| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (polynomial/rational) |
| Difficulty | Standard +0.8 Part (i) requires computing dy/dx using the quotient rule on both parametric equations then applying the chain rule - a multi-step calculus procedure with algebraic complexity. Part (ii) requires algebraic manipulation to eliminate the parameter t from rational expressions involving cubics, which demands insight into the relationship x³ + y³ versus xy and careful algebraic verification. This is more demanding than standard parametric differentiation questions. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dx}{dt} = \frac{(1+t^3) - t(3t^2)}{(1+t^3)^2}\) | M1 | AO 1.1a |
| \(\frac{dx}{dt} = \frac{(1-2t^3)}{(1+t^3)^2}\) o.e. | A1 | AO 1.1 |
| \(\frac{dy}{dt} = \frac{2t(1+t^3) - t^2(3t^2)}{(1+t^3)^2}\) | ||
| \(\frac{dy}{dt} = \frac{(2t - t^4)}{(1+t^3)^2}\) o.e. | A1 | AO 1.1 |
| \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = \frac{2t-t^4}{1-2t^3}\) | M1 | AO 1.1a |
| \(t=1 \Rightarrow \frac{dy}{dx} = -1\) | A1 [5] | AO 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| LHS \(= \frac{t^3 + t^6}{(1+t^3)^3}\) o.e. | M1 | AO 1.1a |
| \(= \frac{t^3(1+t^3)}{(1+t^3)^3}\) | M1 | AO 1.1 |
| RHS \(= \frac{t^3}{(1+t^3)^2}\) = LHS | A1 [3] | AO 2.1 |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dx}{dt} = \frac{(1+t^3) - t(3t^2)}{(1+t^3)^2}$ | M1 | AO 1.1a | DR Use of quotient rule; may be gained for $\frac{dy}{dt}$ if $\frac{dx}{dt}$ not seen (allow $\pm$) |
| $\frac{dx}{dt} = \frac{(1-2t^3)}{(1+t^3)^2}$ o.e. | A1 | AO 1.1 | |
| $\frac{dy}{dt} = \frac{2t(1+t^3) - t^2(3t^2)}{(1+t^3)^2}$ | | | |
| $\frac{dy}{dt} = \frac{(2t - t^4)}{(1+t^3)^2}$ o.e. | A1 | AO 1.1 | |
| $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = \frac{2t-t^4}{1-2t^3}$ | M1 | AO 1.1a | Substitution into their $\frac{dy}{dx}$, dep on earlier M1 |
| $t=1 \Rightarrow \frac{dy}{dx} = -1$ | A1 [5] | AO 2.1 | |
## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| LHS $= \frac{t^3 + t^6}{(1+t^3)^3}$ o.e. | M1 | AO 1.1a | AG Expression for LHS |
| $= \frac{t^3(1+t^3)}{(1+t^3)^3}$ | M1 | AO 1.1 | Factorising |
| RHS $= \frac{t^3}{(1+t^3)^2}$ = LHS | A1 [3] | AO 2.1 | Completion of argument |8 A curve has parametric equations $x = \frac { t } { 1 + t ^ { 3 } } , y = \frac { t ^ { 2 } } { 1 + t ^ { 3 } }$, where $t \neq - 1$.\\
(i) In this question you must show detailed reasoning.
Determine the gradient of the curve at the point where $t = 1$.\\
(ii) Verify that the cartesian equation of the curve is $x ^ { 3 } + y ^ { 3 } = x y$.
\hfill \mbox{\textit{OCR MEI Paper 3 2018 Q8 [8]}}