| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2012 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find gradient at given parameter |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring the chain rule (dy/dx = dy/dt รท dx/dt) with quotient rule and logarithm differentiation, followed by finding the parameter value when x=1 and substituting. Standard C4/P3 technique with no conceptual challenges, slightly above average due to the algebraic manipulation involved. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Either: Use correct quotient rule or equivalent to obtain | ||
| \(\frac{dx}{dt} = \frac{4(2t+3)-8t}{(2t+3)^2}\) or equivalent | B1 | |
| Obtain \(\frac{dy}{dt} = \frac{4}{2t+3}\) or equivalent | B1 | |
| Use \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\) or equivalent | M1 | |
| Obtain \(\frac{1}{3}(2t+3)\) or similarly simplified equivalent | A1 | |
| Or: Express \(t\) in terms of \(x\) or \(y\) e.g. \(t = \frac{3x}{4-2x}\) | B1 | |
| Obtain Cartesian equation e.g. \(y = 2\ln\!\left(\frac{6}{2-x}\right)\) | B1 | |
| Differentiate and obtain \(\frac{dy}{dx} = \frac{2}{2-x}\) | M1 | |
| Obtain \(\frac{1}{3}(2t+3)\) or similarly simplified equivalent | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Obtain \(2t = 3\) or \(t = \frac{3}{2}\) | B1 | |
| Substitute in expression for \(\frac{dy}{dx}\) and obtain \(2\) | B1 | [2] |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Either:** Use correct quotient rule or equivalent to obtain | | |
| $\frac{dx}{dt} = \frac{4(2t+3)-8t}{(2t+3)^2}$ or equivalent | B1 | |
| Obtain $\frac{dy}{dt} = \frac{4}{2t+3}$ or equivalent | B1 | |
| Use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ or equivalent | M1 | |
| Obtain $\frac{1}{3}(2t+3)$ or similarly simplified equivalent | A1 | |
| **Or:** Express $t$ in terms of $x$ or $y$ e.g. $t = \frac{3x}{4-2x}$ | B1 | |
| Obtain Cartesian equation e.g. $y = 2\ln\!\left(\frac{6}{2-x}\right)$ | B1 | |
| Differentiate and obtain $\frac{dy}{dx} = \frac{2}{2-x}$ | M1 | |
| Obtain $\frac{1}{3}(2t+3)$ or similarly simplified equivalent | A1 | [4] |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Obtain $2t = 3$ or $t = \frac{3}{2}$ | B1 | |
| Substitute in expression for $\frac{dy}{dx}$ and obtain $2$ | B1 | [2] |
---3 The parametric equations of a curve are
$$x = \frac { 4 t } { 2 t + 3 } , \quad y = 2 \ln ( 2 t + 3 )$$
(i) Express $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$, simplifying your answer.\\
(ii) Find the gradient of the curve at the point for which $x = 1$.
\hfill \mbox{\textit{CAIE P3 2012 Q3 [6]}}