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I have nowhere else to ask right now and I'm having some trouble, so answer tonight and you get a cookie.

I'm double posting to let you know that you've all failed.

It's 12:05 and no one's earned their cookie.

Such laziness. And I expected more of you all.

You totally shouldn't have XD I'm only in Advanced Algebra, man.

I'm only in Algebra. ._.;

Could have done calculus. I hate math. So I didn't take it. Sorry man. :>

>v< YOu and your silly american math.

But if calculus is about infitve ,derivates and integrals then I'm your man cause it's something I can understand at least X> I'd be only confused about american notation then~

Alright, well it's not that much help now since I had the homework quiz today, but it's good practice. Hopefully you'll know the notation.

So the given is that there's a differential equation:

Then it asks you to calculate the slope at the point (0,3). So you get this:

So the slope at (0,3) is zero. Fantastic.

Then it says "assume that the curve is linear between x=0 and x=0.5." Well we know that at x=0 the slope is zero, because at (0,3) the slope is zero. And I assume that when it says the curve is linear between those two points, it means the slope stays the same between those two points. Am I right?

Then it asks to calculate the y-value when x=0.5. That's what I don't get. I don't know the slope AT x=0.5, only BETWEEN x=0 and x=0.5. So then I'd end up with this:

With no way to solve for y. I was thinking I could find the particular solution of the graph that goes through (0,3) and then somehow turn that around, but I think I tried that and something went wrong. I might've made a mistake though.

Does that make sense, Crow?

oh you notate it the old fashioned way with dy/dx and stuff...we just take the function which is mostly f(x) or g(x) and put a ' to it showing we differentiated it (though that only works if you differentiate it after the variable x) so you get f'(x) for example and that's the y-value just in another words..just as a side note.

Ok the first differentiation always gives you the slope at a particular place of the original graphic. So if you get a 0 for the x-value it means that the graphic at the particular y-value has no slope.....ugh sorry but I really don't get your notation. Why is the x and y value on the same side of the equoation?
DOesn't it have to be something like? Since if you have 2 variables in one equoation you could never solve it.

f'(x)=-x/2 that would make much more sense to me as in these (and would still have the same solution to it since 0/2 = 0)

and the solution to your second one would be :

and that's your slope


Though on retrospective the problem doesn't make much sense to me either. 1.) if the function would continue to go on linearly ,just like that, then the function as a whole has to be a partly differentiale one because a function doesn't change it's whole course in such a drastic way that's why there's no reason for any value between 0 and 5 to have the same slope there.

I don't really know what's going on , I'm sorry >v<

This is pretty easy but the whole problem in itself is just confusing...

Crow, doesn't the problem ask for the Y value of the curve at .5, not the slope at .5? With an unchanging slope between 0 and .5, and a Y value of 3 at 0, then the Y value at .5 would remain 3. If you think about it, it's the most logical way, since it's rather difficult to find a variable when you have 2 variables.

Well but that's the thing. If dy/dx (or f'(x), whatever you want to call it) represents the slope at a given point, the slope CAN'T be zero at x=0.5, because the numerator will be nonzero and no matter what the denominator is (which depends on the y-value) the slope will always either be a constant or undefined (the latter of which I know it isn't, since the graph isn't vertical at that point).

They're on the same side because it's a differential equation. They don't have to be separated. The only way to solve for y is through a lot of integration.

As a note, integ() means I'm taking the integral and sqrt() means I'm taking the square root. Also, +/- is "plus or minus."

Which looks decidedly funky.

The variable c represents the constant that the solution could differ by, so you have to find the constant. To do that you just use whatever point they gave you, which is (0,3). Also, I'm taking the negative answer of the root because the graph is in quadrant IV.

Okay, so that makes our particular equation:

Umm...oh fuck I might have done it. So then does that mean that I just put 0.5 in for x in that equation?

EDIT: Fucking Invisionfree. The copyright symbol is c in parentheses XD