FP3 June 2016 Q2
2. The curve \(C\) has equation \[y = \frac{x^2}{8} - \ln x, \quad 2 \leqslant x \leqslant 3\] Find the length of the curve \(C\) giving your answer in the form \(p + \ln q\), where \(p\) and \(q\) are rational numbers to be found. (7)
| Scheme | Marks |
|---|---|
| \(y = \dfrac{x^2}{8} - \ln x,\quad 2 \leqslant x \leqslant 3\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{x}{4} - \dfrac{1}{x}\) | B1 |
| \(\displaystyle L = \int\sqrt{\left(1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2\right)}\,\mathrm{d}x = \int\sqrt{\left(1 + \left(\frac{x}{4} - \frac{1}{x}\right)^2\right)}\,\mathrm{d}x\) | M1 |
| \(\displaystyle = \int\sqrt{\left(1 + \frac{x^2}{16} - \frac{1}{2} + \frac{1}{x^2}\right)}\,\mathrm{d}x = \int\sqrt{\left(\frac{x^2}{16} + \frac{1}{2} + \frac{1}{x^2}\right)}\,\mathrm{d}x = \int\sqrt{\left(\frac{x}{4} + \frac{1}{x}\right)^2}\,\mathrm{d}x = \int\left(\frac{x}{4} + \frac{1}{x}\right)\mathrm{d}x\) | M1 A1 |
| \(= \dfrac{x^2}{8} + \ln kx\) | A1 |
| \(\left[\dfrac{x^2}{8} + \ln x\right]_2^3 = \left(\dfrac{3^2}{8} + \ln 3\right) - \left(\dfrac{2^2}{8} + \ln 2\right)\) | M1 |
| \(\dfrac{5}{8} + \ln\dfrac{3}{2}\) | A1 |
| (7) | |
| (7 marks) |
Notes
B1: Correct derivative. Allow any correct equivalent e.g. \(\dfrac{2x}{8} - \dfrac{1}{x}\)
M1: Use of a correct formula using their derivative and not the given \(y\).
M1: Squares their derivative to obtain \(ax^2 + bx^{-2} + c\), where none of \(a\), \(b\) or \(c\) are zero – this may be implied by e.g. \(\dfrac{ax^4 + bx^2 + c}{dx^2}\) and adds 1 to their constant term.
A1: Correct integrand \(\dfrac{x}{4} + \dfrac{1}{x}\) or equivalent e.g. \(\dfrac{x^2 + 4}{4x}\) (integral sign not needed)
A1: Correct integration
M1: Substitutes 2 and 3 into an expression of the form \(px^2 + q\ln x\ (p, q \neq 0)\) and subtracts the right way round. Must be seen explicitly or may be implied by a correct exact answer for their integration. If the candidate gives the final single answer in decimals with no substitution shown, e.g. 1.030…this is M0.
A1: Cao and cso (oe e.g \(0.625 + \ln\dfrac{3}{2}\))