Thread: Generic Calculus Questions

Originally Posted by Invader

An electron moving along the x axis has a position given by x = 12te^(-2.1 t) m, where t is in seconds. How far is the electron from the origin when it momentarily stops?



The answer is 2.102m, but I only know that because I understand that I need to find the derivative of x = 0 for t (x's slope will be 0 when it 'stops moving'). Time t will give me the position in the original equation. The actual values I got only because I know a thing or two about getting around my calculator, but I want to know how to take the derivative of the original equation on paper. Can anyone show me how to do this?


[EDIT]: dx/dt = (12 - 25.2t)e^(-2.1t) is that correct? If so, then my next problem involves getting t onto one side of the equation so that I can solve for it.

You shouldn't have to solve for t here if you only want the zeros. 0 is special because if f(x)y(x) = 0 then you can be sure that f(x)=0 or g(x) = 0. so you need to find the zeros of f(x) = 12 - 25.2t and g(x) = exp(-2.1t)

g doesn't have any roots and f is linear so it only has one which is 12/25.2 to whatever accuracy you have. I never can remember the significant digits thing.

That would be a very hard problem to solve if you needed to get a function for t. You'd just want to use newtons method probably.

Heh, that's right, I could have done it that way. Though I think you misread the
constant e in the original equation as exp^ (there is an exponent in
the original too, don't get me wrong).

I'll show you the way I see this kind of problem, and why I went about it the
way I did. I drew a graph for your viewing pleasure.



You can see that where X is zero in the derivative, I get a t to plug into my
main equation that gives me the value of X that I'm looking for (where the
slope is zero). Your way would have been faster right off the bat, but I have a
habit of seeing the problem laid out like the graph above, except that I don't
draw it out.

(should probably point out that I only used the green line to match up the
f(t) and dx/dt appropriately)

That's a really good way to look at it. As the functions get more difficult though, one needs to learn tricks too But keep looking at it geometrically as long as you can. Going between the two without having to rely on either is key to being good at math as it gets more advanced.

BTW, exp(x) = e^x but I don't have to look around for the carrot if I'm feeling lazy. It gets used if you need to put something big in the exponent as well. You see it around more and more as you deal with bigger equations. Try sticking an integral in the exponent

Originally Posted by PhilosopherStoned

BTW, exp(x) = e^x but I don't have to look around for the carrot if I'm feeling lazy.

See, that was never explained to me, even through calc 1. I always assumed
exp was equivalent to the carrot, in that it meant "exponent". Good Game.
I'll have to pass on the integral in the exponent for now, but I'm sure I'm going
to need a decent explanation eventually.

On that note, I'll be using this thread as I encounter new problems I have
difficulty with.

Thank you for volunteering to be my tutor.

No problem. I love this stuff. I could use a review on all this stuff anyway. I quit math for years so I'm rusty on all levels. I've been working on it again for the past month or so though so I've got a lot of the kinks worked back out.

Yeah just treat exp(x) as a constant and get rid of it, it's never 0.

it's more general than treating one of them as a constant though. it works any way that you can factor a function. take sin^5(x)cos^2(x)x^2/(x^2 + 1).

We can factor this into sin^5, cos^2 and y=(x^2)/(x^2 + 1).

then the whole function can only be zero where at least one of those three is.

It's very nice for finding zeros of a function to not have to invert it.