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Nov 2010 p31 q10
2307
A certain substance is formed in a chemical reaction. The mass of substance formed t seconds after the start of the reaction is x grams. At any time the rate of formation of the substance is proportional to \((20 - x)\). When \(t = 0\), \(x = 0\) and \(\frac{dx}{dt} = 1\).
(i) Show that x and t satisfy the differential equation \(\frac{dx}{dt} = 0.05(20 - x)\). [2]
(ii) Find, in any form, the solution of this differential equation. [5]
(iii) Find x when \(t = 10\), giving your answer correct to 1 decimal place. [2]
(iv) State what happens to the value of x as t becomes very large. [1]
Solution
(i) The rate of formation is proportional to \((20 - x)\), so \(\frac{dx}{dt} = k(20 - x)\). Given \(\frac{dx}{dt} = 1\) when \(t = 0\) and \(x = 0\), we have \(1 = k(20 - 0)\), thus \(k = 0.05\).
(ii) Separate variables: \(\frac{dx}{20 - x} = 0.05 \, dt\). Integrate both sides: \(-\ln|20 - x| = 0.05t + C\). Using \(t = 0\), \(x = 0\), we find \(C = -\ln 20\). Thus, \(\ln(20 - x) = \frac{1}{20}t\).