Money Revisited

https://isaacphysics.org/questions/maths_ch7_3_q9?stage=a_level
(I did the previous question: https://isaacphysics.org/questions/maths_ch7_3_q4?stage=a_level)

Im stuck on part f - I don't really know how to integrate the new function - I tried to employ a similar method but cannot get it to work.

Does anyone have any ideas?
Help much appreciated.

Working:

Original post by mosaurlodon

Working:

Seperation of variables which requires the right hand side to be in the form
f(m)g(t)
so what are f and g? The latter is very simple and the working is similar to your first column.

Reply 3

mosaurlodon

Is this the right start?

Or do you have to take out p/100 i.e. dm/dt = p/100(m-100q/p)

ALSO additional question do my images that I post come out normally or do they come out with some weird long text.

Reply 4

mqb2766

Original post by mosaurlodon

Is this the right start?

Or do you have to take out p/100 i.e. dm/dt = p/100(m-100q/p)
ALSO additional question do my images that I post come out normally or do they come out with some weird long text.

You could do it that way, but surely
(mp/100-q)*1
or
(m - 100q/p)*(p/100)
is a bit simpler and the latter closer to your column 1? Though like your column 1, you could consider your lump of constants on either side of the equation and youll end up with the same answer. Without giving the game away, youd expect something like a log(m) on the left and a t on the right, then take exponentials.

Your images are fine.

(edited 1 month ago)

Reply 5

mosaurlodon

Oh yeah that actually makes more sense.

Have I made a mistake around the constant - I just assumed it would equal m_0 in the end but not too sure.

Reply 6

mqb2766

Original post by mosaurlodon

Oh yeah that actually makes more sense.

Have I made a mistake around the constant - I just assumed it would equal m_0 in the end but not too sure.

Id do
dm/dt = (m - 100q/p)*(p/100)
as that will keep the coefficient of m as 1, so simplify the log integral (if you can do it a few different ways, try and think a couple of steps ahead) so
Int 1/(m - 100q/p) dm = Int (p/100) dt
m - 100q/p = e^(pt/100 + c) = Ae^(pt/100)
where A = e^c. Then when t=0 ...

A bit like your integration problem yesterday, dont take short cuts with integration, work it through clearly. If you subbed t=0 into your final expression for m(t), you dont get m0. Similarly, your q would increase the balance, but it represents payments.