OCR MEI C4 — Question 3 5 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric differentiation
TypeFind gradient at given parameter
DifficultyModerate -0.8 This is a straightforward application of the parametric differentiation formula dy/dx = (dy/dt)/(dx/dt). The derivatives are simple (1 - 1/t and 1 + 1/t), and substituting t=2 requires only basic arithmetic. It's easier than average as it's purely procedural with no problem-solving element.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation
3 A curve is defined parametrically by the equations $$x = t - \ln t , \quad y = t + \ln t \quad ( t > 0 )$$ Find the gradient of the curve at the point where \(t = 2\).
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{dx}{dt} = 1 - \frac{1}{t}\), \(\frac{dy}{dt} = 1 + \frac{1}{t}\)B1 Either \(\frac{dx}{dt}\) or \(\frac{dy}{dt}\) soi
\(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\)M1
\(= \frac{1+\frac{1}{t}}{1-\frac{1}{t}}\)A1
When \(t=2\), \(\frac{dy}{dx} = \frac{1+\frac{1}{2}}{1-\frac{1}{2}} = 3\)M1, A1 [5] www 
# Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dx}{dt} = 1 - \frac{1}{t}$, $\frac{dy}{dt} = 1 + \frac{1}{t}$ | B1 | Either $\frac{dx}{dt}$ or $\frac{dy}{dt}$ soi |
| $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ | M1 | |
| $= \frac{1+\frac{1}{t}}{1-\frac{1}{t}}$ | A1 | |
| When $t=2$, $\frac{dy}{dx} = \frac{1+\frac{1}{2}}{1-\frac{1}{2}} = 3$ | M1, A1 [5] | www |

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3 A curve is defined parametrically by the equations

$$x = t - \ln t , \quad y = t + \ln t \quad ( t > 0 )$$

Find the gradient of the curve at the point where $t = 2$.

\hfill \mbox{\textit{OCR MEI C4  Q3 [5]}}
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Q1 19 Q2 18 Q3 5 Q4 19