CAIE P2 2013 November — Question 5 8 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2013
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicParametric differentiation
TypeFind tangent equation at parameter
DifficultyModerate -0.3 This is a straightforward parametric differentiation question requiring standard techniques: find dy/dx using the chain rule, determine the parameter value from a given y-coordinate, then verify a tangent passes through a point. All steps are routine with no conceptual challenges beyond basic parametric calculus.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.06d Natural logarithm: ln(x) function and properties1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation

5 The parametric equations of a curve are $$x = 1 + \sqrt { } t , \quad y = 3 \ln t$$
  1. Find the exact value of the gradient of the curve at the point \(P\) where \(y = 6\).
  2. Show that the tangent to the curve at \(P\) passes through the point \(( 1,0 )\).

AnswerMarks Guidance
(i) State \(\frac{dx}{dt} = \frac{1}{2}t^{-\frac{1}{2}}\) or \(\frac{dy}{dt} = \frac{3}{t}\)B1
Use \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}\)M1
Use \(y = 6\) to find \(t\)M1
Obtain \(t = e^2\)A1
Obtain \(\frac{dy}{dx} = \frac{6}{e}\)A1 [5]
(ii) Obtain \(x\) and form equation of the tangent at their pointM1
Obtain correct equation for tangent \(y - 6 = -\frac{6}{e}(x - (1 + e))\)A1
Show that tangent passes through \((1, 0)\) by substitutionA1 [3]
(i) State $\frac{dx}{dt} = \frac{1}{2}t^{-\frac{1}{2}}$ or $\frac{dy}{dt} = \frac{3}{t}$ | B1 |
Use $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ | M1 |
Use $y = 6$ to find $t$ | M1 |
Obtain $t = e^2$ | A1 |
Obtain $\frac{dy}{dx} = \frac{6}{e}$ | A1 | [5]

(ii) Obtain $x$ and form equation of the tangent at their point | M1 |
Obtain correct equation for tangent $y - 6 = -\frac{6}{e}(x - (1 + e))$ | A1 |
Show that tangent passes through $(1, 0)$ by substitution | A1 | [3]
5 The parametric equations of a curve are

$$x = 1 + \sqrt { } t , \quad y = 3 \ln t$$

(i) Find the exact value of the gradient of the curve at the point $P$ where $y = 6$.\\
(ii) Show that the tangent to the curve at $P$ passes through the point $( 1,0 )$.

\hfill \mbox{\textit{CAIE P2 2013 Q5 [8]}}