| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find parameter value given gradient condition |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring the chain rule (dy/dx = (dy/dt)/(dx/dt)) followed by solving a simple equation involving the normal gradient. The calculus is routine (derivatives of √t and ln t are standard), and finding the normal gradient condition (-1/m = -2) leads to elementary algebra. Slightly easier than average due to the mechanical nature of all steps. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State correct derivative of \(x\) or \(y\) with respect to \(t\) | B1 | \(\dfrac{dx}{dt} = \dfrac{1}{2}t^{-\frac{1}{2}},\quad \dfrac{dy}{dt} = \dfrac{1}{t}\) |
| Use \(\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx}\) | M1 | Use correct chain rule |
| Obtain answer \(\dfrac{dy}{dx} = \dfrac{2}{\sqrt{t}}\) | A1 | Or simplified equivalent e.g. \(2t^{-\frac{3}{2}}\) or \(\dfrac{2\sqrt{t}}{t}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State or imply their \(\dfrac{dy}{dx} = \dfrac{1}{2}\) | M1 | |
| Obtain \(\sqrt{t} = 4\) | A1 | Or equivalent |
| Obtain answer \((7,\ \ln 16)\) | A1 | Or exact equivalent. Can state the two components separately. |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State correct derivative of $x$ or $y$ with respect to $t$ | B1 | $\dfrac{dx}{dt} = \dfrac{1}{2}t^{-\frac{1}{2}},\quad \dfrac{dy}{dt} = \dfrac{1}{t}$ |
| Use $\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx}$ | M1 | Use correct chain rule |
| Obtain answer $\dfrac{dy}{dx} = \dfrac{2}{\sqrt{t}}$ | A1 | Or simplified equivalent e.g. $2t^{-\frac{3}{2}}$ or $\dfrac{2\sqrt{t}}{t}$ |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply their $\dfrac{dy}{dx} = \dfrac{1}{2}$ | M1 | |
| Obtain $\sqrt{t} = 4$ | A1 | Or equivalent |
| Obtain answer $(7,\ \ln 16)$ | A1 | Or exact equivalent. Can state the two components separately. |6 The parametric equations of a curve are
$$x = \sqrt { t } + 3 , \quad y = \ln t$$
for $t > 0$.
\begin{enumerate}[label=(\alph*)]
\item Obtain a simplified expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$.
\item Hence find the exact coordinates of the point on the curve at which the gradient of the normal is - 2 .
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q6 [6]}}