| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (applied/contextual) |
| Difficulty | Standard +0.3 This is a standard differential equations question requiring separation of variables and integration. The setup is given clearly, part (a) involves simple substitution to find the constant, part (b) is routine separation and integration of √x, and part (c) is straightforward substitution. Slightly easier than average due to the scaffolded structure and standard technique application. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dx}{dt} = k\sqrt{x}\) | B1* | Set up a correct differential equation. Allow \(-k\) |
| \(-0.0032 = k\sqrt{0.64}\) so \(k = -0.004\), hence \(\frac{dx}{dt} = -0.004\sqrt{x}\) A.G. | B1d* | Obtain correct differential equation www. Must use \(-0.0032\) when finding \(k\). B0 if \(k = 0.004\) even if then comment about 'decreasing' and sign changed |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int -0.004\, dt = \int x^{-\frac{1}{2}}\, dx\) | M1* | Separate variables and attempt integration. Condone \(dx\), \(dt\) and/or integral sign not being explicit. Could instead invert both sides. Increase by 1 in both powers |
| \(-0.004t = 2x^{\frac{1}{2}} + c\) | A1 | Correct integral — could still be in terms of \(k\). Condone no \(+ c\). Any correct equation e.g. \(t = -500\sqrt{x} + c\) |
| \(-0.004 \times 100 = 2\sqrt{0.64} + c\), \(c = -2\) | M1d* | Use \(t = 100\), \(x = 0.64\) to find \(c\). Substitute given values into their general solution and attempt \(c\). NB check method carefully for their equation and position of \(c\) |
| \(2\sqrt{x} = 2 - 0.004t\), \(x = (1-0.002t)^2\) | A1 | Correct equation. Any equiv as long as \(x\) in terms of \(t\). ISW if correct equation subsequently spoilt |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| when \(x = 0\), \(t = 500\) | M1 | Use \(x = 0\) to attempt a value for \(t\). Must be using \(x = 0\) in their particular solution, with a non-zero value for \(c\) |
| so tank will be empty after 500 seconds | A1 | Units needed. Must come from using a correct equation. Allow 8 mins 20 secs, \(8.\dot{3}\) mins or 8.33 mins |
| [2] |
## Question 8:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dx}{dt} = k\sqrt{x}$ | B1* | Set up a correct differential equation. Allow $-k$ |
| $-0.0032 = k\sqrt{0.64}$ so $k = -0.004$, hence $\frac{dx}{dt} = -0.004\sqrt{x}$ **A.G.** | B1d* | Obtain correct differential equation www. Must use $-0.0032$ when finding $k$. B0 if $k = 0.004$ even if then comment about 'decreasing' and sign changed |
| **[2]** | | |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int -0.004\, dt = \int x^{-\frac{1}{2}}\, dx$ | M1* | Separate variables and attempt integration. Condone $dx$, $dt$ and/or integral sign not being explicit. Could instead invert both sides. Increase by 1 in both powers |
| $-0.004t = 2x^{\frac{1}{2}} + c$ | A1 | Correct integral — could still be in terms of $k$. Condone no $+ c$. Any correct equation e.g. $t = -500\sqrt{x} + c$ |
| $-0.004 \times 100 = 2\sqrt{0.64} + c$, $c = -2$ | M1d* | Use $t = 100$, $x = 0.64$ to find $c$. Substitute given values into their general solution and attempt $c$. NB check method carefully for their equation and position of $c$ |
| $2\sqrt{x} = 2 - 0.004t$, $x = (1-0.002t)^2$ | A1 | Correct equation. Any equiv as long as $x$ in terms of $t$. ISW if correct equation subsequently spoilt |
| **[4]** | | |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| when $x = 0$, $t = 500$ | M1 | Use $x = 0$ to attempt a value for $t$. Must be using $x = 0$ in their particular solution, with a non-zero value for $c$ |
| so tank will be empty after 500 seconds | A1 | Units needed. Must come from using a correct equation. Allow 8 mins 20 secs, $8.\dot{3}$ mins or 8.33 mins |
| **[2]** | | |8 A cylindrical tank is initially full of water. There is a small hole at the base of the tank out of which the water leaks.
The height of water in the tank is $x \mathrm {~m}$ at time $t$ seconds. The rate of change of the height of water may be modelled by the assumption that it is proportional to the square root of the height of water.
When $t = 100 , x = 0.64$ and, at this instant, the height is decreasing at a rate of $0.0032 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } x } { \mathrm {~d} t } = - 0.004 \sqrt { x }$.
\item Find an expression for $x$ in terms of $t$.
\item Hence determine at what time, according to this model, the tank will be empty.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2019 Q8 [8]}}