Moderate -0.3 This is a straightforward parametric differentiation question requiring the chain rule (dy/dx = dy/dθ ÷ dx/dθ) and standard derivatives of sin θ and cos 2θ. The simplification using the double angle formula (sin 2θ = 2sin θ cos θ) makes it slightly easier than average, as it's a routine technique that students practice regularly. No novel insight or complex multi-step reasoning required.
4 The parametric equations of a curve are
$$x = 4 \sin \theta , \quad y = 3 - 2 \cos 2 \theta$$
where \(- \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi\). Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\), simplifying your answer as far as possible.
State \(\frac{dy}{d\theta} = 4\sin 2\theta\), or equivalent
B1
Use \(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta}\)
M1
Obtain \(\frac{dy}{dx}\) in any correct form, e.g. \(\frac{\sin 2\theta}{\cos\theta}\)
A1
Simplify and obtain answer \(2\sin\theta\)
A1√
[5]
[The f.t. is on gradients of the form \(k\sin 2\theta / \cos\theta\), or equivalent.]
State $\frac{dx}{d\theta} = 4\cos\theta$ | B1 |
State $\frac{dy}{d\theta} = 4\sin 2\theta$, or equivalent | B1 |
Use $\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta}$ | M1 |
Obtain $\frac{dy}{dx}$ in any correct form, e.g. $\frac{\sin 2\theta}{\cos\theta}$ | A1 |
Simplify and obtain answer $2\sin\theta$ | A1√ | [5]
[The f.t. is on gradients of the form $k\sin 2\theta / \cos\theta$, or equivalent.]
4 The parametric equations of a curve are
$$x = 4 \sin \theta , \quad y = 3 - 2 \cos 2 \theta$$
where $- \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi$. Express $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $\theta$, simplifying your answer as far as possible.
\hfill \mbox{\textit{CAIE P2 2009 Q4 [5]}}