| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Related rates |
| Difficulty | Standard +0.3 This is a straightforward C4 differential equations question requiring students to form a differential equation from a worded scenario (dP/dt = k/P), separate variables and integrate to get P² = 2kt + c, then use initial conditions to find constants. Part (ii) involves simple substitution to check model predictions. The inverse proportionality setup and variable separation are standard C4 techniques with no novel problem-solving required, making this 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dP}{dt} = \frac{k}{P}\) | B1 | Or \(\frac{dP}{dt} = \frac{1}{kP}\); \(k\) should be unspecified at this stage |
| \(k = 100\) from \(\frac{dP}{dt} = \frac{k}{P}\) | B1 | Or \(k=0.01\) from \(\frac{dP}{dt} = \frac{1}{kP}\); may be seen later |
| \(\int P\,dP = \int(\text{their }k)\,dt\) | M1* | Allow \(k=1\); allow omission of \(\int\) and recovery of omission of one operator for M1*A1 |
| \(\frac{P^2}{2} = kt + c\) | A1 | Or \(t = \frac{P^2}{2k} + d\); if M0, SC2 for \(\ln P = kt + c\) thereafter only M1 may be earned |
| Substitution of \(t=0\) and \(P=100\) | M1dep* | May follow incorrect algebraic manipulation, but equation must include \(c\) (or \(d\)); NB \(c=5000\) or \(d=-50\) |
| \(P = \sqrt{10000 + 200t}\) or \(10\sqrt{100+2t}\) or \(P=\sqrt{200(50+t)}\) isw cao | A1 [6] | Allow recovery from e.g. use of \(x\) for \(P\) throughout, but withhold final A1 for e.g. \(x=\sqrt{10000+200t}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(t=8\), \(P=107.7\) or \(108\) so model was a good fit in 2008 oe | B1 | Or \(t=8.3(2)\) when \(P=108\) + comment; value of \(P\) or \(t\) must be found and correct comment made in each case; comments may be in same sentence |
| \(t=13\), \(P=112(.2)\), so model was not appropriate in 2013 oe | B1 [2] | Or \(t=31.9(2)\) or \(32\) when \(P=128\) + comment; if B0B0, SC1 for both values found no FT marks available; comments on trends, extrapolation etc do not score; just ticks/crosses etc do not score |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dP}{dt} = \frac{k}{P}$ | B1 | Or $\frac{dP}{dt} = \frac{1}{kP}$; $k$ should be unspecified at this stage |
| $k = 100$ from $\frac{dP}{dt} = \frac{k}{P}$ | B1 | Or $k=0.01$ from $\frac{dP}{dt} = \frac{1}{kP}$; may be seen later |
| $\int P\,dP = \int(\text{their }k)\,dt$ | M1* | Allow $k=1$; allow omission of $\int$ and recovery of omission of one operator for M1*A1 |
| $\frac{P^2}{2} = kt + c$ | A1 | Or $t = \frac{P^2}{2k} + d$; if M0, SC2 for $\ln P = kt + c$ thereafter only M1 may be earned |
| Substitution of $t=0$ and $P=100$ | M1dep* | May follow incorrect algebraic manipulation, but equation must include $c$ (or $d$); NB $c=5000$ or $d=-50$ |
| $P = \sqrt{10000 + 200t}$ or $10\sqrt{100+2t}$ or $P=\sqrt{200(50+t)}$ isw cao | A1 **[6]** | Allow recovery from e.g. use of $x$ for $P$ throughout, but withhold final A1 for e.g. $x=\sqrt{10000+200t}$ |
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## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $t=8$, $P=107.7$ or $108$ so model was a good fit in 2008 oe | B1 | Or $t=8.3(2)$ when $P=108$ + comment; value of $P$ or $t$ must be found and correct comment made in each case; comments may be in same sentence |
| $t=13$, $P=112(.2)$, so model was not appropriate in 2013 oe | B1 **[2]** | Or $t=31.9(2)$ or $32$ when $P=128$ + comment; if B0B0, SC1 for both values found no FT marks available; comments on trends, extrapolation etc do not score; just ticks/crosses etc do not score |
---8 In the year 2000 the population density, $P$, of a village was 100 people per $\mathrm { km } ^ { 2 }$, and was increasing at the rate of 1 person per $\mathrm { km } ^ { 2 }$ per year. The rate of increase of the population density is thought to be inversely proportional to the size of the population density. The time in years after the year 2000 is denoted by $t$.\\
(i) Write down a differential equation to model this situation, and solve it to express $P$ in terms of $t$.\\
(ii) In 2008 the population density of the village was 108 people per $\mathrm { km } ^ { 2 }$ and in 2013 it was 128 people per $\mathrm { km } ^ { 2 }$. Determine how well the model fits these figures.
\hfill \mbox{\textit{OCR C4 2015 Q8 [8]}}