| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Substitution method |
| Difficulty | Standard +0.3 This is a structured multi-part C4 differential equations question with clear scaffolding. Parts (a) and (b) involve straightforward modeling and substitution. Part (c) requires standard separation of variables. Part (d) provides the substitution needed for integration. While it requires multiple techniques (modeling, separation, substitution integration), each step is guided and uses standard C4 methods, making it slightly easier than average. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dV}{dt} = 1600 - c\sqrt{h}\) or \(\frac{dV}{dt}=1600-k\sqrt{h}\) | M1 | Either of these statements |
| \((V=4000h) \Rightarrow \frac{dV}{dh}=4000\) | M1 | \(\frac{dV}{dh}=4000\) or \(\frac{dh}{dV}=\frac{1}{4000}\) |
| \(\frac{dh}{dt}=\frac{dh}{dV}\times\frac{dV}{dt} = \frac{\frac{dV}{dt}}{\frac{dV}{dh}}\) | ||
| \(\frac{dh}{dt}=\frac{1600-c\sqrt{h}}{4000}=\frac{1600}{4000}-\frac{c\sqrt{h}}{4000}=0.4-k\sqrt{h}\) | A1 AG | Convincing proof of \(\frac{dh}{dt}\) [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| When \(h=25\), \(\frac{dV}{dt}=400\): \(400=c\sqrt{25}=c(5) \Rightarrow c=80\) | ||
| \(k=\frac{c}{4000}=\frac{80}{4000}=0.02\) | B1 AG | Proof that \(k=0.02\) [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(400=4000k\sqrt{25} \Rightarrow 400=k(20000) \Rightarrow k=\frac{400}{20000}=0.02\) | B1 AG | Using 400, 4000 and \(h=25\) or \(\sqrt{h}=5\). Proof that \(k=0.02\) [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dh}{dt}=0.4-k\sqrt{h} \Rightarrow \int\frac{dh}{0.4-k\sqrt{h}}=\int dt\) | M1 oe | Separates the variables with \(\int\frac{dh}{0.4-k\sqrt{h}}\) and \(\int dt\) on either side, integral signs not necessary |
| \(\therefore \text{time required} = \int_0^{100}\frac{1}{0.4-0.02\sqrt{h}}dh\ (\div 0.02)\) | ||
| \(\text{time required} = \int_0^{100}\frac{50}{20-\sqrt{h}}dh\) | A1 AG | Correct proof [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\int_0^{100}\frac{50}{20-\sqrt{h}}dh\) with substitution \(h=(20-x)^2\) | ||
| \(\frac{dh}{dx}=2(20-x)(-1)\) or \(\frac{dh}{dx}=-2(20-x)\) | B1 aef | Correct \(\frac{dh}{dx}\) |
| \(h=(20-x)^2 \Rightarrow \sqrt{h}=20-x \Rightarrow x=20-\sqrt{h}\) | ||
| \(\int\frac{50}{20-\sqrt{h}}dh = \int\frac{50}{x}\cdot -2(20-x)dx\) | M1 | \(\pm\lambda\int\frac{20-x}{x}dx\) or \(\pm\lambda\int\frac{20-x}{20-(20-x)}dx\) where \(\lambda\) is a constant |
| \(= 100\int\frac{x-20}{x}dx = 100\int\left(1-\frac{20}{x}\right)dx\) | ||
| \(= 100(x-20\ln x)\ (+c)\) | M1 | \(\pm\alpha x \pm \beta\ln x;\ \alpha,\beta\neq 0\) |
| A1 | \(100x - 2000\ln x\) | |
| Change limits: \(h=0\Rightarrow x=20\); \(h=100\Rightarrow x=10\) | ||
| \(\int_0^{100}\frac{50}{20-\sqrt{h}}dh = \left[100x-2000\ln x\right]_{20}^{10}\) | ddM1 | Correct use of limits, putting them in the correct way round. Either \(x=10\) and \(x=20\) or \(h=100\) and \(h=0\) |
| \(= (1000-2000\ln 10)-(2000-2000\ln 20)\) | ||
| \(= 2000\ln 20 - 2000\ln 10 - 1000\) | Combining logs to give \(2000\ln 2 - 1000\) | |
| \(= 2000\ln 2 - 1000\) | A1 aef | or \(-2000\ln\left(\frac{1}{2}\right)-1000\) [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Time required \(= 2000\ln 2 - 1000 = 386.2943611\ldots\) sec \(= 386\) seconds (nearest second) \(= 6\) minutes and 26 seconds | B1 | 6 minutes, 26 seconds [1] |
# Question 8:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dV}{dt} = 1600 - c\sqrt{h}$ or $\frac{dV}{dt}=1600-k\sqrt{h}$ | M1 | Either of these statements |
| $(V=4000h) \Rightarrow \frac{dV}{dh}=4000$ | M1 | $\frac{dV}{dh}=4000$ or $\frac{dh}{dV}=\frac{1}{4000}$ |
| $\frac{dh}{dt}=\frac{dh}{dV}\times\frac{dV}{dt} = \frac{\frac{dV}{dt}}{\frac{dV}{dh}}$ | | |
| $\frac{dh}{dt}=\frac{1600-c\sqrt{h}}{4000}=\frac{1600}{4000}-\frac{c\sqrt{h}}{4000}=0.4-k\sqrt{h}$ | A1 AG | Convincing proof of $\frac{dh}{dt}$ **[3]** |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| When $h=25$, $\frac{dV}{dt}=400$: $400=c\sqrt{25}=c(5) \Rightarrow c=80$ | | |
| $k=\frac{c}{4000}=\frac{80}{4000}=0.02$ | B1 AG | Proof that $k=0.02$ **[1]** |
**Aliter Way 2:**
| Working | Mark | Guidance |
|---------|------|----------|
| $400=4000k\sqrt{25} \Rightarrow 400=k(20000) \Rightarrow k=\frac{400}{20000}=0.02$ | B1 AG | Using 400, 4000 and $h=25$ or $\sqrt{h}=5$. Proof that $k=0.02$ **[1]** |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dh}{dt}=0.4-k\sqrt{h} \Rightarrow \int\frac{dh}{0.4-k\sqrt{h}}=\int dt$ | M1 oe | Separates the variables with $\int\frac{dh}{0.4-k\sqrt{h}}$ and $\int dt$ on either side, integral signs not necessary |
| $\therefore \text{time required} = \int_0^{100}\frac{1}{0.4-0.02\sqrt{h}}dh\ (\div 0.02)$ | | |
| $\text{time required} = \int_0^{100}\frac{50}{20-\sqrt{h}}dh$ | A1 AG | Correct proof **[2]** |
## Part (d):
| Working | Mark | Guidance |
|---------|------|----------|
| $\int_0^{100}\frac{50}{20-\sqrt{h}}dh$ with substitution $h=(20-x)^2$ | | |
| $\frac{dh}{dx}=2(20-x)(-1)$ or $\frac{dh}{dx}=-2(20-x)$ | B1 aef | Correct $\frac{dh}{dx}$ |
| $h=(20-x)^2 \Rightarrow \sqrt{h}=20-x \Rightarrow x=20-\sqrt{h}$ | | |
| $\int\frac{50}{20-\sqrt{h}}dh = \int\frac{50}{x}\cdot -2(20-x)dx$ | M1 | $\pm\lambda\int\frac{20-x}{x}dx$ or $\pm\lambda\int\frac{20-x}{20-(20-x)}dx$ where $\lambda$ is a constant |
| $= 100\int\frac{x-20}{x}dx = 100\int\left(1-\frac{20}{x}\right)dx$ | | |
| $= 100(x-20\ln x)\ (+c)$ | M1 | $\pm\alpha x \pm \beta\ln x;\ \alpha,\beta\neq 0$ |
| | A1 | $100x - 2000\ln x$ |
| Change limits: $h=0\Rightarrow x=20$; $h=100\Rightarrow x=10$ | | |
| $\int_0^{100}\frac{50}{20-\sqrt{h}}dh = \left[100x-2000\ln x\right]_{20}^{10}$ | ddM1 | Correct use of limits, putting them in the correct way round. Either $x=10$ and $x=20$ or $h=100$ and $h=0$ |
| $= (1000-2000\ln 10)-(2000-2000\ln 20)$ | | |
| $= 2000\ln 20 - 2000\ln 10 - 1000$ | | Combining logs to give $2000\ln 2 - 1000$ |
| $= 2000\ln 2 - 1000$ | A1 aef | or $-2000\ln\left(\frac{1}{2}\right)-1000$ **[6]** |
## Part (e):
| Working | Mark | Guidance |
|---------|------|----------|
| Time required $= 2000\ln 2 - 1000 = 386.2943611\ldots$ sec $= 386$ seconds (nearest second) $= 6$ minutes and 26 seconds | B1 | 6 minutes, 26 seconds **[1]** |
**Total: 13 marks**8. Liquid is pouring into a large vertical circular cylinder at a constant rate of $1600 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$ and is leaking out of a hole in the base, at a rate proportional to the square root of the height of the liquid already in the cylinder. The area of the circular cross section of the cylinder is $4000 \mathrm {~cm} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that at time $t$ seconds, the height $h \mathrm {~cm}$ of liquid in the cylinder satisfies the differential equation
$$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.4 - k \sqrt { } h \text {, where } k \text { is a positive constant. }$$
When $h = 25$, water is leaking out of the hole at $400 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$.
\item Show that $k = 0.02$
\item Separate the variables of the differential equation
$$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.4 - 0.02 \sqrt { } h$$
to show that the time taken to fill the cylinder from empty to a height of 100 cm is given by
$$\int _ { 0 } ^ { 100 } \frac { 50 } { 20 - \sqrt { } h } \mathrm {~d} h$$
Using the substitution $h = ( 20 - x ) ^ { 2 }$, or otherwise,
\item find the exact value of $\int _ { 0 } ^ { 100 } \frac { 50 } { 20 - \sqrt { h } } \mathrm {~d} h$.
\item Hence find the time taken to fill the cylinder from empty to a height of 100 cm , giving your answer in minutes and seconds to the nearest second.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2008 Q8 [13]}}