| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| 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 straightforward implicit differentiation question with standard steps: substitute to find p, differentiate implicitly using chain rule, find gradient at the given point, and write tangent equation. The algebra is clean (sin(π)=0 simplifies nicely) and all techniques are routine C3 material with no conceptual surprises. |
| 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 | Mark | Guidance |
| \(p = 4\pi^2\) or \((2\pi)^2\) | B1 | Also allow \(x = 4\pi^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = (4y - \sin 2y)^2 \Rightarrow \frac{dx}{dy} = 2(4y - \sin 2y)(4 - 2\cos 2y)\) | M1A1 | M1: Uses chain rule to get form \(A(4y-\sin 2y)(B \pm C\cos 2y)\), \(A,B,C \neq 0\). A1: Fully correct |
| Sub \(y = \frac{\pi}{2}\): \(\frac{dx}{dy} = 24\pi\) (= 75.4), \(\frac{dy}{dx} = \frac{1}{24\pi}\) (= 0.013) | M1 | Sub \(y=\frac{\pi}{2}\) into their \(\frac{dx}{dy}\) or inverted form |
| Equation of tangent: \(y - \frac{\pi}{2} = \frac{1}{24\pi}(x - 4\pi^2)\) | M1 | Score for correct method finding tangent at \((4\pi^2, \frac{\pi}{2})\) |
| Using \(x=0\): \(y = \frac{\pi}{3}\) | M1, A1 | M1: Setting \(x=0\) and solving for \(y\). A1: cso \(y = \frac{\pi}{3}\); need not see \((0, \frac{\pi}{3})\) |
## Question 5:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $p = 4\pi^2$ or $(2\pi)^2$ | B1 | Also allow $x = 4\pi^2$ |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = (4y - \sin 2y)^2 \Rightarrow \frac{dx}{dy} = 2(4y - \sin 2y)(4 - 2\cos 2y)$ | M1A1 | M1: Uses chain rule to get form $A(4y-\sin 2y)(B \pm C\cos 2y)$, $A,B,C \neq 0$. A1: Fully correct |
| Sub $y = \frac{\pi}{2}$: $\frac{dx}{dy} = 24\pi$ (= 75.4), $\frac{dy}{dx} = \frac{1}{24\pi}$ (= 0.013) | M1 | Sub $y=\frac{\pi}{2}$ into their $\frac{dx}{dy}$ or inverted form |
| Equation of tangent: $y - \frac{\pi}{2} = \frac{1}{24\pi}(x - 4\pi^2)$ | M1 | Score for correct method finding tangent at $(4\pi^2, \frac{\pi}{2})$ |
| Using $x=0$: $y = \frac{\pi}{3}$ | M1, A1 | M1: Setting $x=0$ and solving for $y$. A1: cso $y = \frac{\pi}{3}$; need not see $(0, \frac{\pi}{3})$ |
---5. The point $P$ lies on the curve with equation
$$x = ( 4 y - \sin 2 y ) ^ { 2 }$$
Given that $P$ has $( x , y )$ coordinates $\left( p , \frac { \pi } { 2 } \right)$, where $p$ is a constant,
\begin{enumerate}[label=(\alph*)]
\item find the exact value of $p$.
The tangent to the curve at $P$ cuts the $y$-axis at the point $A$.
\item Use calculus to find the coordinates of $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2015 Q5 [7]}}