Standard +0.3 This is a straightforward parametric differentiation question requiring the chain rule formula dy/dx = (dy/dθ)/(dx/dθ). While it involves quotient rule for dx/dθ and some algebraic manipulation to reach the given form, it follows a standard procedure with no conceptual surprises—slightly easier than average for A-level.
\(-\sin\theta(2-\sin\theta)^{-1} - \cos\theta(2-\sin\theta)^{-2}(-\cos\theta)\) or equivalent
Use \(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta}\)
M1
\(\left(\frac{dy}{dx} = (1-2\sin\theta) \div \frac{1-2\sin\theta}{(2-\sin\theta)^2}\right)\); allow M1 even if errors in both derivatives
Obtain \(\frac{dy}{dx} = (2-\sin\theta)^2\)
A1
AG – must see working in above cell; allow \(\cos^2\theta + \sin^2\theta = 1\) to be implied; \(x\) instead of \(\theta\) or missing \(\theta\) more than twice on right side then A0
## Question 4:
| Answer | Mark | Guidance |
|--------|------|----------|
| State $\frac{dy}{d\theta} = 1 - 2\sin\theta$ | B1 | Ignore left side throughout $dx/dt$, $dy/dt$, $dx$, $dy$; but must see $\frac{dy}{dx}$ for final A1 |
| Use correct quotient rule, or product rule if rewrite $x$ as $\cos\theta(2-\sin\theta)^{-1}$ | M1 | Incorrect formula seen M0 A0 otherwise BOD |
| Obtain $\frac{dx}{d\theta} = \frac{-(2-\sin\theta)\sin\theta + \cos^2\theta}{(2-\sin\theta)^2}$ o.e. | A1 | $-\sin\theta(2-\sin\theta)^{-1} - \cos\theta(2-\sin\theta)^{-2}(-\cos\theta)$ or equivalent |
| Use $\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta}$ | M1 | $\left(\frac{dy}{dx} = (1-2\sin\theta) \div \frac{1-2\sin\theta}{(2-\sin\theta)^2}\right)$; allow M1 even if errors in both derivatives |
| Obtain $\frac{dy}{dx} = (2-\sin\theta)^2$ | A1 | AG – must see working in above cell; allow $\cos^2\theta + \sin^2\theta = 1$ to be implied; $x$ instead of $\theta$ or missing $\theta$ more than twice on right side then A0 |
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