| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (applied/contextual) |
| Difficulty | Standard +0.3 This is a standard separable differential equation with straightforward integration. Part (a) requires simple substitution to find k, and part (b) involves routine separation of variables and integration of a square root function. The context is familiar (tank draining) and the mathematical steps are all standard C4 techniques with no novel problem-solving required. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dh}{dt} = k\sqrt{(h-9)}\), \(9 < h \leq 200\); \(h=130\), \(\frac{dh}{dt}=-1.1\) | — | — |
| \(-1.1 = k\sqrt{(130-9)} \Rightarrow k = ...\) | M1 | Substitutes \(h=130\) and either \(\frac{dh}{dt}=-1.1\) or \(\frac{dh}{dt}=1.1\) into printed equation and rearranges to give \(k=...\) |
| \(k = -\frac{1}{10}\) or \(-0.1\) | A1 | \(k = -\frac{1}{10}\) or \(-0.1\) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\frac{dh}{\sqrt{(h-9)}} = \int k\,dt\) | B1 | Separates variables correctly. \(dh\) and \(dt\) should not be in wrong positions (can be implied by later working). Ignore integral signs. |
| \(\int(h-9)^{-\frac{1}{2}}\,dh = \int k\,dt\) | — | — |
| \(\frac{(h-9)^{\frac{1}{2}}}{\left(\frac{1}{2}\right)} = kt\,(+c)\) | M1 | Integrates \(\frac{\pm\lambda}{\sqrt{(h-9)}}\) to give \(\pm\mu\sqrt{(h-9)}\); \(\lambda,\mu\neq 0\) |
| \(\frac{(h-9)^{\frac{1}{2}}}{\left(\frac{1}{2}\right)} = kt\) or \(\frac{(h-9)^{\frac{1}{2}}}{\left(\frac{1}{2}\right)} = (\text{their }k)t\), with/without \(+c\) | A1 | Or equivalent, simplified or unsimplified |
| \(\{t=0, h=200\} \Rightarrow 2\sqrt{(200-9)} = k(0)+c\) | M1 | Some evidence of applying both \(t=0\) and \(h=200\) to changed equation containing constant of integration |
| \(\Rightarrow c = 2\sqrt{191} \Rightarrow 2(h-9)^{\frac{1}{2}} = -0.1t+2\sqrt{191}\) | dM1 | Dependent on previous M mark. Applies \(h=50\) and their value of \(c\) to changed equation and rearranges to find \(t=...\) |
| \(\{h=50\} \Rightarrow 2\sqrt{(50-9)} = -0.1t+2\sqrt{191}\) | — | — |
| \(t = 20\sqrt{191}-20\sqrt{41}\) or \(t=148.3430145... = 148\) (minutes) (nearest minute) | A1 cso | \(t = 20\sqrt{191}-20\sqrt{41}\) isw or awrt 148 |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_{200}^{50}\frac{dh}{\sqrt{(h-9)}} = \int_0^T k\,dt\) | B1 | Separates variables correctly. \(dh\) and \(dt\) should not be in wrong positions (can be implied by later working). Integral signs and limits not necessary. |
| \(\int_{200}^{50}(h-9)^{-\frac{1}{2}}\,dh = \int_0^T k\,dt\) | — | — |
| \(\left[\frac{(h-9)^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}\right]_{200}^{50} = [kt]_0^T\) | M1 | Integrates \(\frac{\pm\lambda}{\sqrt{(h-9)}}\) to give \(\pm\mu\sqrt{(h-9)}\); \(\lambda,\mu\neq 0\) |
| \(\frac{(h-9)^{\frac{1}{2}}}{\left(\frac{1}{2}\right)} = kt\) or equivalent, with/without limits | A1 | Or equivalent, simplified or unsimplified |
| \(2\sqrt{41}-2\sqrt{191} = kt\) or \(kT\) | M1 | Attempts to apply limits \(h=200\), \(h=50\) and (can be implied) \(t=0\) to changed equation |
| \(t = \frac{2\sqrt{41}-2\sqrt{191}}{-0.1}\) | dM1 | Dependent on previous M mark. Rearranges to find \(t=...\) |
| \(t = 20\sqrt{191}-20\sqrt{41}\) or \(t=148.3430145... = 148\) (minutes) (nearest minute) | A1 cso | \(t = 20\sqrt{191}-20\sqrt{41}\) or awrt 148 or 2 hours and awrt 28 minutes |
| [6] | ||
| Total: 8 |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Note | Guidance |
| 7(b) | Allow first B1 for writing \(\frac{dt}{dh} = \frac{1}{k\sqrt{(h-9)}}\) or \(\frac{dt}{dh} = \frac{1}{(\text{their }k)\sqrt{(h-9)}}\) or equivalent | |
| 7(b) | \(\frac{dt}{dh} = \frac{1}{k\sqrt{(h-9)}}\) leading to \(t = \frac{2}{k}\sqrt{(h-9)}\,(+c)\) with/without \(+c\) is B1M1A1 | |
| 7(b) | After finding \(k=0.1\) in part (a), it is only possible to gain full marks in part (b) by initially writing \(\frac{dh}{dt} = -k\sqrt{(h-9)}\) or \(\int\frac{dh}{\sqrt{(h-9)}} = \int -k\,dt\) or \(\frac{dh}{dt} = -0.1\sqrt{(h-9)}\) or \(\int\frac{dh}{\sqrt{(h-9)}} = \int -0.1\,dt\). Otherwise those candidates who find \(k=0.1\) in part (a) should lose at least the final A1 mark in part (b). |
# Question 7:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dh}{dt} = k\sqrt{(h-9)}$, $9 < h \leq 200$; $h=130$, $\frac{dh}{dt}=-1.1$ | — | — |
| $-1.1 = k\sqrt{(130-9)} \Rightarrow k = ...$ | M1 | Substitutes $h=130$ and either $\frac{dh}{dt}=-1.1$ or $\frac{dh}{dt}=1.1$ into printed equation and rearranges to give $k=...$ |
| $k = -\frac{1}{10}$ or $-0.1$ | A1 | $k = -\frac{1}{10}$ or $-0.1$ |
| **[2]** | | |
---
## Part (b) Way 1
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\frac{dh}{\sqrt{(h-9)}} = \int k\,dt$ | B1 | Separates variables correctly. $dh$ and $dt$ should not be in wrong positions (can be implied by later working). Ignore integral signs. |
| $\int(h-9)^{-\frac{1}{2}}\,dh = \int k\,dt$ | — | — |
| $\frac{(h-9)^{\frac{1}{2}}}{\left(\frac{1}{2}\right)} = kt\,(+c)$ | M1 | Integrates $\frac{\pm\lambda}{\sqrt{(h-9)}}$ to give $\pm\mu\sqrt{(h-9)}$; $\lambda,\mu\neq 0$ |
| $\frac{(h-9)^{\frac{1}{2}}}{\left(\frac{1}{2}\right)} = kt$ or $\frac{(h-9)^{\frac{1}{2}}}{\left(\frac{1}{2}\right)} = (\text{their }k)t$, with/without $+c$ | A1 | Or equivalent, simplified or unsimplified |
| $\{t=0, h=200\} \Rightarrow 2\sqrt{(200-9)} = k(0)+c$ | M1 | Some evidence of applying both $t=0$ and $h=200$ to changed equation containing constant of integration |
| $\Rightarrow c = 2\sqrt{191} \Rightarrow 2(h-9)^{\frac{1}{2}} = -0.1t+2\sqrt{191}$ | dM1 | **Dependent on previous M mark.** Applies $h=50$ and their value of $c$ to changed equation and rearranges to find $t=...$ |
| $\{h=50\} \Rightarrow 2\sqrt{(50-9)} = -0.1t+2\sqrt{191}$ | — | — |
| $t = 20\sqrt{191}-20\sqrt{41}$ or $t=148.3430145... = 148$ (minutes) (nearest minute) | A1 cso | $t = 20\sqrt{191}-20\sqrt{41}$ isw or awrt 148 |
| **[6]** | | |
---
## Part (b) Way 2
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_{200}^{50}\frac{dh}{\sqrt{(h-9)}} = \int_0^T k\,dt$ | B1 | Separates variables correctly. $dh$ and $dt$ should not be in wrong positions (can be implied by later working). Integral signs and limits not necessary. |
| $\int_{200}^{50}(h-9)^{-\frac{1}{2}}\,dh = \int_0^T k\,dt$ | — | — |
| $\left[\frac{(h-9)^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}\right]_{200}^{50} = [kt]_0^T$ | M1 | Integrates $\frac{\pm\lambda}{\sqrt{(h-9)}}$ to give $\pm\mu\sqrt{(h-9)}$; $\lambda,\mu\neq 0$ |
| $\frac{(h-9)^{\frac{1}{2}}}{\left(\frac{1}{2}\right)} = kt$ or equivalent, with/without limits | A1 | Or equivalent, simplified or unsimplified |
| $2\sqrt{41}-2\sqrt{191} = kt$ or $kT$ | M1 | Attempts to apply limits $h=200$, $h=50$ and (can be implied) $t=0$ to changed equation |
| $t = \frac{2\sqrt{41}-2\sqrt{191}}{-0.1}$ | dM1 | **Dependent on previous M mark.** Rearranges to find $t=...$ |
| $t = 20\sqrt{191}-20\sqrt{41}$ or $t=148.3430145... = 148$ (minutes) (nearest minute) | A1 cso | $t = 20\sqrt{191}-20\sqrt{41}$ or awrt 148 or 2 hours and awrt 28 minutes |
| **[6]** | | |
| **Total: 8** | | |
---
## Question 7 Notes
| Part | Note | Guidance |
|---|---|---|
| 7(b) | Allow first B1 for writing $\frac{dt}{dh} = \frac{1}{k\sqrt{(h-9)}}$ or $\frac{dt}{dh} = \frac{1}{(\text{their }k)\sqrt{(h-9)}}$ or equivalent | |
| 7(b) | $\frac{dt}{dh} = \frac{1}{k\sqrt{(h-9)}}$ leading to $t = \frac{2}{k}\sqrt{(h-9)}\,(+c)$ with/without $+c$ is B1M1A1 | |
| 7(b) | After finding $k=0.1$ in part (a), it is only possible to gain full marks in part (b) by **initially writing** $\frac{dh}{dt} = -k\sqrt{(h-9)}$ **or** $\int\frac{dh}{\sqrt{(h-9)}} = \int -k\,dt$ **or** $\frac{dh}{dt} = -0.1\sqrt{(h-9)}$ **or** $\int\frac{dh}{\sqrt{(h-9)}} = \int -0.1\,dt$. Otherwise those candidates who find $k=0.1$ in part (a) should lose at least the final A1 mark in part (b). | |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cd958ff3-ed4e-4bd7-aa4b-339da6d618a6-24_835_1160_255_529}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a vertical cylindrical tank of height 200 cm containing water. Water is leaking from a hole $P$ on the side of the tank.
At time $t$ minutes after the leaking starts, the height of water in the tank is $h \mathrm {~cm}$.
The height $h \mathrm {~cm}$ of the water in the tank satisfies the differential equation
$$\frac { \mathrm { d } h } { \mathrm {~d} t } = k ( h - 9 ) ^ { \frac { 1 } { 2 } } , \quad 9 < h \leqslant 200$$
where $k$ is a constant.
Given that, when $h = 130$, the height of the water is falling at a rate of 1.1 cm per minute,
\begin{enumerate}[label=(\alph*)]
\item find the value of $k$.
Given that the tank was full of water when the leaking started,
\item solve the differential equation with your value of $k$, to find the value of $t$ when $h = 50$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2017 Q7 [8]}}