Standard +0.3 This is a straightforward implicit differentiation question requiring students to differentiate both sides with respect to x (using product rule for 4x sin x), find dy/dx at the given point, then find the normal gradient and equation. While it involves multiple techniques (implicit differentiation, product rule, point-normal form), these are standard C3/C4 procedures with no conceptual surprises, making it slightly easier than average.
2. The point \(P\) with coordinates \(\left( \frac { \pi } { 2 } , 1 \right)\) lies on the curve with equation
$$4 x \sin x = \pi y ^ { 2 } + 2 x , \quad \frac { \pi } { 6 } \leqslant x \leqslant \frac { 5 \pi } { 6 }$$
Find an equation of the normal to the curve at \(P\).
Alternative (explicit differentiation): \(y = \frac{1}{\sqrt{\pi}}(4x\sin x - 2x)^{\frac{1}{2}}\); \(\frac{dy}{dx} = \left(\frac{1}{2\sqrt{\pi}}\right)(4x\sin x - 2x)^{-\frac{1}{2}}(4x\cos x + 4\sin x - 2)\)
M1 M1
M1: \(\frac{d(4x\sin x)}{dx} = \pm 4x\cos x + 4\sin x\); M1: \((4x\sin x - 2x)^{\frac{1}{2}} \to k(4x\sin x - 2x)^{-\frac{1}{2}}\)
\(\frac{dy}{dx} = \frac{1}{2\sqrt{\pi}}(4x\sin x - 2x)^{-\frac{1}{2}}(4x\cos x + 4\sin x - 2)\)
Uses \(x = \frac{\pi}{2}\) and \(y = 1\) to obtain value for \(\frac{dy}{dx}\); for implicit, must have \(dy/dx\) terms with \(x\)'s and \(y\)'s; explicit just requires \(x = \frac{\pi}{2}\)
\(y - 1 = -\pi\!\left(x - \frac{\pi}{2}\right)\) or \(y = -\pi x + c \Rightarrow c = 1 + \frac{\pi^2}{2}\); uses normal gradient \(-1\big/\frac{dy}{dx}\) and \(x = \frac{\pi}{2},\ y = 1\)
M1
Must use \(-1/\left(\text{their }\frac{dy}{dx}\right)\) with \(x = \frac{\pi}{2}\) and \(y = 1\) correctly placed
\(y - 1 = -\pi\!\left(x - \frac{\pi}{2}\right)\)
A1cso
Allow 3sf+ decimal equivalents e.g. \(y = -3.14x + 5.93\)
(6 marks)
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d(4x\sin x)}{dx} = 4x\cos x + 4\sin x$ | M1 | Applies product rule to $4x\sin x$; gives $\frac{d(4x\sin x)}{dx} = \pm 4x\cos x + 4\sin x$ |
| $\frac{d(\pi y^2)}{dy} = 2\pi y\frac{dy}{dx}$ | M1 | Applies chain rule to $\pi y^2$; gives $\frac{d(\pi y^2)}{dy} = Ay\frac{dy}{dx}$ |
| $4x\sin x = \pi y^2 + 2x \Rightarrow 4x\cos x + 4\sin x = 2\pi y\frac{dy}{dx} + 2$ | A1 | Fully correct differentiation; accept $4x\cos x\,dx + 4\sin x\,dx = 2\pi y\,dy + 2\,dx$ |
| **Alternative (explicit differentiation):** $y = \frac{1}{\sqrt{\pi}}(4x\sin x - 2x)^{\frac{1}{2}}$; $\frac{dy}{dx} = \left(\frac{1}{2\sqrt{\pi}}\right)(4x\sin x - 2x)^{-\frac{1}{2}}(4x\cos x + 4\sin x - 2)$ | M1 M1 | M1: $\frac{d(4x\sin x)}{dx} = \pm 4x\cos x + 4\sin x$; M1: $(4x\sin x - 2x)^{\frac{1}{2}} \to k(4x\sin x - 2x)^{-\frac{1}{2}}$ |
| $\frac{dy}{dx} = \frac{1}{2\sqrt{\pi}}(4x\sin x - 2x)^{-\frac{1}{2}}(4x\cos x + 4\sin x - 2)$ | A1 | |
| $x = \frac{\pi}{2},\ y = 1 \Rightarrow 4 = 2\pi\frac{dy}{dx} + 2 \Rightarrow \frac{dy}{dx} = \left(\frac{1}{\pi}\right)$ | M1 | Uses $x = \frac{\pi}{2}$ and $y = 1$ to obtain value for $\frac{dy}{dx}$; for implicit, must have $dy/dx$ terms with $x$'s and $y$'s; explicit just requires $x = \frac{\pi}{2}$ |
| $y - 1 = -\pi\!\left(x - \frac{\pi}{2}\right)$ or $y = -\pi x + c \Rightarrow c = 1 + \frac{\pi^2}{2}$; uses normal gradient $-1\big/\frac{dy}{dx}$ and $x = \frac{\pi}{2},\ y = 1$ | M1 | Must use $-1/\left(\text{their }\frac{dy}{dx}\right)$ with $x = \frac{\pi}{2}$ and $y = 1$ correctly placed |
| $y - 1 = -\pi\!\left(x - \frac{\pi}{2}\right)$ | A1cso | Allow 3sf+ decimal equivalents e.g. $y = -3.14x + 5.93$ |
**(6 marks)**
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2. The point $P$ with coordinates $\left( \frac { \pi } { 2 } , 1 \right)$ lies on the curve with equation
$$4 x \sin x = \pi y ^ { 2 } + 2 x , \quad \frac { \pi } { 6 } \leqslant x \leqslant \frac { 5 \pi } { 6 }$$
Find an equation of the normal to the curve at $P$.\\
\hfill \mbox{\textit{Edexcel C34 2016 Q2 [6]}}