| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - partial fractions |
| Difficulty | Standard +0.3 This is a structured differential equations question with clear scaffolding through multiple parts. Part (i) requires simple algebraic manipulation, part (ii) is verification by substitution (routine), part (iii) involves standard separation of variables with partial fractions (a core C4 technique), and parts (iv)-(v) require interpretation of solutions. While it covers several skills, each step is guided and uses standard methods without requiring novel insight or complex problem-solving. |
| Spec | 1.02y Partial fractions: decompose rational functions1.06a Exponential function: a^x and e^x graphs and properties1.08k Separable differential equations: dy/dx = f(x)g(y)4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dh}{dt} = 0 \Rightarrow 20-h=0 \Rightarrow h=20\) | B1 | \(h=20\); tree stops growing / reaches maximum height of 20m |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h = 20(1-e^{-0.1t})\), \(\frac{dh}{dt} = 20 \times 0.1e^{-0.1t} = 2e^{-0.1t}\) | M1 | Differentiate |
| \(10\frac{dh}{dt} = 20e^{-0.1t}\) | A1 | |
| \(20 - h = 20 - 20(1-e^{-0.1t}) = 20e^{-0.1t}\) ✓ | A1 | Verify RHS = LHS |
| When \(t=0\): \(h = 20(1-1) = 0\) ✓ | B1 B1 | Initial condition verified; conclusion stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(200\frac{dh}{dt} = 400 - h^2 \Rightarrow \frac{200}{400-h^2}\,dh = dt\) | M1 | Separate variables |
| \(\frac{1}{400-h^2} = \frac{1}{(20-h)(20+h)} = \frac{A}{20-h} + \frac{B}{20+h}\) | M1 | Partial fractions |
| \(A = \frac{1}{40}\), \(B = \frac{1}{40}\) | A1 | Correct values |
| \(\frac{200}{40}\int\left(\frac{1}{20-h}+\frac{1}{20+h}\right)dh = \int dt\) | ||
| \(5\left[-\ln(20-h)+\ln(20+h)\right] = t + c\) | M1 A1 | Integrate correctly |
| \(5\ln\frac{20+h}{20-h} = t + c\) | ||
| When \(t=0\), \(h=0\): \(c=0\) | M1 | Apply initial condition |
| \(\frac{20+h}{20-h} = e^{0.2t}\) | M1 | Rearrange |
| \(20+h = (20-h)e^{0.2t}\); \(h(1+e^{0.2t}) = 20(e^{0.2t}-1)\) | M1 | Solve for \(h\) |
| \(h = \frac{20(e^{0.2t}-1)}{e^{0.2t}+1} = \frac{20(1-e^{-0.2t})}{1+e^{-0.2t}}\) | A1 | Multiply numerator and denominator by \(e^{-0.2t}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| As \(t\to\infty\), \(e^{-0.2t}\to 0\), so \(h\to\frac{20(1)}{1} = 20\) | B1 | Long-term height is 20m |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Model 1: \(t=1\), \(h=20(1-e^{-0.1}) \approx 1.903\) m | M1 | Calculate \(h\) for \(t=1\) in both models |
| Model 2: \(t=1\), \(h=\frac{20(1-e^{-0.2})}{1+e^{-0.2}} \approx 1.987\) m | A1 | Both values correct |
| Model 2 is closer to 2m, so Model 2 fits better | A1 | Correct conclusion with justification |
## Question 8(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dh}{dt} = 0 \Rightarrow 20-h=0 \Rightarrow h=20$ | B1 | $h=20$; tree stops growing / reaches maximum height of 20m |
---
## Question 8(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h = 20(1-e^{-0.1t})$, $\frac{dh}{dt} = 20 \times 0.1e^{-0.1t} = 2e^{-0.1t}$ | M1 | Differentiate |
| $10\frac{dh}{dt} = 20e^{-0.1t}$ | A1 | |
| $20 - h = 20 - 20(1-e^{-0.1t}) = 20e^{-0.1t}$ ✓ | A1 | Verify RHS = LHS |
| When $t=0$: $h = 20(1-1) = 0$ ✓ | B1 B1 | Initial condition verified; conclusion stated |
---
## Question 8(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $200\frac{dh}{dt} = 400 - h^2 \Rightarrow \frac{200}{400-h^2}\,dh = dt$ | M1 | Separate variables |
| $\frac{1}{400-h^2} = \frac{1}{(20-h)(20+h)} = \frac{A}{20-h} + \frac{B}{20+h}$ | M1 | Partial fractions |
| $A = \frac{1}{40}$, $B = \frac{1}{40}$ | A1 | Correct values |
| $\frac{200}{40}\int\left(\frac{1}{20-h}+\frac{1}{20+h}\right)dh = \int dt$ | | |
| $5\left[-\ln(20-h)+\ln(20+h)\right] = t + c$ | M1 A1 | Integrate correctly |
| $5\ln\frac{20+h}{20-h} = t + c$ | | |
| When $t=0$, $h=0$: $c=0$ | M1 | Apply initial condition |
| $\frac{20+h}{20-h} = e^{0.2t}$ | M1 | Rearrange |
| $20+h = (20-h)e^{0.2t}$; $h(1+e^{0.2t}) = 20(e^{0.2t}-1)$ | M1 | Solve for $h$ |
| $h = \frac{20(e^{0.2t}-1)}{e^{0.2t}+1} = \frac{20(1-e^{-0.2t})}{1+e^{-0.2t}}$ | A1 | Multiply numerator and denominator by $e^{-0.2t}$ |
---
## Question 8(iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| As $t\to\infty$, $e^{-0.2t}\to 0$, so $h\to\frac{20(1)}{1} = 20$ | B1 | Long-term height is 20m |
---
## Question 8(v):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Model 1: $t=1$, $h=20(1-e^{-0.1}) \approx 1.903$ m | M1 | Calculate $h$ for $t=1$ in both models |
| Model 2: $t=1$, $h=\frac{20(1-e^{-0.2})}{1+e^{-0.2}} \approx 1.987$ m | A1 | Both values correct |
| Model 2 is closer to 2m, so Model 2 fits better | A1 | Correct conclusion with justification |8 The growth of a tree is modelled by the differential equation
$$10 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 20 - h ,$$
where $h$ is its height in metres and the time $t$ is in years. It is assumed that the tree is grown from seed, so that $h = 0$ when $t = 0$.\\
(i) Write down the value of $h$ for which $\frac { \mathrm { d } h } { \mathrm {~d} t } = 0$, and interpret this in terms of the growth of the tree.\\
(ii) Verify that $h = 20 \left( 1 - \mathrm { e } ^ { - 0.1 t } \right)$ satisfies this differential equation and its initial condition.
The alternative differential equation
$$200 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 400 - h ^ { 2 }$$
is proposed to model the growth of the tree. As before, $h = 0$ when $t = 0$.\\
(iii) Using partial fractions, show by integration that the solution to the alternative differential equation is
$$h = \frac { 20 \left( 1 - \mathrm { e } ^ { - 0.2 t } \right) } { 1 + \mathrm { e } ^ { - 0.2 t } } .$$
(iv) What does this solution indicate about the long-term height of the tree?\\
(v) After a year, the tree has grown to a height of 2 m . Which model fits this information better?
\hfill \mbox{\textit{OCR MEI C4 2013 Q8 [19]}}