| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| 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 and power rule. Part (i) involves routine differentiation of fractional powers and algebraic manipulation to reach the given form. Part (ii) adds a simple chain rule application with substitution of given values. While it requires careful handling of fractional indices, it follows standard C3 techniques without requiring problem-solving insight or novel approaches. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0\) | M1 | Implicit differentiation |
| (must show = 0) | A1 | |
| \(\frac{dy}{dx} = -\frac{\frac{2}{3}x^{-1/3}}{\frac{2}{3}y^{-1/3}}\) | M1 | Solving for \(\frac{dy}{dx}\) |
| \(= -\frac{y^{1/3}}{x^{1/3}} = -\left(\frac{y}{x}\right)^{\frac{1}{3}}\) | E1 [4] | www. Must show, or explain, one more step |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\) | M1 | Any correct form of chain rule |
| \(= -\left(\frac{8}{1}\right)^{\frac{1}{3}} \cdot 6\) | A1 | |
| \(= -12\) | A1cao [3] |
## Question 2:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0$ | M1 | Implicit differentiation |
| (must show = 0) | A1 | |
| $\frac{dy}{dx} = -\frac{\frac{2}{3}x^{-1/3}}{\frac{2}{3}y^{-1/3}}$ | M1 | Solving for $\frac{dy}{dx}$ |
| $= -\frac{y^{1/3}}{x^{1/3}} = -\left(\frac{y}{x}\right)^{\frac{1}{3}}$ | E1 [4] | www. Must show, or explain, one more step |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$ | M1 | Any correct form of chain rule |
| $= -\left(\frac{8}{1}\right)^{\frac{1}{3}} \cdot 6$ | A1 | |
| $= -12$ | A1cao [3] | |
---2 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 $\frac { \mathrm { d } x } { \mathrm {~d} t } = 6$.
\hfill \mbox{\textit{OCR MEI C3 Q2 [7]}}