| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Exponential growth/decay - approach to limit (dN/dt = k(N - N₀)) |
| Difficulty | Moderate -0.3 This is a standard separable differential equations question with straightforward integration and interpretation. Part (i) guides students through the integration, parts (ii)-(iii) are routine substitutions, and part (iv) requires only recognizing the limiting behavior. The algebra is simple and the question type is common in C4 textbooks, making it slightly easier than average. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Sep variables eg \(\int \frac{1}{6-h}(dh) = \int \frac{1}{20}(dt)\) | *M1 | s.o.i. Or \(\frac{dt}{dh} = \frac{20}{6-h} \to M1\) |
| LHS \(= -\ln(6-h)\) | A1 | & then \(t = -20\ln(6-h) (+c) \to A1+A1\) |
| RHS \(= \frac{1}{20}t\) \((+c)\) | A1 | |
| Subst \(t = 0, h = 1\) into equation containing 'c' | dep*M1 | |
| Correct value of their \(c = -(20)\ln 5\) WWW | A1 | or \((20)\ln 5\) if on LHS |
| Produce \(t = 20\ln\frac{5}{6-h}\) WWW AG | A1 | Must see \(\ln 5 - \ln(6-h)\) |
| (ii) When \(h = 2\), \(t = 20\ln\frac{5}{4} = 4.46(2871)\) | B1 | Accept 4.5, \(4\frac{1}{2}\) |
| (iii) Solve \(10 = 20\ln\frac{5}{6-h}\) to \(\frac{5}{6-h} = e^{0.5}\) | M1 | or \(6 - h = 5e^{-0.5}\) or suitable \(\frac{1}{2}\)-way stage |
| \(h = 2.97(2.9673467...)\) | A1 | \(6 - 5e^{-0.5}\) or \(6 - e^{1.109}\) |
| [In (ii),(iii) accept non-decimal (exact) answers but \(-1\) once.] Accept truncated values in (ii),(iii). | ||
| (iv) Any indication of (approximately) 6 (m) | B1 |
(i) Sep variables eg $\int \frac{1}{6-h}(dh) = \int \frac{1}{20}(dt)$ | *M1 | s.o.i. Or $\frac{dt}{dh} = \frac{20}{6-h} \to M1$
LHS $= -\ln(6-h)$ | A1 | & then $t = -20\ln(6-h) (+c) \to A1+A1$
RHS $= \frac{1}{20}t$ $(+c)$ | A1 |
Subst $t = 0, h = 1$ into equation containing 'c' | dep*M1 |
Correct value of their $c = -(20)\ln 5$ WWW | A1 | or $(20)\ln 5$ if on LHS
Produce $t = 20\ln\frac{5}{6-h}$ WWW AG | A1 | Must see $\ln 5 - \ln(6-h)$ | 6
(ii) When $h = 2$, $t = 20\ln\frac{5}{4} = 4.46(2871)$ | B1 | Accept 4.5, $4\frac{1}{2}$
(iii) Solve $10 = 20\ln\frac{5}{6-h}$ to $\frac{5}{6-h} = e^{0.5}$ | M1 | or $6 - h = 5e^{-0.5}$ or suitable $\frac{1}{2}$-way stage
$h = 2.97(2.9673467...)$ | A1 | $6 - 5e^{-0.5}$ or $6 - e^{1.109}$
[In (ii),(iii) accept non-decimal (exact) answers but $-1$ once.] Accept truncated values in (ii),(iii). | | | | 2
(iv) Any indication of (approximately) 6 (m) | B1 | | 1
---8 The height, $h$ metres, of a shrub $t$ years after planting is given by the differential equation
$$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { 6 - h } { 20 }$$
A shrub is planted when its height is 1 m .\\
(i) Show by integration that $t = 20 \ln \left( \frac { 5 } { 6 - h } \right)$.\\
(ii) How long after planting will the shrub reach a height of 2 m ?\\
(iii) Find the height of the shrub 10 years after planting.\\
(iv) State the maximum possible height of the shrub.
\hfill \mbox{\textit{OCR C4 2007 Q8 [10]}}