| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 8 |
| 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 with clear geometric setup. Part (i) involves basic coordinate geometry (distance formula), part (ii) is standard implicit differentiation of a simple equation, and part (iii) applies the chain rule. All steps are routine C3 techniques with no novel insight required, making it slightly easier than average. |
| 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 |
| \(QA = 3 - y\) | B1 | |
| \(PA = 6 - (3-y) = 3 + y\) | B1 | |
| By Pythagoras, \(PA^2 = OP^2 + OA^2\) | E1 | Must show some working to indicate Pythagoras (e.g. \(x^2 + 3^2\)) |
| \((3+y)^2 = x^2 + 3^2 = x^2 + 9\) | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Differentiating implicitly: \(2(y+3)\frac{dy}{dx} = 2x\) | M1 | Allow errors in RHS derivative (but not LHS) - notation should be correct |
| \(\Rightarrow \frac{dy}{dx} = \frac{x}{y+3}\) | E1 | Brackets must be used |
| *Or*: \(9 + 6y + y^2 = x^2 + 9 \Rightarrow 6y + y^2 = x^2\) | ||
| \(\Rightarrow (6+2y)\frac{dy}{dx} = 2x\) | M1 | Allow errors in RHS derivative (but not LHS) - notation should be correct |
| \(\Rightarrow \frac{dy}{dx} = \frac{x}{y+3}\) | E1 | Brackets must be used |
| *Or*: \(y = \sqrt{(x^2+9)} - 3 \Rightarrow \frac{dy}{dx} = \frac{1}{2}(x^2+9)^{-1/2} \cdot 2x\) | M1 | (cao) |
| \(= \frac{x}{\sqrt{x^2+9}} = \frac{x}{y+3}\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\) | M1 | Chain rule (soi) |
| \(= \frac{4}{2+3} \times 2\) | A1 | |
| \(= \frac{8}{5}\) | A1 [3] |
## Question 3:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $QA = 3 - y$ | B1 | |
| $PA = 6 - (3-y) = 3 + y$ | B1 | |
| By Pythagoras, $PA^2 = OP^2 + OA^2$ | E1 | Must show some working to indicate Pythagoras (e.g. $x^2 + 3^2$) |
| $(3+y)^2 = x^2 + 3^2 = x^2 + 9$ | [3] | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Differentiating implicitly: $2(y+3)\frac{dy}{dx} = 2x$ | M1 | Allow errors in RHS derivative (but not LHS) - notation should be correct |
| $\Rightarrow \frac{dy}{dx} = \frac{x}{y+3}$ | E1 | Brackets must be used |
| *Or*: $9 + 6y + y^2 = x^2 + 9 \Rightarrow 6y + y^2 = x^2$ | | |
| $\Rightarrow (6+2y)\frac{dy}{dx} = 2x$ | M1 | Allow errors in RHS derivative (but not LHS) - notation should be correct |
| $\Rightarrow \frac{dy}{dx} = \frac{x}{y+3}$ | E1 | Brackets must be used |
| *Or*: $y = \sqrt{(x^2+9)} - 3 \Rightarrow \frac{dy}{dx} = \frac{1}{2}(x^2+9)^{-1/2} \cdot 2x$ | M1 | (cao) |
| $= \frac{x}{\sqrt{x^2+9}} = \frac{x}{y+3}$ | E1 | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$ | M1 | Chain rule (soi) |
| $= \frac{4}{2+3} \times 2$ | A1 | |
| $= \frac{8}{5}$ | A1 [3] | |
---3 Fig. 6 shows the triangle OAP , where O is the origin and A is the point $( 0,3 )$. The point $\mathrm { P } ( x , 0 )$ moves on the positive $x$-axis. The point $\mathrm { Q } ( 0 , y )$ moves between O and A in such a way that $\mathrm { AQ } + \mathrm { AP } = 6$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ce82bfc4-90dd-4127-a11c-281cdcca70cf-2_488_848_514_640}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
(i) Write down the length AQ in terms of $y$. Hence find AP in terms of $y$, and show that
$$( y + 3 ) ^ { 2 } = x ^ { 2 } + 9$$
(ii) Use this result to show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x } { y + 3 }$.\\
(iii) When $x = 4$ and $y = 2 , \frac { \mathrm {~d} x } { \mathrm {~d} t } = 2$. Calculate $\frac { \mathrm { d } y } { \mathrm {~d} t }$ at this time.
\hfill \mbox{\textit{OCR MEI C3 Q3 [8]}}