Challenging +1.2 This is a standard arc length calculation using the parametric formula, requiring differentiation of power functions and simplification of a square root expression. While it involves multiple steps and algebraic manipulation to reach the given answer, the technique is routine for Further Maths students and the integrand simplifies cleanly. The 'show that' format removes uncertainty about the final answer, making it moderately above average difficulty but not requiring novel insight.
A curve is defined parametrically by the equations
$$x = \frac{3}{2}t^3 + 5$$
$$y = t^{\frac{3}{2}} \quad (t \geq 0)$$
Show that the arc length of the curve from \(t = 0\) to \(t = 2\) is equal to 26 units.
[5 marks]
Question 15:
15 | Obtains
7
9 9
x = t 2 and y = t2
2 2 | 1.1b | B1 | 72
9 9
x = t 2 and y = t
2 2
8 1
x 2 + y 2 = ( t 4 + t 7 )
4
Arc length
2 8 1
s = t ( 4 + 7 t d ) t
0 4
9 2
= 2 t ( 1 + t 3 ) d t
2 0
2
3
= + 1 ( 3 2 t )
0
32 32
= 9 – 1
= 2 6
x2 + y2
Obtains their | 1.1a | M1
Writes integrand in the form
k t 2 ( 1 + t 3 ) | 3.1a | M1
32
Obtains ( 1 + t 3 ) | 2.2a | A1
Completes a reasoned
argument to obtain 26
AG | 2.1 | R1
Question total | 5
Q | Marking instructions | AO | Marks | Typical solution
A curve is defined parametrically by the equations
$$x = \frac{3}{2}t^3 + 5$$
$$y = t^{\frac{3}{2}} \quad (t \geq 0)$$
Show that the arc length of the curve from $t = 0$ to $t = 2$ is equal to 26 units.
[5 marks]
\hfill \mbox{\textit{AQA Further Paper 1 2024 Q15 [5]}}