| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2021 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find gradient at given parameter |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring standard techniques: finding dy/dθ and dx/dθ, then dividing them, followed by finding θ when y=8 and substituting. The trigonometric differentiation is routine (chain rule for cosec³θ and sin2θ), and solving cosec³θ=8 gives θ=π/6 directly. Slightly above average due to the multi-step nature and need to handle cosec carefully, but no novel insight required. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \cosec^3\theta \Rightarrow \frac{dy}{d\theta} = -3\cosec^2\theta\cosec\theta\cot\theta\) | B1 | Correct expression for \(\frac{dy}{d\theta}\); seen or implied in any form e.g. \(\frac{-3\cos\theta}{\sin^4\theta}\) |
| \(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta}\) | M1 | Obtains \(\frac{dx}{d\theta} = k\cos 2\theta\) or \(\alpha\cos^2\theta + \beta\sin^2\theta\) (from product rule on \(\sin\theta\cos\theta\)) and attempts \(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta}\) |
| \(\frac{dy}{dx} = \frac{-3\cosec^3\theta\cot\theta}{2\cos 2\theta}\) | A1 | Correct expression in any form; may see e.g. \(\frac{-3\cos\theta}{2\sin^4\theta\cos 2\theta}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 8 \Rightarrow \cosec^3\theta = 8 \Rightarrow \sin^3\theta = \frac{1}{8} \Rightarrow \sin\theta = \frac{1}{2}\) | M1 | Recognises need to find \(\sin\theta\) or \(\theta\) when \(y=8\); correct work leading to \(\sin\theta = \frac{1}{2}\) or \(\theta = \frac{\pi}{6}\) or \(30°\) |
| \(\theta = \frac{\pi}{6} \Rightarrow \frac{dy}{dx} = \frac{-3\cosec^3\left(\frac{\pi}{6}\right)\cot\left(\frac{\pi}{6}\right)}{2\cos\left(\frac{2\pi}{6}\right)} = \ldots\) or \(\sin\theta = \frac{1}{2} \Rightarrow \frac{dy}{dx} = \frac{\frac{-3}{\sin^3\theta} \times \frac{\cos\theta}{\sin\theta}}{2(1-2\sin^2\theta)}\) | M1 | Uses value of \(\sin\theta\) or \(\theta\) in their \(\frac{dy}{dx}\) from part (a); working in exact form to obtain exact value |
| \(= -24\sqrt{3}\) | A1 | Correct gradient deduced |
## Question 13:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \cosec^3\theta \Rightarrow \frac{dy}{d\theta} = -3\cosec^2\theta\cosec\theta\cot\theta$ | B1 | Correct expression for $\frac{dy}{d\theta}$; seen or implied in any form e.g. $\frac{-3\cos\theta}{\sin^4\theta}$ |
| $\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta}$ | M1 | Obtains $\frac{dx}{d\theta} = k\cos 2\theta$ or $\alpha\cos^2\theta + \beta\sin^2\theta$ (from product rule on $\sin\theta\cos\theta$) and attempts $\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta}$ |
| $\frac{dy}{dx} = \frac{-3\cosec^3\theta\cot\theta}{2\cos 2\theta}$ | A1 | Correct expression in any form; may see e.g. $\frac{-3\cos\theta}{2\sin^4\theta\cos 2\theta}$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 8 \Rightarrow \cosec^3\theta = 8 \Rightarrow \sin^3\theta = \frac{1}{8} \Rightarrow \sin\theta = \frac{1}{2}$ | M1 | Recognises need to find $\sin\theta$ or $\theta$ when $y=8$; correct work leading to $\sin\theta = \frac{1}{2}$ or $\theta = \frac{\pi}{6}$ or $30°$ |
| $\theta = \frac{\pi}{6} \Rightarrow \frac{dy}{dx} = \frac{-3\cosec^3\left(\frac{\pi}{6}\right)\cot\left(\frac{\pi}{6}\right)}{2\cos\left(\frac{2\pi}{6}\right)} = \ldots$ **or** $\sin\theta = \frac{1}{2} \Rightarrow \frac{dy}{dx} = \frac{\frac{-3}{\sin^3\theta} \times \frac{\cos\theta}{\sin\theta}}{2(1-2\sin^2\theta)}$ | M1 | Uses value of $\sin\theta$ or $\theta$ in their $\frac{dy}{dx}$ from part (a); working in exact form to obtain exact value |
| $= -24\sqrt{3}$ | A1 | Correct gradient deduced |
---\begin{enumerate}
\item The curve $C$ has parametric equations
\end{enumerate}
$$x = \sin 2 \theta \quad y = \operatorname { cosec } ^ { 3 } \theta \quad 0 < \theta < \frac { \pi } { 2 }$$
(a) Find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $\theta$\\
(b) Hence find the exact value of the gradient of the tangent to $C$ at the point where $y = 8$
\hfill \mbox{\textit{Edexcel Paper 2 2021 Q13 [6]}}