| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Iterative method for parameter value |
| Difficulty | Standard +0.3 This is a straightforward parametric equations question with standard techniques: part (i) requires simple algebraic manipulation of y=1, part (ii) is routine iteration with a calculator (no convergence analysis needed), and part (iii) uses the standard dy/dx formula for parametric curves. All steps are procedural with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equate \(4t^2e^{-t}\) to 1, rearrange to \(t^2 = \ldots\) and hence \(t = \ldots\) | M1 | Allow M1 for \(t = \sqrt{\frac{1}{4}e^{-t}}\) |
| Confirm \(t = \frac{1}{2}e^{\frac{1}{2}t}\) with necessary detail needed as answer is given | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use iterative process correctly at least once | M1 | |
| Obtain final answer \(t = 0.715\) | A1 | |
| Show sufficient iterations to 5 sf to justify answer or show a sign change in the interval \([0.7145, 0.7155]\) | A1 | SC: M1A1 from iterations to 4sf resulting in 0.71 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Obtain \(\frac{dx}{dt} = 3 + 12e^{-2t}\) | B1 | |
| Use product rule to find \(\frac{dy}{dt}\) | M1 | |
| Obtain \(8te^{-t} - 4t^2e^{-t}\) | A1 | |
| Divide correctly to obtain \(\frac{dy}{dx}\) | M1 | |
| Substitute value from part (ii) to obtain 0.31 | A1 | Allow greater accuracy |
## Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Equate $4t^2e^{-t}$ to 1, rearrange to $t^2 = \ldots$ and hence $t = \ldots$ | M1 | Allow M1 for $t = \sqrt{\frac{1}{4}e^{-t}}$ |
| Confirm $t = \frac{1}{2}e^{\frac{1}{2}t}$ with necessary detail needed as answer is given | A1 | |
---
## Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use iterative process correctly at least once | M1 | |
| Obtain final answer $t = 0.715$ | A1 | |
| Show sufficient iterations to 5 sf to justify answer or show a sign change in the interval $[0.7145, 0.7155]$ | A1 | **SC**: M1A1 from iterations to 4sf resulting in 0.71 |
---
## Question 6(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain $\frac{dx}{dt} = 3 + 12e^{-2t}$ | B1 | |
| Use product rule to find $\frac{dy}{dt}$ | M1 | |
| Obtain $8te^{-t} - 4t^2e^{-t}$ | A1 | |
| Divide correctly to obtain $\frac{dy}{dx}$ | M1 | |
| Substitute value from part (ii) to obtain 0.31 | A1 | Allow greater accuracy |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{f5e0b088-73db-405b-a832-aa01d9fcba64-08_396_716_260_712}
The diagram shows the curve with parametric equations
$$x = 3 t - 6 \mathrm { e } ^ { - 2 t } , \quad y = 4 t ^ { 2 } \mathrm { e } ^ { - t }$$
for $0 \leqslant t \leqslant 2$. At the point $P$ on the curve, the $y$-coordinate is 1 .\\
(i) Show that the value of $t$ at the point $P$ satisfies the equation $t = \frac { 1 } { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } t }$.\\
(ii) Use the iterative formula $t _ { n + 1 } = \frac { 1 } { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } t _ { n } }$ with $t _ { 1 } = 0.7$ to find the value of $t$ at $P$ correct to 3 significant figures. Give the result of each iteration to 5 significant figures.\\
(iii) Find the gradient of the curve at $P$, giving the answer correct to 2 significant figures.\\
\hfill \mbox{\textit{CAIE P2 2019 Q6 [10]}}