Standard +0.3 This is a straightforward application of implicit differentiation with a clear structure: find the x-coordinate at the given y-value, differentiate to find dy/dx (or dx/dy), find the gradient of the normal, then write the equation. The trigonometry is standard (tan of a simple angle), and the method is a routine textbook exercise for C3 level, making it slightly easier than average.
4. The point \(P\) is the point on the curve \(x = 2 \tan \left( y + \frac { \pi } { 12 } \right)\) with \(y\)-coordinate \(\frac { \pi } { 4 }\).
Find an equation of the normal to the curve at \(P\).
Differentiation of \(2\tan\!\left(y+\frac{\pi}{12}\right) \rightarrow 2\sec^2\!\left(y+\frac{\pi}{12}\right)\); no need to identify with \(\frac{dx}{dy}\)
Accept if \(\frac{dx}{dy}\) is inverted and \(y=\frac{\pi}{4}\) substituted into \(\frac{dy}{dx}\); \(\frac{dx}{dy}=8\) or \(\frac{dy}{dx}=\frac{1}{8}\)
When \(y=\frac{\pi}{4}\), \(x = 2\sqrt{3}\) (awrt \(3.46\))
B1
Value of \(x=2\sqrt{3}\) corresponding to \(y=\frac{\pi}{4}\)
Requires all necessary elements for finding numerical equation of normal; either invert \(\frac{dx}{dy}\) to find \(\frac{dy}{dx}\) then use \(m_1 \times m_2 = -1\), or use \(-\frac{dx}{dy}\)
Any correct form; need not be simplified; e.g. \(y = -8x + \frac{\pi}{4} + 16\sqrt{3}\), or \(y=-8x+\) (awrt) \(28.5\)
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dy} = 2\sec^2\!\left(y + \frac{\pi}{12}\right)$ | M1, A1 | Differentiation of $2\tan\!\left(y+\frac{\pi}{12}\right) \rightarrow 2\sec^2\!\left(y+\frac{\pi}{12}\right)$; no need to identify with $\frac{dx}{dy}$ |
| Substitute $y = \frac{\pi}{4}$: $\frac{dx}{dy} = 2\sec^2\!\left(\frac{\pi}{4}+\frac{\pi}{12}\right) = 8$ | M1, A1 | Accept if $\frac{dx}{dy}$ is inverted and $y=\frac{\pi}{4}$ substituted into $\frac{dy}{dx}$; $\frac{dx}{dy}=8$ or $\frac{dy}{dx}=\frac{1}{8}$ |
| When $y=\frac{\pi}{4}$, $x = 2\sqrt{3}$ (awrt $3.46$) | B1 | Value of $x=2\sqrt{3}$ corresponding to $y=\frac{\pi}{4}$ |
| $\left(y - \frac{\pi}{4}\right) = \text{their } m\left(x - \text{their } 2\sqrt{3}\right)$ | M1 | Requires all necessary elements for finding numerical equation of normal; either invert $\frac{dx}{dy}$ to find $\frac{dy}{dx}$ then use $m_1 \times m_2 = -1$, or use $-\frac{dx}{dy}$ |
| $\left(y - \frac{\pi}{4}\right) = -8\left(x - 2\sqrt{3}\right)$ | A1 | Any correct form; need not be simplified; e.g. $y = -8x + \frac{\pi}{4} + 16\sqrt{3}$, or $y=-8x+$ (awrt) $28.5$ |
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4. The point $P$ is the point on the curve $x = 2 \tan \left( y + \frac { \pi } { 12 } \right)$ with $y$-coordinate $\frac { \pi } { 4 }$.
Find an equation of the normal to the curve at $P$.\\
\hfill \mbox{\textit{Edexcel C3 2012 Q4 [7]}}