AQA FP2 2013 January — Question 6 8 marks

Exam BoardAQA
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionJanuary
Marks8
PaperDownload PDF ↗
TopicParametric curves and Cartesian conversion
TypeArc length of parametric curve
DifficultyStandard +0.8 This is a standard FP2 arc length question requiring parametric differentiation and integration by substitution. Part (a) involves routine application of the arc length formula with algebraic manipulation to factor the integrand. Part (b) requires recognizing and executing a u-substitution (u = t² + 16), which is a standard technique. While it involves multiple steps and careful algebra, it follows a well-established procedure without requiring novel insight, making it moderately above average difficulty for A-level but typical for Further Maths.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.08h Integration by substitution4.08d Volumes of revolution: about x and y axes

A curve is defined parametrically by $$x = t^3 + 5, \quad y = 6t^2 - 1$$ The arc length between the points where \(t = 0\) and \(t = 3\) on the curve is \(s\).
  1. Show that \(s = \int_{0}^{3} 3t\sqrt{t^2 + A} \, \text{d}t\), stating the value of the constant \(A\). [4 marks]
  2. Hence show that \(s = 61\). [4 marks]

A curve is defined parametrically by
$$x = t^3 + 5, \quad y = 6t^2 - 1$$

The arc length between the points where $t = 0$ and $t = 3$ on the curve is $s$.

\begin{enumerate}[label=(\alph*)]
\item Show that $s = \int_{0}^{3} 3t\sqrt{t^2 + A} \, \text{d}t$, stating the value of the constant $A$. [4 marks]

\item Hence show that $s = 61$. [4 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA FP2 2013 Q6 [8]}}