I have some homework/study questions I've got for maths, and I'd like a few pointers on graphing things please. Find the slope of y = x^2 -1 at (2,3). I got 15 for the slope. But how does that number translate into a graphed line? What does that mean when sketched? Remember, I never really studied any maths, just getting into it.
to get the slope of a function at any particular point one has to differentiate the function. here y = x^2 -1; then the diffentiated function is dx/dy = 2*x (sometimes notation is y'=2*x). Now, how you get from the original function y=x^2-1 to the function y'=2x, is a bit too complicated to write down here, without knowledge of what you know already. if you like, I can tell you later. Anyway, at point (2,3) the x-value is x=2, so the slope has a value of y'=dx/dyx = 2*2 = 4. (not 15 ... can you tell me how you got that?) this value (y'=4) gives the amount of increase in value y per increase of value x. Thus it is a measure of how steep the line is. (a value of 0 means it is a horizontal line)
"""how you get from the original function y=x^2-1 to the function y'=2x, is a bit too complicated to write down here""" it's actually not complicated at all what you do is n.X^m ------> (m.n) X^(m-1) at least that's hat I remember of those integrales
FWIW, it's a differenatial, not an integral. The integral of 0->∞ y = x^2 -1 dx = x^3/3-x. Th"being pedantic" ed
Here's how you get from y=x^2 - 1 to the slope: y'(x) = dy/dx = limit(h->0) {[y(x+h) - y(x)]/h} so... y' = limit(h->0) {[(x+h)^2 - 1 - (x^2-1)]/h]} y' = limit(h->0) {[x^2 +2hx +h^2 - 1 - x^2 + 1]/h} y' = limit(h->0) [2x + h] y' = 2x
New question. There are two ball bearings. One has radius "r", the other has radius "r + 1.2mm". I'm to use the chain method to find total volume when "r = 3.3mm". Apparently I should start with this: Total Volume = 4/3 pi r^3 + 4/3 pi (r + 1.2)^3 Where, if everyone uses the same notation as I am being taught, 4/3 pi is "u'.
Please Register or Log in to view the hidden image! I think he meant REALLY get from one to the other.
HAHAHA that integral, as you defined it on the left hand side of the equality, is infinity ... lmfao....
I know. I am a mathematics teacher, so I am not satisfied with just saying how things are done. I also want to tell why they are. I thought you would all understand that. What is the use of knowledge if there is no insight? (PS is that what you meant Mallory?)