Wikipedia:Reference desk/Archives/Mathematics/2008 October 18

= October 18 =

Equations of the form $$a\cos{\theta}=b\sin{\theta}$$
How do I solve such an equation for $$\theta$$? Thanks --Colonel Cow (talk) 02:02, 18 October 2008 (UTC)


 * Try using the arctangent.  siℓℓy rabbit  (  talk  ) 02:11, 18 October 2008 (UTC)


 * Ah of course, I knew it was easier than it looked. I can't believe this had me wrapped around for so long, thanks. --Colonel Cow (talk) 02:15, 18 October 2008 (UTC)

Integration question...
OK for disclaimers this IS a homework question, but I've worked through it and ended up an answer that differs in sign. Can anyone point out what I did wrong? I can't see how an exponent can change sign in the steps that I did. Thanks.

Given that,$$v=\frac{ds}{dt}=\frac{10m}{k}\left(1-e^{\frac{kt}{m}}\right)$$, where k and m are constants, find s in relation to t.

Rearranging and ntegrating both sides gives $$s=\frac{10m}{k}\left(t-\frac{m}{k}e^{\frac{kt}{m}}\right)+c$$

When t=0, s=0, $$0=\frac{10m}{k}\left(0-\frac{m}{k}e^{0}\right)+c=\frac{-10m^{2}}{k^{2}}+c$$ so $$c=\frac{10m^{2}}{k^{2}}$$

So final answer is $$s=\frac{10m}{k}\left(t-\frac{m}{k}\left(e^{\frac{kt}{m}}-1\right)\right)$$ But it is incorrect as the correct answer has $$e^{\frac{-kt}{m}}$$ instead. --antilivedT 05:51, 18 October 2008 (UTC)

In fact, your answer is correct. In any case you can immediately observe that $$e^{\frac{-kt}{m}}$$ could not be the right answer, for it goes to 0 exponentially, whereas your velocity diverges. Maybe a print error has occurred, or a fly dropped a small thing, like in the movie Brazil. Probably there was a negative exponent in the velocity at the beginning, because is phisically more natural than the positive; this fits with $$e^{\frac{-kt}{m}}$$ I guess. PMajer (talk) 08:39, 18 October 2008 (UTC)


 * Yes, there is either a mistake in the question or in the "correct" answer. I think the question being wrong is more likely since as it currently stands the object spends most of the time travelling backwards at ever increasing speed, which would be an odd choice of co-ordinates and an unusual question. An object starting off stationary and asymptotically approaching a given velocity would be a more usual question. --Tango (talk) 12:37, 18 October 2008 (UTC)

The question is not a dimensionally balanced equation. The constant 10 has the dimension of $$st^2.$$ Insert  $$c=10k^2/m^2.$$
 * $$\frac{d\frac sc}{d\frac{kt}m}=1-e^{\frac{kt}m}$$

Make your life comfortable by rescaling the variables such that the constants become equal to one. Substitute $$x=kt/m$$ and $$y=s/c$$
 * $$\frac{dy}{dx}=1-e^x$$

Tango is right that there should be a minus sign in the exponential in the question
 * $$\frac{dy}{dx}=1-e^{-x}$$

That describes a moving particle with friction under constant external force.
 * $$\frac{d^2y}{dx^2}+\frac{dy}{dx}=1$$

Integrate.
 * $$y=y_0+x+e^{-x} \,$$

Putting the constants back:
 * $$\frac s c=\frac {s_0}c+\frac k m t+e^{-\frac k m t} \,$$

Bo Jacoby (talk) 18:32, 18 October 2008 (UTC).
 * It's dangerous to try and guess the dimensions when all you have to go on is the choice of variable names. --Tango (talk) 18:57, 18 October 2008 (UTC)
 * We don't go on the variabel names. Substituting s=cy changes the constants, so 10 has got a kind of dimension. Bo Jacoby (talk) 20:32, 18 October 2008 (UTC).


 * Yes the original equation must've been $$v=\frac{ds}{dt}=\frac{10m}{k}\left(1-e^{\frac{-kt}{m}}\right)$$. It is part of a bigger question on a moving object with friction under constant force, given that $$m\frac{dv}{dt}=10m-kv$$, find A and B in $$v=A(1-e^{Bt})$$ and then find s in relation to t. The answer says $$B=\frac{k}{m}$$, and I got that same incorrect answer by forgetting that v was negative while integrating the first part of the question. Hopefully this doesn't happen in the actual exam. Thanks for your help guys. --antilivedT 23:30, 18 October 2008 (UTC)

Fourier transform in f instead of w (angular frequency) in Maple
Maple, with its FourierTransform and fourier functions calculates the Fourier transform of a function, but using e^(-I*w) (angular frequency) instead of normal frequency (e^(-I*2*Pi*f)). I could manually define the Fourier transform integral as I wished but, obviously, if I did that it won't calculate the FTs of non-square integrable functions (it will just say undefined). How can I get the program to do that? Thanks. --Taraborn (talk) 12:14, 18 October 2008 (UTC)