| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - partial fractions |
| Difficulty | Challenging +1.2 This is a structured multi-part differential equations question with clear signposting through each step. Part (i) requires chain rule differentiation of the given volume formula, (ii) is routine partial fractions, (iii) is standard separation of variables with integration (though the algebra is slightly involved), and (iv) is numerical substitution. While it requires multiple techniques and careful algebraic manipulation, each step is clearly guided and uses standard C4 methods without requiring novel insight or problem-solving beyond following the prescribed path. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08l Interpret differential equation solutions: in context |
| Answer | Marks |
|---|---|
| \(\frac{dV}{dt} = -kV,\ \frac{dV}{dh} = 10\pi h - \pi h^2\) | B2 |
| \(\frac{dV}{dt} = \frac{dV}{dh}\times\frac{dh}{dt} \therefore -kV = (10\pi h - \pi h^2)\frac{dh}{dt}\) | M1 |
| \(-\frac{1}{3}k\pi h^2(15-h) = \pi h(10-h)\frac{dh}{dt}\) | |
| \(-kh(15-h) = 3(10-h)\frac{dh}{dt} \therefore \frac{dh}{dt} = -\frac{kh(15-h)}{3(10-h)}\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(\frac{3(10-h)}{h(15-h)} \equiv \frac{A}{h} + \frac{B}{15-h},\ 3(10-h) \equiv A(15-h) + Bh\) | M1 |
| \(h=0 \Rightarrow A=2,\ h=15 \Rightarrow B=-1 \therefore \frac{3(10-h)}{h(15-h)} \equiv \frac{2}{h} - \frac{1}{15-h}\) | A2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int\frac{3(10-h)}{h(15-h)}\ dh = \int -k\ dt,\ \int\left(\frac{2}{h} - \frac{1}{15-h}\right)dh = \int -k\ dt\) | M1 | |
| \(2\ln | h | + \ln |
| \(t=0,\ h=5 \Rightarrow 2\ln 5 + \ln 10 = c,\ c = \ln 250\) | M1 | |
| \(\ln\frac{h^2(15-h)}{250} = -kt,\ \frac{h^2(15-h)}{250} = e^{-kt},\ h^2(15-h) = 250e^{-kt}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(t=2,\ h=4 \Rightarrow 176 = 250e^{-2k}\) | ||
| \(k = -\frac{1}{2}\ln\frac{176}{250} = 0.175\) (3sf) | M1 A1 | (16) |
# Question 8:
## Part (i):
| $\frac{dV}{dt} = -kV,\ \frac{dV}{dh} = 10\pi h - \pi h^2$ | B2 | |
|---|---|---|
| $\frac{dV}{dt} = \frac{dV}{dh}\times\frac{dh}{dt} \therefore -kV = (10\pi h - \pi h^2)\frac{dh}{dt}$ | M1 | |
| $-\frac{1}{3}k\pi h^2(15-h) = \pi h(10-h)\frac{dh}{dt}$ | | |
| $-kh(15-h) = 3(10-h)\frac{dh}{dt} \therefore \frac{dh}{dt} = -\frac{kh(15-h)}{3(10-h)}$ | M1 A1 | |
## Part (ii):
| $\frac{3(10-h)}{h(15-h)} \equiv \frac{A}{h} + \frac{B}{15-h},\ 3(10-h) \equiv A(15-h) + Bh$ | M1 | |
|---|---|---|
| $h=0 \Rightarrow A=2,\ h=15 \Rightarrow B=-1 \therefore \frac{3(10-h)}{h(15-h)} \equiv \frac{2}{h} - \frac{1}{15-h}$ | A2 | |
## Part (iii):
| $\int\frac{3(10-h)}{h(15-h)}\ dh = \int -k\ dt,\ \int\left(\frac{2}{h} - \frac{1}{15-h}\right)dh = \int -k\ dt$ | M1 | |
|---|---|---|
| $2\ln|h| + \ln|15-h| = -kt + c$ | M1 A1 | |
| $t=0,\ h=5 \Rightarrow 2\ln 5 + \ln 10 = c,\ c = \ln 250$ | M1 | |
| $\ln\frac{h^2(15-h)}{250} = -kt,\ \frac{h^2(15-h)}{250} = e^{-kt},\ h^2(15-h) = 250e^{-kt}$ | M1 A1 | |
## Part (iv):
| $t=2,\ h=4 \Rightarrow 176 = 250e^{-2k}$ | | |
|---|---|---|
| $k = -\frac{1}{2}\ln\frac{176}{250} = 0.175$ (3sf) | M1 A1 | **(16)** |
**Total: (72)**8.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{72221d03-8a4e-49d6-b5f9-cdcb4c9cbf1a-3_252_757_267_484}
\end{center}
The diagram 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 $h \mathrm {~cm}$ and the volume of water in the bowl is $V \mathrm {~cm} ^ { 3 }$, 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 $V$.\\
(i) Show that
$$\frac { \mathrm { d } h } { \mathrm {~d} t } = - \frac { k h ( 15 - h ) } { 3 ( 10 - h ) } ,$$
where $k$ is a positive constant.\\
(ii) Express $\frac { 3 ( 10 - h ) } { h ( 15 - h ) }$ in partial fractions.
Given that when $t = 0 , h = 5$,\\
(iii) show that
$$h ^ { 2 } ( 15 - h ) = 250 \mathrm { e } ^ { - k t } .$$
Given also that when $t = 2 , h = 4$,\\
(iv) find the value of $k$ to 3 significant figures.
\hfill \mbox{\textit{OCR C4 Q8 [16]}}