Maths Question

[Callum]

Ok... If at all possible I need this answered by about 4 am my time [img]tongue.gif[/img] If not... then I can check before about 9 30 and again before 2 30... Otherwise I'm screwed [img]tongue.gif[/img]

Ok, and here we go.

I have a graph with the function Y = 10(0.84^X), where Y represents the level of a drug in the bloodstream of a patient, X hours after the first dosage, where the original dosage was 10 micrograms, and the amount decreases so that 84% of the previous hour's amount is left every hour. All well and good, its an exponential decay curve... (at least I think so... that makes sense right?)

However, then the question says that the patient takes another dose of 10 micrograms every hour. And we have to draw a diagram. Still simple...

But then it asks for a function for this new graph... And as far as I know... there is no way to draw a graph that suddenly jumps up 10 before continuing as normal. Is there? So the function is the same, except every six hours, it increases by 10...

Is there a way to obtain a function for this?

How bout if instead of the function resuming, and the levels decreasing at 83% per hour as before... the part of the graph after six hours simply translates upwards ten... and doesn't decrease by 83 of the value... but by 83 percent of the value minus ten... Is this making any sense?

Picture to come soon to help clarify...

[Callum]

As promised... an image...

I need to find a formula for one of the red or blue graphs...

The formula for the green graph is y = 10(0.84^x)

The graphs repeat in the same manner (every six hours, y increases by 10)

Anyone? I'm stumped...

[shamrock_uk]

Just so you know that IW isn't ignoring you, I've asked a couple of my friends but they've all been stumped too.

Good luck!

[ 02-09-2006, 07:20 PM: Message edited by: shamrock_uk ]

[Ilander]

Piecewise is the only way you can represent such a function.

[Assassin]

Would a series work? I'm not sure if you could call it a function per say though...

[Ilander]

The net effect is another exponential curve, but unless the medication is given CONTINUOUSLY, you don't get a curve you can represent in any way besides piecewise :S

[JrKASperov]

Originally posted by Callum: Y = 10(0.84^X), But then it asks for a function for this new graph... And as far as I know... there is no way to draw a graph that suddenly jumps up 10 before continuing as normal. Is there? So the function is the same, except every six hours, it increases by 10... Is there a way to obtain a function for this?

Yes. It needs to be done piecewise, so that for every six hours you have a new graph. But finding this graph is surprisingly easy:

1. Find the function value of Y at the time the dose of 10 is administered. (NOT AFTER it has been administered but BEFORE)

2. Add this value to the dose you added, being 10.

3. Assuming the drop rate is the same (it should be unvariable under the taken dose) you take new formula: Y' = (10 + value before new dose) (0.84^x)

4. For the next jump of six hours, repeat, only this time you will most likely find a different value when administering new dose.

[Callum]

OK... Turns out I was just wrong... it didn't say "find a single function"... it said represent this on your GDC or graphing program... Which means that piecewise it worked [img]tongue.gif[/img]

0<=x<6 y=10[0.84^x]
6<=x<12 y=10[0.84^x]+10[0.84^[x-6]]
12<=x<18 y=10[0.84^x]+10[0.84^[x-6]]+10[0.84^[x-12]]
12<=x<18 y=10[0.84^x]+10[0.84^[x-6]]+10[0.84^[x-12]]+10[0.84^[x-24]]

For those of you interested... [img]tongue.gif[/img]

[shadeoracle]

If you wanted to put it into a single function you could use Heaviside functions which pretty much replicate piecewise functions. Therefore your function is:
y = H[x]*10[0.84^x] - H[x-6]*10[0.84^x] + H[x-6]*10[0.84^x]+10[0.84^[x-6]] - H[x-12]*10[0.84^x]+10[0.84^[x-6]] + H[x-12]*10[0.84^x]+10[0.84^[x-6]]+10[0.84^[x-12]]+10[0.84^[x-24]] - H[x-18]*10[0.84^x]+10[0.84^[x-6]]+10[0.84^[x-12]]+10[0.84^[x-24]] + ...

For those of you who are interested, a Heaviside function, denoted H[x], has the following properties:
H[x] = 0 for x<0
H[x] = 1 for x>=0
So above where I have, say, H[x-6], the Heaviside function 'turns on' at x=6.

Note: I learnt this in my second year at uni so it isn't a good idea to use this in class if you haven't covered it.

[ZFR]

By the way Heaviside function can be represented analitically:

H(x)= ( sqrt(x²) + x ) / 2x

(except in this case it's undefined for 0)