DualReaver Wrote:[COLOR="Green"]I got this type of problem in calc today. I know the formula to use, but can't figure out what to do.
y=X^3, X[SIZE="1"]0[/SIZE]=3
Find the instantaneous velocity of the object at X[SIZE="1"]0[/SIZE].
The formula is:
lim...y={f(X) - f(X[SIZE="1"]0[/SIZE])} / (X - X[SIZE="1"]0[/SIZE])
X->X[SIZE="1"]0[/SIZE]
The numbers and equation for y are just placeholders, because they were on a test and I don't have it back yet. What exactly am I supposed to be doing here? :f6:[/COLOR]
If you haven't learned differentiation, it literally represents the 'rate' at which a function changes.
Say you have y = x^3 to represent an object's motion, position, as measured in x and y axes. The differentiation (or derivative, whatever) of that function will give a function which describes the "rate" at which that motion is changing, in other words, its "speed" (because movement/time = speed).
Applying the same understanding again, the differentiation of a speed is the object's acceleration (or how much speed it changes in x time).
As for the problem itself, I haven't done Math in years and even when I did, I wasn't really focused but I guess I can give it a try.
lim y = [f(x) - f(xo)]/(x - xo)
x->xo
Suggests the "limitation" that the function reaches as x approaches xo, or simply understood as the value of y at xo. They prefer to use "limitation" since more than often there isn't a specific value for y at x = xo.
lim y = [f(x) - f(xo)]/(x - xo) =
x->xo
lim [(x^3 - 27)/(x - 3)] =
x->3
lim (x^2 + 3x + 9) = 27.
x->3
I have a feeling I'm wrong on many levels but I'd like to use this opportunity to know where I'm wrong lol.