| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - partial fractions |
| Difficulty | Standard +0.8 This is a substantial multi-part differential equations problem requiring implicit differentiation of a volume formula, partial fractions decomposition, separation of variables, and integration involving logarithms. While the techniques are all C4 standard (related rates, partial fractions, separable DEs), the problem requires careful algebraic manipulation across multiple steps and integration of a non-trivial expression. The 17 total marks and extended chain of reasoning place it above average difficulty, though it remains a structured question with clear signposting rather than requiring novel insight. |
| Spec | 1.02y Partial fractions: decompose rational functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{dV}{dt} = -kV, \quad \frac{dV}{dh} = 10\pi h - \pi h^2\) | B2 | |
| \(\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt} \Rightarrow -kV = (10\pi h - \pi h^2)\frac{dh}{dt}\) | M1 | |
| \(-\frac{1}{3}k\pi h^2(15-h) = \pi h(10-h)\frac{dh}{dt}\) | ||
| \(-kh(15-h) = 3(10-h)\frac{dh}{dt} \Rightarrow \frac{dh}{dt} = -\frac{kh(15-h)}{3(10-h)}\) | M1 A1 | |
| (b) \(\frac{3(10-h)}{h(15-h)} = \frac{A}{h} + \frac{B}{15-h}\) | M1 | |
| \(3(10-h) \equiv A(15-h) + Bh\) | ||
| \(h = 0 \Rightarrow A = 2, \quad h = 15 \Rightarrow B = -1\) | A2 | |
| \(\therefore \frac{3(10-h)}{h(15-h)} = \frac{2}{h} - \frac{1}{15-h}\) | ||
| (c) \(\int \frac{3(10-h)}{h(15-h)} \, dh = \int -k \, dt, \quad \int\left(\frac{2}{h} - \frac{1}{15-h}\right) dh = \int -k \, dt\) | M1 | |
| \(2\ln | h | + \ln |
| \(t = 0, h = 5 \Rightarrow c = \ln 250\) | M1 | |
| \(2\ln | h | + \ln |
| \(\ln\frac{h^2(15-h)}{250} = -kt, \quad \frac{h^2(15-h)}{250} = e^{-kt}\) | M1 A1 | |
| \(\Rightarrow h^2(15-h) = 250e^{-kt}\) | ||
| (d) \(t = 2, h = 4 \Rightarrow 176 = 250e^{-2k}\) | M1 | |
| \(k = -\frac{1}{2}\ln\frac{176}{250} = 0.175\) (3sf) | M1 A1 | (17 marks total) |
**(a)** $\frac{dV}{dt} = -kV, \quad \frac{dV}{dh} = 10\pi h - \pi h^2$ | B2 |
$\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt} \Rightarrow -kV = (10\pi h - \pi h^2)\frac{dh}{dt}$ | M1 |
$-\frac{1}{3}k\pi h^2(15-h) = \pi h(10-h)\frac{dh}{dt}$ | |
$-kh(15-h) = 3(10-h)\frac{dh}{dt} \Rightarrow \frac{dh}{dt} = -\frac{kh(15-h)}{3(10-h)}$ | M1 A1 |
**(b)** $\frac{3(10-h)}{h(15-h)} = \frac{A}{h} + \frac{B}{15-h}$ | M1 |
$3(10-h) \equiv A(15-h) + Bh$ | |
$h = 0 \Rightarrow A = 2, \quad h = 15 \Rightarrow B = -1$ | A2 |
$\therefore \frac{3(10-h)}{h(15-h)} = \frac{2}{h} - \frac{1}{15-h}$ | |
**(c)** $\int \frac{3(10-h)}{h(15-h)} \, dh = \int -k \, dt, \quad \int\left(\frac{2}{h} - \frac{1}{15-h}\right) dh = \int -k \, dt$ | M1 |
$2\ln|h| + \ln|15-h| = -kt + c$ | M1 A1 |
$t = 0, h = 5 \Rightarrow c = \ln 250$ | M1 |
$2\ln|h| + \ln|15-h| - \ln 250 = -kt$ | |
$\ln\frac{h^2(15-h)}{250} = -kt, \quad \frac{h^2(15-h)}{250} = e^{-kt}$ | M1 A1 |
$\Rightarrow h^2(15-h) = 250e^{-kt}$ | |
**(d)** $t = 2, h = 4 \Rightarrow 176 = 250e^{-2k}$ | M1 |
$k = -\frac{1}{2}\ln\frac{176}{250} = 0.175$ (3sf) | M1 A1 | (17 marks total) |
**Total: 75 marks**\includegraphics{figure_2}
Figure 2 shows a hemispherical bowl of radius 5 cm.
The bowl is filled with water but the water leaks from a hole at the base of the bowl. At time $t$ minutes, the depth of water is $h$ cm and the volume of water in the bowl is $V$ cm³, where
$$V = \frac{1}{3}\pi h^2(15 - h).$$
In a model it is assumed that the rate at which the volume of water in the bowl decreases is proportional to $V$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac{dh}{dt} = -\frac{kh(15-h)}{3(10-h)},$$
where $k$ is a positive constant. [5]
\item Express $\frac{3(10-h)}{h(15-h)}$ in partial fractions. [3]
\end{enumerate}
Given that when $t = 0$, $h = 5$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item show that
$$h^2(15-h) = 250e^{-kt}.$$ [6]
\end{enumerate}
Given also that when $t = 2$, $h = 4$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item find the value of $k$ to 3 significant figures. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q7 [17]}}