In a chemical reaction, two substances $A$ and $B$ are combined to form a new substance $C$. T

Question

In a chemical reaction, two substances $A$ and $B$ are combined to form a new substance $C$. The initial masses of $A$, $B$and $C$ are $8$, $6$ and $0$ units respectively. After time $t$ seconds, the masses of $A$ and $B$ are each reduced by $x$ units, and the mass of $C$ increases by $2x$ units. The rate of change of the mass of $C$ with respect to $t$ is proportional to the product of the masses of $A$ and $B$ at any time $t$. Write down a differential equation relating $x$ and $t$.[1] It is observed that when $t=5$, $x=5$. By solving the differential equation in part (i), show that $x=\frac{24{{\left( \frac{9}{4} \right)}^{-\frac{t}{5}}}-24}{3{{\left( \frac{9}{4} \right)}^{-\frac{t}{5}}}-4}$.[7] Sketch the graph of $x$ against $t$.[2] In another experiment with the same initial masses of substances $A$ and $B$ as before, the increase in the mass of $C$ is modelled by the differential equation $\frac{{{\mathrm{d}}^{2}}x}{\mathrm{d}{{t}^{2}}}=-8{{\mathrm{e}}^{-2t}}-\frac{16}{9}{{\left( t+1 \right)}^{-\frac{7}{3}}}$ Given that $x$ eventually stabilises at $6$ units after a long time, find $x$ in terms of $t$.[4]

Accepted Answer

(i) $\frac{\mathrm{d}x}{\mathrm{d}t}=k(x-8)(x-6)$, where $k>0$. (ii) $x=\frac{24\left(\frac94\right)^{-t/5}-24}{3\left(\frac94\right)^{-t/5}-4}$. (iii) The curve increases from $(0,0)$ towards $x=6$. (iv) $x=-2e^{-2t}-4(t+1)^{-1/3}+6$.

2022 NYJC P1 Q11

2022 NYJC P1 Q11

Video Solution:

Solution:

Solution locked