| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (applied/contextual) |
| Difficulty | Standard +0.3 This is a structured, heavily scaffolded differential equations question where students are guided through each step with 'show that' prompts. Parts (i)-(iii) involve standard chain rule and separation of variables. Parts (iv)-(vi) require more careful algebraic manipulation and interpretation, but the algebraic identity in (v) is given to help with integration. While it's a longer question requiring sustained work, the scaffolding and standard techniques make it slightly easier than average for A-level. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V = \frac{1}{3}x^3 \Rightarrow \frac{dV}{dx} = x^2\) | B1 | |
| \(\frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt} = x^2\frac{dx}{dt}\) | M1 | oe eg \(dx/dt = dx/dV \cdot dV/dt = 1/x^2 \cdot -kx = -k/x\) |
| \(\Rightarrow x^2\frac{dx}{dt} = -kx\) | ||
| \(\Rightarrow x\frac{dx}{dt} = -k\) * | A1 [3] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x\frac{dx}{dt} = -k \Rightarrow \int x\,dx = \int -k\,dt\) | M1 | separating variables and intention to integrate |
| \(\Rightarrow \frac{1}{2}x^2 = -kt + c\) | A1 | condone absence of \(c\) |
| When \(t=0\), \(x=10 \Rightarrow 50 = c\) | B1 | finding \(c\) correctly ft their integral of form \(ax^2 = bt+c\) where \(a,b\) non zero constants |
| \(\Rightarrow \frac{1}{2}x^2 = 50 - kt\) | ||
| \(\Rightarrow x = \sqrt{(100-2kt)}\) * | A1 [4] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(t=50\), \(x=0\) | M1 | |
| \(0 = 100 - 100k \Rightarrow k=1\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(dV/dt = 1 - kx = 1 - x\) | M1 | for \(dV/dt = 1-kx\) or better |
| \(\Rightarrow x^2\frac{dx}{dt} = 1-x\) | ||
| \(\Rightarrow \frac{dx}{dt} = \frac{1-x}{x^2}\) * | A1 [2] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{1-x} - x - 1 = \frac{1-(1-x)x-(1-x)}{1-x}\) | M1 | combining to single fraction |
| \(= \frac{1-x+x^2-1+x}{1-x} = \frac{x^2}{1-x}\) * | A1 | AG |
| \(\int\frac{x^2}{1-x}\,dx = \int dt \Rightarrow \int\left(\frac{1}{1-x} - x - 1\right)dx = t+c\) | M1 | separating variables & subst for \(x^2/(1-x)\) and intending to integrate |
| \(\Rightarrow -\ln(1-x) - \frac{1}{2}x^2 - x = t + c\) | A1 | condone absence of \(c\) |
| When \(t=0\), \(x=0 \Rightarrow c = -\ln 1 - 0 - 0 = 0\) | B1 | finding \(c\) for equation of correct form eg \(c=0\), or \(\pm\ln 1\) (allow \(c=0\) without evaluation here) |
| \(\Rightarrow t = \ln\left(\frac{1}{1-x}\right) - \frac{1}{2}x^2 - x\) * | A1 [6] | cao AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| understanding that \(\ln(1/0)\) or \(1/0\) is undefined oe | B1 [1] | www |
## Question 2:
### Part 2(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V = \frac{1}{3}x^3 \Rightarrow \frac{dV}{dx} = x^2$ | B1 | |
| $\frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt} = x^2\frac{dx}{dt}$ | M1 | oe eg $dx/dt = dx/dV \cdot dV/dt = 1/x^2 \cdot -kx = -k/x$ |
| $\Rightarrow x^2\frac{dx}{dt} = -kx$ | | |
| $\Rightarrow x\frac{dx}{dt} = -k$ * | A1 [3] | **AG** |
### Part 2(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x\frac{dx}{dt} = -k \Rightarrow \int x\,dx = \int -k\,dt$ | M1 | separating variables and intention to integrate |
| $\Rightarrow \frac{1}{2}x^2 = -kt + c$ | A1 | condone absence of $c$ |
| When $t=0$, $x=10 \Rightarrow 50 = c$ | B1 | finding $c$ correctly ft their integral of form $ax^2 = bt+c$ where $a,b$ non zero constants |
| $\Rightarrow \frac{1}{2}x^2 = 50 - kt$ | | |
| $\Rightarrow x = \sqrt{(100-2kt)}$ * | A1 [4] | **AG** |
### Part 2(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $t=50$, $x=0$ | M1 | |
| $0 = 100 - 100k \Rightarrow k=1$ | A1 [2] | |
### Part 2(iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $dV/dt = 1 - kx = 1 - x$ | M1 | for $dV/dt = 1-kx$ or better |
| $\Rightarrow x^2\frac{dx}{dt} = 1-x$ | | |
| $\Rightarrow \frac{dx}{dt} = \frac{1-x}{x^2}$ * | A1 [2] | **AG** |
### Part 2(v):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{1-x} - x - 1 = \frac{1-(1-x)x-(1-x)}{1-x}$ | M1 | combining to single fraction | or long division or cross multiplying |
| $= \frac{1-x+x^2-1+x}{1-x} = \frac{x^2}{1-x}$ * | A1 | **AG** | check signs |
| $\int\frac{x^2}{1-x}\,dx = \int dt \Rightarrow \int\left(\frac{1}{1-x} - x - 1\right)dx = t+c$ | M1 | separating variables & subst for $x^2/(1-x)$ and intending to integrate | need both sides of integral |
| $\Rightarrow -\ln(1-x) - \frac{1}{2}x^2 - x = t + c$ | A1 | condone absence of $c$ | accept $\ln(1/(1-x))$ as $-\ln(1-x)$ www |
| When $t=0$, $x=0 \Rightarrow c = -\ln 1 - 0 - 0 = 0$ | B1 | finding $c$ for equation of correct form eg $c=0$, or $\pm\ln 1$ (allow $c=0$ without evaluation here) | ie $a\ln(1-x)+bx^2+dx=et+c$ $a,b,d,e$ non zero constants; do not allow if $c=0$ without evaluation |
| $\Rightarrow t = \ln\left(\frac{1}{1-x}\right) - \frac{1}{2}x^2 - x$ * | A1 [6] | cao **AG** | |
### Part 2(vi):
| Answer/Working | Mark | Guidance |
|---|---|---|
| understanding that $\ln(1/0)$ or $1/0$ is undefined oe | B1 [1] | www | $\ln(1/0) = \ln 0$, $1/0 = \infty$ and $\ln(1/0) = \infty$ are all B0 |
---2 Water is leaking from a container. After $t$ seconds, the depth of water in the container is $x \mathrm {~cm}$, and the volume of water is $V \mathrm {~cm} ^ { 3 }$, where $V = \frac { 1 } { 3 } x ^ { 3 }$. The rate at which water is lost is proportional to $x$, so that $\frac { \mathrm { d } V } { \mathrm {~d} t } = - k x$, where $k$ is a constant.\\
(i) Show that $x \frac { \mathrm {~d} x } { \mathrm {~d} t } = - k$.
Initially, the depth of water in the container is 10 cm .\\
(ii) Show by integration that $x = \sqrt { 100 - 2 k t }$.\\
(iii) Given that the container empties after 50 seconds, find $k$.
Once the container is empty, water is poured into it at a constant rate of $1 \mathrm {~cm} ^ { 3 }$ per second. The container continues to lose water as before.\\
(iv) Show that, $t$ seconds after starting to pour the water in, $\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 1 - x } { x ^ { 2 } }$.\\
(v) Show that $\frac { 1 } { 1 - x } - x - 1 = \frac { x ^ { 2 } } { 1 - x }$.
Hence solve the differential equation in part (iv) to show that
$$t = \ln \left( \frac { 1 } { 1 - x } \right) - \frac { 1 } { 2 } x ^ { 2 } - x$$
(vi) Show that the depth cannot reach 1 cm .
\hfill \mbox{\textit{OCR MEI C4 Q2 [18]}}