at it and know what the problem is asking.
ReplyDeleteThe first thing you need to know is the word approximate. That's the key word, every time you see it know it's a linearization problem.
The formula is f(x) = f(c)+ f'(c) (x-c)
Example:
Approximate the tangent line to y=x^2 at x=1.
dy/dx=2x
dy/dx=2 at x =1
y=(1)^2 = 1
f(x) = 1 + 2 (x - 1)
We also learned about differentials which is when something is solved for dy or dx.
Example:
Use differentials to approximate the square root of 16.5
1. Identify an equation : f(x)= square root of x
2. f(x) + f'(x)dx: square root of x + (1)/(2)(square root of x)(dx)
3. Determine dx (the decimal and to the right): .5
4. Determine x(left of the decimal): 16
5. Plug in: square root of 16 + (1)/(2)(square root of 16)(.5)
= 4.0625
error=.0005