The mass MMM grams of a crystal growing in a chemical bath at time ttt hours is modelled by the differential equation
dMdt=10tMe2t,M≥0,t≥0 \frac{\text{d}M}{\text{d}t} = \frac{10t \sqrt{M}}{\text{e}^{2t}}, \quad M \ge 0, \quad t \ge 0 dtdM=e2t10tM,M≥0,t≥0Given that the initial mass of the crystal is 444 g, solve this differential equation to find an expression for M12M^{\frac{1}{2}}M21 in the form M12=f(t)M^{\frac{1}{2}} = f(t)M21=f(t).
Hence determine the equation of the horizontal asymptote to the curve with equation M12=f(t)M^{\frac{1}{2}} = f(t)M21=f(t).
92 exam-style questions on WJEC A Level Maths 4.6 Integration (A-level only), covering 4.6.1 Integration (A-level only) and 4.6.2 Integration (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.