| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Show dy/dx equals given expression |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring standard application of the chain rule to fractional powers, followed by a routine application of the chain rule with a parameter. The 'show that' format provides the target answer, making it slightly easier than average, though the fractional powers require careful handling. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{2}{3}x^{1/3} - \frac{2}{3}x^{-1/3}\frac{dy}{dx} = 0\) | M1, A1 | [5] Implicit differentiation (must show \(= 0\)) |
| \(\frac{dy}{dx} = \frac{\frac{2}{3}x^{1/3}}{\frac{2}{3}x^{-1/3}}\) | M1 | solving for \(dy/dx\) |
| \(= -\frac{x^{1/3}}{x^{-1/3}} = -\left(\frac{x}{1}\right)^* = -\left(\frac{y}{x}\right)^*\) | E1 | [4] www. Must show, or explain, one more step. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\) | M1 | any correct form of chain rule |
| \(= -\left(\frac{8}{1}\right)^* \cdot 6\) | A1 | |
| \(= -12\) | A1cao | [3] |
## Part (i)
$\frac{2}{3}x^{1/3} - \frac{2}{3}x^{-1/3}\frac{dy}{dx} = 0$ | M1, A1 | [5] Implicit differentiation (must show $= 0$)
$\frac{dy}{dx} = \frac{\frac{2}{3}x^{1/3}}{\frac{2}{3}x^{-1/3}}$ | M1 | solving for $dy/dx$
$= -\frac{x^{1/3}}{x^{-1/3}} = -\left(\frac{x}{1}\right)^* = -\left(\frac{y}{x}\right)^*$ | E1 | [4] www. Must show, or explain, one more step.
## Part (ii)
$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$ | M1 | any correct form of chain rule
$= -\left(\frac{8}{1}\right)^* \cdot 6$ | A1 |
$= -12$ | A1cao | [3]
---7 The variables $x$ and $y$ satisfy the equation $x ^ { \frac { 2 } { 3 } } + y ^ { \frac { 2 } { 3 } } = 5$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \left( \frac { y } { x } \right) ^ { \frac { 1 } { 3 } }$.
Both $x$ and $y$ are functions of $t$.\\
(ii) Find the value of $\frac { \mathrm { d } y } { \mathrm {~d} t }$ when $x = 1 , y = 8$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 6$.
Section B (36 marks)\\
\hfill \mbox{\textit{OCR MEI C3 2009 Q7 [7]}}