| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find tangent equation at point |
| Difficulty | Standard +0.3 This is a slightly above-average C3 question requiring implicit differentiation with the product rule and chain rule for tan(2y), then finding a tangent equation. The verification in part (a) is routine substitution. While it involves multiple techniques, the structure is standard and the algebraic manipulation is straightforward once dy/dx is found. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 8\frac{\pi}{8}\tan\left(2\times\frac{\pi}{8}\right) = \pi\) | B1* | Proof — at least one intermediate line must be seen including a term in tangent. Accept as minimum \(y=\frac{\pi}{8} \Rightarrow x=\pi\tan\left(\frac{\pi}{4}\right)=\pi\). No errors permitted as this is a given answer. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\mathrm{d}x}{\mathrm{d}y} = 8\tan 2y + 16y\sec^2(2y)\) | M1 | Applies product rule to \(8y\tan 2y\) achieving \(A\tan 2y + By\sec^2(2y)\) |
| One term correct: either \(8\tan 2y\) or \(+16y\sec^2(2y)\) | A1 | No requirement for \(\frac{\mathrm{d}x}{\mathrm{d}y}=\) |
| Both sides correct: \(\frac{\mathrm{d}x}{\mathrm{d}y} = 8\tan 2y + 16y\sec^2(2y)\) | A1 | Accept implicit differentiation forms |
| At \(P\): \(\frac{\mathrm{d}x}{\mathrm{d}y} = 8\tan\frac{2\pi}{8} + 16\frac{\pi}{8}\sec^2\left(2\times\frac{\pi}{8}\right) = \{8+4\pi\}\) | M1 | Fully substituting \(y=\frac{\pi}{8}\); accept \(\frac{\mathrm{d}x}{\mathrm{d}y}\approx 20.6\) or \(\frac{\mathrm{d}y}{\mathrm{d}x}\approx 0.05\) as evidence |
| \(\frac{y-\frac{\pi}{8}}{x-\pi} = \frac{1}{8+4\pi}\), accept \(y-\frac{\pi}{8}=0.049(x-\pi)\) | M1A1 | Correct attempt at tangent equation at \(\left(\pi,\frac{\pi}{8}\right)\); gradient must be inverted numerical value of \(\frac{\mathrm{d}x}{\mathrm{d}y}\). Watch for negative reciprocals (M0) |
| \((8+4\pi)y = x + \frac{\pi^2}{2}\) | A1 | Correct answer and solution only. Accept \(4(2+\pi)y=x+0.5\pi^2\) and unsimplified forms |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 8\frac{\pi}{8}\tan\left(2\times\frac{\pi}{8}\right) = \pi$ | B1* | Proof — at least one intermediate line must be seen including a term in tangent. Accept as minimum $y=\frac{\pi}{8} \Rightarrow x=\pi\tan\left(\frac{\pi}{4}\right)=\pi$. No errors permitted as this is a given answer. |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\mathrm{d}x}{\mathrm{d}y} = 8\tan 2y + 16y\sec^2(2y)$ | M1 | Applies product rule to $8y\tan 2y$ achieving $A\tan 2y + By\sec^2(2y)$ |
| One term correct: either $8\tan 2y$ or $+16y\sec^2(2y)$ | A1 | No requirement for $\frac{\mathrm{d}x}{\mathrm{d}y}=$ |
| Both sides correct: $\frac{\mathrm{d}x}{\mathrm{d}y} = 8\tan 2y + 16y\sec^2(2y)$ | A1 | Accept implicit differentiation forms |
| At $P$: $\frac{\mathrm{d}x}{\mathrm{d}y} = 8\tan\frac{2\pi}{8} + 16\frac{\pi}{8}\sec^2\left(2\times\frac{\pi}{8}\right) = \{8+4\pi\}$ | M1 | Fully substituting $y=\frac{\pi}{8}$; accept $\frac{\mathrm{d}x}{\mathrm{d}y}\approx 20.6$ or $\frac{\mathrm{d}y}{\mathrm{d}x}\approx 0.05$ as evidence |
| $\frac{y-\frac{\pi}{8}}{x-\pi} = \frac{1}{8+4\pi}$, accept $y-\frac{\pi}{8}=0.049(x-\pi)$ | M1A1 | Correct attempt at tangent equation at $\left(\pi,\frac{\pi}{8}\right)$; gradient must be inverted numerical value of $\frac{\mathrm{d}x}{\mathrm{d}y}$. Watch for negative reciprocals (M0) |
| $(8+4\pi)y = x + \frac{\pi^2}{2}$ | A1 | Correct answer and solution only. Accept $4(2+\pi)y=x+0.5\pi^2$ and unsimplified forms |
---3. The curve $C$ has equation $x = 8 y \tan 2 y$
The point $P$ has coordinates $\left( \pi , \frac { \pi } { 8 } \right)$
\begin{enumerate}[label=(\alph*)]
\item Verify that $P$ lies on $C$.
\item Find the equation of the tangent to $C$ at $P$ in the form $a y = x + b$, where the constants $a$ and $b$ are to be found in terms of $\pi$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2014 Q3 [8]}}