| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - partial fractions |
| Difficulty | Standard +0.3 This is a standard C4 differential equations question with clear scaffolding through five parts. Part (i) requires straightforward integration of an exponential, parts (iii-iv) follow a routine separable variables method with partial fractions (a core C4 technique), and limiting behavior questions are standard. While multi-step, each component is textbook-standard with no novel insight required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08d Evaluate definite integrals: between limits1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v = \int 10e^{-\frac{1}{2}t}\,dt\) | M1 | separate variables and intend to integrate |
| \(= -20e^{-\frac{1}{2}t} + c\) | A1 | \(-20e^{-\frac{1}{2}t}\) |
| when \(t=0\), \(v=0\) | M1 | finding \(c\) |
| \(0 = -20 + c \Rightarrow c = 20\) | ||
| so \(v = 20 - 20e^{-\frac{1}{2}t}\) | A1 [4] | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| As \(t \to \infty\), \(e^{-1/2t} \to 0\) | M1 | |
| \(\Rightarrow v \to 20\) | A1 [2] | ft (for their \(c>0\), found) |
| So long term speed is \(20\text{ m s}^{-1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{(w-4)(w+5)} = \frac{A}{w-4} + \frac{B}{w+5}\) | M1 | |
| \(= \frac{A(w+5)+B(w-4)}{(w-4)(w+5)}\) | ||
| \(1 \equiv A(w+5)+B(w-4)\) | ||
| \(w=4\): \(1 = 9A \Rightarrow A = 1/9\) | M1 | cover up, substitution or equating coeffs |
| \(w=-5\): \(1 = -9B \Rightarrow B = -1/9\) | A1 | \(1/9\) |
| \(\Rightarrow \frac{1}{(w-4)(w+5)} = \frac{1/9}{w-4} - \frac{1/9}{w+5}\) | A1 | \(-1/9\) |
| \(= \frac{1}{9(w-4)} - \frac{1}{9(w+5)}\) | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dw}{dt} = -\frac{1}{2}(w-4)(w+5)\) | ||
| \(\Rightarrow \int\frac{dw}{(w-4)(w+5)} = \int-\frac{1}{2}\,dt\) | M1 | separating variables |
| \(\Rightarrow \int\left[\frac{1}{9(w-4)} - \frac{1}{9(w+5)}\right]dw = \int-\frac{1}{2}\,dt\) | M1 | substituting their partial fractions |
| \(\Rightarrow \frac{1}{9}\ln(w-4) - \frac{1}{9}\ln(w+5) = -\frac{1}{2}t + c\) | A1ft | integrating correctly (condone absence of \(c\)) |
| \(\Rightarrow \frac{1}{9}\ln\frac{w-4}{w+5} = -\frac{1}{2}t + c\) | ||
| When \(t=0\), \(w=10 \Rightarrow c = \frac{1}{9}\ln\frac{6}{15} = \frac{1}{9}\ln\frac{2}{5}\) | M1 | correctly evaluating \(c\) (at any stage) |
| \(\Rightarrow \ln\frac{w-4}{w+5} = -\frac{9}{2}t + \ln\frac{2}{5}\) | M1 | combining lns (at any stage) |
| \(\Rightarrow \frac{w-4}{w+5} = e^{-\frac{9}{2}t+\ln\frac{2}{5}} = \frac{2}{5}e^{-\frac{9}{2}t} = 0.4e^{-4.5t}\) * | E1 [6] | www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| As \(t \to \infty\), \(e^{-4.5t} \to 0\) | M1 | |
| \(\Rightarrow w - 4 \to 0\) | A1 [2] | |
| So long term speed is \(4\text{ m s}^{-1}\) |
## Question 4:
### Part 4(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = \int 10e^{-\frac{1}{2}t}\,dt$ | M1 | separate variables and intend to integrate |
| $= -20e^{-\frac{1}{2}t} + c$ | A1 | $-20e^{-\frac{1}{2}t}$ |
| when $t=0$, $v=0$ | M1 | finding $c$ |
| $0 = -20 + c \Rightarrow c = 20$ | | |
| so $v = 20 - 20e^{-\frac{1}{2}t}$ | A1 [4] | cao |
### Part 4(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| As $t \to \infty$, $e^{-1/2t} \to 0$ | M1 | |
| $\Rightarrow v \to 20$ | A1 [2] | ft (for their $c>0$, found) |
| So long term speed is $20\text{ m s}^{-1}$ | | |
### Part 4(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{(w-4)(w+5)} = \frac{A}{w-4} + \frac{B}{w+5}$ | M1 | |
| $= \frac{A(w+5)+B(w-4)}{(w-4)(w+5)}$ | | |
| $1 \equiv A(w+5)+B(w-4)$ | | |
| $w=4$: $1 = 9A \Rightarrow A = 1/9$ | M1 | cover up, substitution or equating coeffs |
| $w=-5$: $1 = -9B \Rightarrow B = -1/9$ | A1 | $1/9$ |
| $\Rightarrow \frac{1}{(w-4)(w+5)} = \frac{1/9}{w-4} - \frac{1/9}{w+5}$ | A1 | $-1/9$ |
| $= \frac{1}{9(w-4)} - \frac{1}{9(w+5)}$ | [4] | |
### Part 4(iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dw}{dt} = -\frac{1}{2}(w-4)(w+5)$ | | |
| $\Rightarrow \int\frac{dw}{(w-4)(w+5)} = \int-\frac{1}{2}\,dt$ | M1 | separating variables |
| $\Rightarrow \int\left[\frac{1}{9(w-4)} - \frac{1}{9(w+5)}\right]dw = \int-\frac{1}{2}\,dt$ | M1 | substituting their partial fractions |
| $\Rightarrow \frac{1}{9}\ln(w-4) - \frac{1}{9}\ln(w+5) = -\frac{1}{2}t + c$ | A1ft | integrating correctly (condone absence of $c$) |
| $\Rightarrow \frac{1}{9}\ln\frac{w-4}{w+5} = -\frac{1}{2}t + c$ | | |
| When $t=0$, $w=10 \Rightarrow c = \frac{1}{9}\ln\frac{6}{15} = \frac{1}{9}\ln\frac{2}{5}$ | M1 | correctly evaluating $c$ (at any stage) |
| $\Rightarrow \ln\frac{w-4}{w+5} = -\frac{9}{2}t + \ln\frac{2}{5}$ | M1 | combining lns (at any stage) |
| $\Rightarrow \frac{w-4}{w+5} = e^{-\frac{9}{2}t+\ln\frac{2}{5}} = \frac{2}{5}e^{-\frac{9}{2}t} = 0.4e^{-4.5t}$ * | E1 [6] | www |
### Part 4(v):
| Answer/Working | Mark | Guidance |
|---|---|---|
| As $t \to \infty$, $e^{-4.5t} \to 0$ | M1 | |
| $\Rightarrow w - 4 \to 0$ | A1 [2] | |
| So long term speed is $4\text{ m s}^{-1}$ | | |
---4 A skydiver drops from a helicopter. Before she opens her parachute, her speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ after time $t$ seconds is modelled by the differential equation
$$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 \mathrm { e } ^ { - \frac { 1 } { 2 } t }$$
When $t = 0 , v = 0$.\\
(i) Find $v$ in terms of $t$.\\
(ii) According to this model, what is the speed of the skydiver in the long term?
She opens her parachute when her speed is $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Her speed $t$ seconds after this is $w \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and is modelled by the differential equation
$$\frac { \mathrm { d } w } { \mathrm {~d} t } = - \frac { 1 } { 2 } ( w - 4 ) ( w + 5 )$$
(iii) Express $\frac { 1 } { ( w - 4 ) ( w + 5 ) }$ in partial fractions.\\
(iv) Using this result, show that $\frac { w - 4 } { w + 5 } = 0.4 \mathrm { e } ^ { - 4.5 t }$.\\
(v) According to this model, what is the speed of the skydiver in the long term?
\hfill \mbox{\textit{OCR MEI C4 Q4 [18]}}