# Differential equations help , rate of change

• Jun 16th 2010, 11:44 AM
Tweety
Differential equations help , rate of change
A mathematician is selling goods at a car boot sale. She believes that the rate at
which she makes sales depends on the length of time since the start of the sale,
t hours, and the total value of sales she has made up to that time, £ x.
She uses the model

$\displaystyle \frac{dx}{dt} = \frac{k(5-t)}{x}$
where k is a constant.

Given that after two hours she has made sales of £96 in total,
(a) solve the differential equation and show that she made £72 in the first hour of
the sale.

The mathematician believes that is it not worth staying at the sale once she is making sales at a rate of less than £10 per hour.
(b) Verify that at 3 hours and 5 minutes after the start of the sale, she should have

$\displaystyle \int x dx = \int k(5-t)$

$\displaystyle \frac{1}{2}x^{2} = 5kt - \frac{1}{2}t^{2} + c$

t= 0, x =0 , c = 0

t = 2 , x = 96 k = 576

I really suck on part'b', how would I go about it?

thank you
• Jun 16th 2010, 11:46 AM
Ackbeet
There's an error in your $\displaystyle t$ integration. I would revisit that.
• Jun 16th 2010, 11:48 AM
Ackbeet
Although, your solution of k comes about as the result of a correct integration.
• Jun 16th 2010, 11:52 AM
Ackbeet
You haven't quite solved part a) yet. What's left?
• Jun 16th 2010, 11:56 AM
Tweety
But I get x = 72, with my equation? Also, I cant seem to spot what is wrong with my integration, could you please tell me?

Thanks.
• Jun 16th 2010, 12:01 PM
Ackbeet
Plugging in the value of $\displaystyle t$ and solving for $\displaystyle x$ at the appropriate time was what was required. A simple step that you've already done in your head, no doubt; however, in order to show it, you just need to write it down.

As to the integral, it's a situation with an unclear notation. The correct $\displaystyle t$ integration is $\displaystyle k(5t-\frac{1}{2}\,t^{2})+C$, not
$\displaystyle 5kt - \frac{1}{2}t^{2} + C$. See the difference?

As for part b), try finding out what her sales rate is at the given time.
• Jun 16th 2010, 12:05 PM
Tweety
Quote:

Originally Posted by Ackbeet
Plugging in the value of $\displaystyle t$ and solving for $\displaystyle x$ at the appropriate time was what was required. A simple step that you've already done in your head, no doubt; however, in order to show it, you just need to write it down.

As to the integral, it's a situation with an unclear notation. The correct $\displaystyle t$ integration is $\displaystyle k(5t-\frac{1}{2}\,t^{2})+C$, not
$\displaystyle 5kt - \frac{1}{2}t^{2} + C$. See the difference?

As for part b), try finding out what her sales rate is at the given time.

Oh I see what you mean, I forgot to distribute the 'k' properly.

Also would I be using my formula and input t = 3.05?
• Jun 16th 2010, 12:08 PM
Ackbeet
Not sure if you should use $\displaystyle 3.05$ or $\displaystyle 3.08\overline{3}$; I'm inclined to the latter, since you're probably using decimals in the DE, and so you'd have to convert minutes to percent of an hour.
• Jun 16th 2010, 12:19 PM
Tweety
Thank you very much, I have it worked it out now, using t= 3.0833