Standard +0.3 This is a straightforward parametric differentiation question requiring chain rule application: find dy/dt and dx/dt, then divide. The derivatives involve standard results (d/dt of tan t and ln(sin 2t) using chain rule), and the algebra simplifies cleanly to the given answer. Slightly above average difficulty due to the logarithmic differentiation and trigonometric simplification required, but follows a standard method with no novel insight needed.
4 The parametric equations of a curve are
$$x = 2 t - \tan t , \quad y = \ln ( \sin 2 t )$$
for \(0 < t < \frac { 1 } { 2 } \pi\).
Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \cot t\).
State or imply \(\frac{dx}{dt} = 2 - \sec^2 t\) or \(\frac{dy}{dt} = 2\dfrac{\cos 2t}{\sin 2t}\)
B1
Use \(\dfrac{dy}{dx} = \dfrac{dy}{dt} \div \dfrac{dx}{dt}\)
M1
Obtain correct answer in any form
A1
Use double angle formula to express derivative in terms of \(\cos x\) and \(\sin x\)
M1
Obtain the given answer correctly
A1
AG
## Question 4:
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $\frac{dx}{dt} = 2 - \sec^2 t$ or $\frac{dy}{dt} = 2\dfrac{\cos 2t}{\sin 2t}$ | B1 | |
| Use $\dfrac{dy}{dx} = \dfrac{dy}{dt} \div \dfrac{dx}{dt}$ | M1 | |
| Obtain correct answer in any form | A1 | |
| Use double angle formula to express derivative in terms of $\cos x$ and $\sin x$ | M1 | |
| Obtain the given answer correctly | A1 | AG |
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4 The parametric equations of a curve are
$$x = 2 t - \tan t , \quad y = \ln ( \sin 2 t )$$
for $0 < t < \frac { 1 } { 2 } \pi$.\\
Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \cot t$.\\
\hfill \mbox{\textit{CAIE P3 2022 Q4 [5]}}