Thread: Generic Calculus Questions

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.

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.

Saw no reason to open another thread for something that fits squarely into this one, so without further ado...

Spoiler for It's nothing really..:

Today was the last day of instruction in my differentials class, and the professor was going over practical applications of the Laplace Transform to model complex currents and voltages over elements in a circuit that can have multiple inputs with varying frequencies. I was so blown away by the ease of it all, especially now that I've had a long-time question answered.

More to the point, I wanted to ask about the origins of the Laplace Transform.. How exactly was it worked out by Mr. Laplace? I've already given a few attempts at finding out on my own but it appears that every description I run into demands I have a more in-depth knowledge of mathematics concepts I've never heard of. Is there an explanation in Layman's terms? I truly appreciate how useful this is. I just can't imagine how the end result (the transform of a function in time domain) was achieved.

Any halps are of course greatly appreciated.

Just so you know, electon positions can't be calculated that way.

Does that mean we don't have an expression that dictates an electron's motion with respect to time, or that the expression can't be transformed back and forth, or that it's useless to do so? :O
Enlighten me, señor.

Broken image methinks.

I haven't actually studied Laplacian Transforms yet, but from what I've seen, you basically take a Taylor series and generalise it to a continuous case. Obviously you can't explicitly give the coefficients any more if there are infinite number of them, so they're written as a function - this is the function that you 'take the transform' of, and it gives you the new function.

It's pretty straightforward working out what the transform should be with this in mind, and then it's also easy to see that this is equivalent to the way that it's normally written (it's in a slightly weird form because it turns out that calculations are just easier when it's written the way that it is).

From my understanding, Euler had been working on using integrals as solutions to differential equations. Laplace was extending the work. The key property of the transform is that it turns L[f'(t)] into sL[f(t)] - f(0). This can be seen by integration by parts. So once you see that, it's only a matter of time before it occurs to you (if you're as smart as Laplace) to use that property to transform differential equations into algebraic equations.

With regards to the electron, I think Ninja was saying that you can't calculate a position for an electron using an equation like that (from one of the earlier questions asked on the thread).