OCR MEI C4 — Question 4 5 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric differentiation
TypeFind tangent equation at parameter
DifficultyModerate -0.5 This is a straightforward parametric differentiation question requiring students to find dy/dx using the chain rule (dy/dx = (dy/dt)/(dx/dt)), substitute t=-2 to find the point and gradient, then write the tangent equation using y-y₁=m(x-x₁). It's slightly easier than average as it involves simple polynomial differentiation and direct substitution with no complications.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation
4 A curve is given by the parametric equations \(x = t ^ { 2 } , y = 3 t\) for all values of \(t\). Find the equation of the tangent to the curve at the point where \(t = - 2\).
Question 4:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{dx}{dt} = 2t,\quad \frac{dy}{dt} = 3\)M1
\(\Rightarrow \frac{dy}{dx} = \frac{3}{2t}\)A1
When \(t = -2\): \(x = 4\), \(y = -6\), \(\frac{dy}{dx} = -\frac{3}{4}\)B1
Equation of tangent: \(y + 6 = -\frac{3}{4}(x-4)\)M1
\(\Rightarrow 4y + 3x + 12 = 0\)A1
Total: 5 marks 
## Question 4:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dx}{dt} = 2t,\quad \frac{dy}{dt} = 3$ | M1 | |
| $\Rightarrow \frac{dy}{dx} = \frac{3}{2t}$ | A1 | |
| When $t = -2$: $x = 4$, $y = -6$, $\frac{dy}{dx} = -\frac{3}{4}$ | B1 | |
| Equation of tangent: $y + 6 = -\frac{3}{4}(x-4)$ | M1 | |
| $\Rightarrow 4y + 3x + 12 = 0$ | A1 | |
| **Total: 5 marks** | | |

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4 A curve is given by the parametric equations $x = t ^ { 2 } , y = 3 t$ for all values of $t$. Find the equation of the tangent to the curve at the point where $t = - 2$.

\hfill \mbox{\textit{OCR MEI C4  Q4 [5]}}
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