| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | March |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Normal passes through specific point verification |
| Difficulty | Moderate -0.3 Part (a) is a straightforward application of the chain rule for parametric differentiation (dy/dx = (dy/dt)/(dx/dt)), requiring product rule for dx/dt and polynomial differentiation for dy/dt, with algebraic simplification. Part (b) involves routine substitution to find a point, calculating the normal gradient, and verifying a point lies on the normal line. Both parts are standard textbook exercises with no novel insight required, making this slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Obtain \(\frac{dx}{dt} = e^{2t} + 2te^{2t}\) | B1 | OE |
| Use \(\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}\) | M1 | \(\frac{dy}{dx} = \frac{2t+1}{e^{2t}(1+2t)}\) |
| Obtain the given answer \(\frac{dy}{dx} = e^{-2t}\) | A1 | AG: need to see \(e^{2t}(1+2t)\) in denominator |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Obtain \(x = -e^{-2}\) or \(-\frac{1}{e^2}\) and \(y = 3\) at \(t = -1\) | B1 | |
| Obtain gradient of normal \(= -e^{-2}\) or \(-\frac{1}{e^2}\) | B1 | |
| \(x = 0\) substituted into equation of normal or use of gradients to give \(y = 3 - \frac{1}{e^4}\) with no errors | B1 | Equation of normal \(y - 3 = -e^{-2}(x - {-e^{-2}})\); AG; SC Decimals B0 B1 B0 \(-0.135\) |
| Total | 3 |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Obtain $\frac{dx}{dt} = e^{2t} + 2te^{2t}$ | **B1** | OE |
| Use $\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}$ | **M1** | $\frac{dy}{dx} = \frac{2t+1}{e^{2t}(1+2t)}$ |
| Obtain the given answer $\frac{dy}{dx} = e^{-2t}$ | **A1** | AG: need to see $e^{2t}(1+2t)$ in denominator |
| **Total** | **3** | |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Obtain $x = -e^{-2}$ or $-\frac{1}{e^2}$ and $y = 3$ at $t = -1$ | **B1** | |
| Obtain gradient of normal $= -e^{-2}$ or $-\frac{1}{e^2}$ | **B1** | |
| $x = 0$ substituted into equation of normal or use of gradients to give $y = 3 - \frac{1}{e^4}$ with no errors | **B1** | Equation of normal $y - 3 = -e^{-2}(x - {-e^{-2}})$; AG; **SC** Decimals **B0 B1 B0** $-0.135$ |
| **Total** | **3** | |
---5 The parametric equations of a curve are
$$x = t \mathrm { e } ^ { 2 t } , \quad y = t ^ { 2 } + t + 3$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { - 2 t }$.
\item Hence show that the normal to the curve, where $t = - 1$, passes through the point $\left( 0,3 - \frac { 1 } { \mathrm { e } ^ { 4 } } \right)$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q5 [6]}}