CAIE P3 2014 June — Question 4 7 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2014
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric differentiation
TypeFind parameter value given gradient condition
DifficultyStandard +0.3 This is a straightforward parametric differentiation question requiring standard techniques: differentiate both parametric equations, apply the chain rule formula dy/dx = (dy/dt)/(dx/dt), and simplify using basic trigonometric identities. Part (ii) involves solving cot t = 2, which is routine. The 5 marks for part (i) reflect algebraic manipulation rather than conceptual difficulty. Slightly above average due to the trigonometric simplification required, but still a standard textbook exercise.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation

The parametric equations of a curve are $$x = t - \tan t, \quad y = \ln(\cos t),$$ for \(-\frac{1}{4}\pi < t < \frac{1}{4}\pi\).
  1. Show that \(\frac{dy}{dx} = \cot t\). [5]
  2. Hence find the \(x\)-coordinate of the point on the curve at which the gradient is equal to 2. Give your answer correct to 3 significant figures. [2]

AnswerMarks
(i) State \(\frac{dx}{dt} = 1 - \sec^2 t\), or equivalentB1
Use chain ruleM1
Obtain \(\frac{dy}{dt} = -\frac{\sin t}{\cos t}\), or equivalentA1
Use \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\)M1
Obtain the given answer correctly.A1
Total: 5 marks
(ii) State or imply \(t = \tan^{-1}(1)\)B1
Obtain answer \(x = -0.0364\)B1
Total: 2 marks
**(i)** State $\frac{dx}{dt} = 1 - \sec^2 t$, or equivalent | B1 |
Use chain rule | M1 |
Obtain $\frac{dy}{dt} = -\frac{\sin t}{\cos t}$, or equivalent | A1 |
Use $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ | M1 |
Obtain the given answer correctly. | A1 |
| Total: 5 marks |

**(ii)** State or imply $t = \tan^{-1}(1)$ | B1 |
Obtain answer $x = -0.0364$ | B1 |
| Total: 2 marks |

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The parametric equations of a curve are
$$x = t - \tan t, \quad y = \ln(\cos t),$$
for $-\frac{1}{4}\pi < t < \frac{1}{4}\pi$.

\begin{enumerate}[label=(\roman*)]
\item Show that $\frac{dy}{dx} = \cot t$. [5]
\item Hence find the $x$-coordinate of the point on the curve at which the gradient is equal to 2. Give your answer correct to 3 significant figures. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE P3 2014 Q4 [7]}}