Mathematics 9709 · AS & A Level · Differential equations
Differential equations — practice question
The diagram shows a water tank with vertical sides and a horizontal rectangular base. The area of the base is $2\text{ m}^2$. When $t = 0$ the tank contains no water, and water starts to enter at a rate of $1\text{ m}^3$ per hour. At the same time, water also begins leaving through the base at a rate of $0.2\sqrt{h}\text{ m}^3$ per hour, where $h$ represents the depth of water in the tank after $t$ hours.
(i)[3]
Form a differential equation linking $h$ and $t$, and show that the time $T$ hours needed for the depth of water to reach $4\text{ m}$ is given by $T = \int_{0}^{4} \frac{10}{5 - \sqrt{h}}\,dh$.
(ii)[6]
Use the substitution $u = 5 - \sqrt{h}$ to find $T$.
Worked solution & mark scheme
This 9-mark question has a full step-by-step worked solution and mark scheme. One marking point: “State or imply the relation $\frac{dV}{dt}=2\frac{dh}{dt}$” …