Finding constants from data

7 In a chemical process, the mass \(M\) grams of a chemical at time \(t\) minutes is modelled by the differential equation $$\frac { \mathrm { d } M } { \mathrm {~d} t } = \frac { M } { t \left( 1 + t ^ { 2 } \right) }$$

1. Find \(\int \frac { t } { 1 + t ^ { 2 } } \mathrm {~d} t\).
2. Find constants \(A , B\) and \(C\) such that $$\frac { 1 } { t \left( 1 + t ^ { 2 } \right) } = \frac { A } { t } + \frac { B t + C } { 1 + t ^ { 2 } } .$$
3. Use integration, together with your results in parts (i) and (ii), to show that $$M = \frac { K t } { \sqrt { 1 + t ^ { 2 } } } ,$$ where \(K\) is a constant.
4. When \(t = 1 , M = 25\). Calculate \(K\). What is the mass of the chemical in the long term?

A market trader notices that daily sales are dependent on two variables: number of hours, \(t\), after the stall opens total sales, \(x\), in pounds since the stall opened. The trader models the rate of sales as directly proportional to \(\frac{8 - t}{x}\) After two hours the rate of sales is £72 per hour and total sales are £336

1. Show that $$x \frac{dx}{dt} = 4032(8 - t)$$ [3 marks]
2. Hence, show that $$x^2 = 4032t(16 - t)$$ [3 marks]
3. The stall opens at 09.30.
   1. The trader closes the stall when the rate of sales falls below £24 per hour. Using the results in parts (a) and (b), calculate the earliest time that the trader closes the stall. [6 marks]
   2. Explain why the model used by the trader is not valid at 09.30. [2 marks]