Now I will do my last post of the holidays. I will start with the limit rules. The limit rules are:
1) if the highest exponent is the same on the top and bottom then the limit is the top coefficient over the bottom coefficient of the highest exponents.
2) If the highest exponent is on the top then the limit is infinity.
3) But if the highest exponent is on the bottom then the limit is 0.
Some examples are:
lim->infinity (5x^2-2x)/(4x^2+4x-1) Then the limit is 5/4 because it follows rule 1.
lim->infinity (x^3+6)/(7x^2-3) Then the limit is infinity because it follows rule 2.
lim->infinity (2x^4-6)/(5x^6+7x+9) Then the limit is 0 because it follows rule 3.
Next I will explain linearization. The steps for working linearization problems are:
1. Identify the equation
2. Use the formula f(x)+f ' (x)dx
3. Determine your dx in the problem
4. Then determine your x in the problem
5. Plug in everything you get
6. Solve the equation
Finally I will talk about the trig inverse intergration formulas. The trig inverse integration formulas are: (sr=square root)
1. S du/sr(a^2-u^2)=-1/sr(u)arcsin u/a +C
2. S du/a^2+u^2=1/du(a)arctan u/a +C
3. S du/u sr(u^2-a^2)=1/du(a)arcsec lul/a +C
For another question I have is instantaneous and average speed. I seem to have lost my notes on this so I cannot review on how to do this.
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Average speed is just the slope.
ReplyDeleteThe example we have in our notes is:
An anvil falls off the roof onto Ryan's head. What is its' average speed during the first two seconds if y=16t^2 describes the fall.
So your point is [0, something] and [2, something] since it is during the first two seconds.
To find your other point, plug each value into the equation.
0: 16(0)^2 = 0
2: 16(2)^2= 64
Next all you have to do is find the slope.
The formula for slope is y2-y1/ x2-x1
So 64-0 / 2-0 = 32
Instantaneous speed is the definition of a derivative which is
ReplyDeleteas the lim h ->0 f(x+h) - f(x)/ h
Example: Find the instantaneous speed at t=2
y=16t^2
f(t) = 16t^2
plugging into the formula you get as the lim h -> 0
16(2+h)^2-16(2)^2/h
Then simplify:
16(4+4h+h^2) - 64 / h
64+64h+16h^2-64/ h
64+16h
Then plug in 0: 64+16(0)= 64
OR the simpler way to do this is since you know it is the definition of the derivative, once you plug into the formula, all you have to do is take the derivative of 16t^2 and plug in two.
f' = 32t
32(2) = 64