| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| 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 differentiation and tangent line question. Part (i) requires chain rule application to differentiate a power of a quadratic (standard C3 technique). Part (ii) involves finding dx/dy at a point, converting to dy/dx, then using point-slope form—all routine procedures with no conceptual challenges beyond careful algebraic manipulation. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Obtain derivative \(k(37 + 10y - 2y^2)^{-\frac{1}{2}} f(y)\) | M1 | any constant \(k\); any linear function for \(f\) |
| Obtain \(\frac{1}{2}(10 - 4y)(37 + 10y - 2y^2)^{-\frac{1}{2}}\) | A1\2 | or equiv |
| Answer | Marks | Guidance |
|---|---|---|
| Either: Sub'te \(y = 3\) in expression for \(\frac{du}{dy}\) | *M1 | |
| Take reciprocal of expression/value | *M1 | and without change of sign |
| Obtain \(-7\) for gradient of tangent | A1 | |
| Attempt equation of tangent | M1 | dep *M *M |
| Obtain \(y = -7x + 52\) | A1\5 | and no second equation |
| Or: Sub'te \(y = 3\) in expression for \(\frac{dx}{dy}\) | M1 | |
| Attempt formation of eq'n \(x = m'y + c\) | M1 | where \(m'\) is attempt at \(\frac{dx}{dy}\) |
| Obtain \(x - 7 = -\frac{1}{2}(y - 3)\) | A1 | or equiv |
| Attempt rearrangement to required form | M1 | |
| Obtain \(y = -7x + 52\) | A1\(5) | and no second equation |
## Part (i)
Obtain derivative $k(37 + 10y - 2y^2)^{-\frac{1}{2}} f(y)$ | M1 | any constant $k$; any linear function for $f$
Obtain $\frac{1}{2}(10 - 4y)(37 + 10y - 2y^2)^{-\frac{1}{2}}$ | A1\2 | or equiv
## Part (ii)
Either: Sub'te $y = 3$ in expression for $\frac{du}{dy}$ | *M1 |
Take reciprocal of expression/value | *M1 | and without change of sign
Obtain $-7$ for gradient of tangent | A1 |
Attempt equation of tangent | M1 | dep *M *M
Obtain $y = -7x + 52$ | A1\5 | and no second equation
Or: Sub'te $y = 3$ in expression for $\frac{dx}{dy}$ | M1 |
Attempt formation of eq'n $x = m'y + c$ | M1 | where $m'$ is attempt at $\frac{dx}{dy}$
Obtain $x - 7 = -\frac{1}{2}(y - 3)$ | A1 | or equiv
Attempt rearrangement to required form | M1 |
Obtain $y = -7x + 52$ | A1\(5) | and no second equation
---\includegraphics{figure_3}
The diagram shows the curve with equation $x = (37 + 10y - 2y^2)^{\frac{1}{2}}$.
\begin{enumerate}[label=(\roman*)]
\item Find an expression for $\frac{dx}{dy}$ in terms of $y$. [2]
\item Hence find the equation of the tangent to the curve at the point $(7, 3)$, giving your answer in the form $y = mx + c$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR C3 2009 Q6 [7]}}