| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2019 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Exponential growth/decay - non-standard rate function |
| Difficulty | Standard +0.3 This is a straightforward separable differential equation with standard exponential decay modeling. Part (i) requires simple substitution to find the constant of proportionality, part (ii) is routine separation of variables and integration (leading to a standard exponential solution), and part (iii) asks for basic limit behavior. The question is slightly easier than average because it's highly scaffolded, uses standard techniques throughout, and the 'show that' in part (i) removes any uncertainty about the model setup. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State \(\frac{dN}{dt} = ke^{-0.02t}N\) and show \(k = -0.01\) | B1 | OE; \((-10 = k \times 1 \times 1000)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Separate variables correctly and integrate at least one side | B1 | \(\int \frac{1}{N}\,dN = \int -0.01e^{-0.02t}\,dt\) |
| Obtain term \(\ln N\) | B1 | OE |
| Obtain term \(0.5e^{-0.02t}\) | B1 | OE |
| Use \(N = 1000\), \(t = 0\) to evaluate a constant, or as limits, in a solution with terms \(a\ln N\) and \(be^{-0.02t}\), where \(ab \neq 0\) | M1 | |
| Obtain correct solution in any form e.g. \(\ln N - \ln 1000 = 0.5\!\left(e^{-0.02t} - 1\right)\) | A1 | \(\ln 1000 - \frac{1}{2} = 6.41\) |
| Substitute \(N = 800\) and obtain \(t = 29.6\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State that \(N\) approaches \(\dfrac{1000}{\sqrt{e}}\) | B1 | Accept 606 or 607 or 606.5 |
## Question 4(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| State $\frac{dN}{dt} = ke^{-0.02t}N$ and show $k = -0.01$ | B1 | OE; $(-10 = k \times 1 \times 1000)$ |
## Question 4(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Separate variables correctly and integrate at least one side | B1 | $\int \frac{1}{N}\,dN = \int -0.01e^{-0.02t}\,dt$ |
| Obtain term $\ln N$ | B1 | OE |
| Obtain term $0.5e^{-0.02t}$ | B1 | OE |
| Use $N = 1000$, $t = 0$ to evaluate a constant, or as limits, in a solution with terms $a\ln N$ and $be^{-0.02t}$, where $ab \neq 0$ | M1 | |
| Obtain correct solution in any form e.g. $\ln N - \ln 1000 = 0.5\!\left(e^{-0.02t} - 1\right)$ | A1 | $\ln 1000 - \frac{1}{2} = 6.41$ |
| Substitute $N = 800$ and obtain $t = 29.6$ | A1 | |
## Question 4(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| State that $N$ approaches $\dfrac{1000}{\sqrt{e}}$ | B1 | Accept 606 or 607 or 606.5 |
---4 The number of insects in a population $t$ weeks after the start of observations is denoted by $N$. The population is decreasing at a rate proportional to $N \mathrm { e } ^ { - 0.02 t }$. The variables $N$ and $t$ are treated as continuous, and it is given that when $t = 0 , N = 1000$ and $\frac { \mathrm { d } N } { \mathrm {~d} t } = - 10$.\\
(i) Show that $N$ and $t$ satisfy the differential equation
$$\frac { \mathrm { d } N } { \mathrm {~d} t } = - 0.01 \mathrm { e } ^ { - 0.02 t } N .$$
(ii) Solve the differential equation and find the value of $t$ when $N = 800$.\\
(iii) State what happens to the value of $N$ as $t$ becomes large.\\
\hfill \mbox{\textit{CAIE P3 2019 Q4 [8]}}