FP3 June 2013 Q4
4.
Figure 1 shows part of the curve with equation \[y = 40\,\mathrm{arcosh}\,x - 9x, \qquad x \geqslant 1\]
Use calculus to find the exact coordinates of the turning point of the curve, giving your answer in the form \(\left(\dfrac{p}{q},\ r\ln 3 + s\right)\), where \(p\), \(q\), \(r\) and \(s\) are integers. (7)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{40}{\sqrt{(x^2 - 1)}} - 9\) M1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{p}{\sqrt{(x^2 - 1)}} - q\) A1: Cao | M1 A1 |
| Put \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathbf{0}\) and obtain \(x^2 = \ldots\) (Allow sign errors only) e.g. \(\left(\dfrac{1681}{81}\right)\) | dM1 |
| \(x = \dfrac{41}{9}\) M1: Square root A1: \(x = \dfrac{41}{9}\) or exact equivalent (not \(\pm\dfrac{41}{9}\)) | M1 A1 |
| \(y = 40\ln\left\{\left(\tfrac{41}{9}\right) + \sqrt{\left(\tfrac{41}{9}\right)^2 - 1}\right\} - \text{"}41\text{"}\) Substitutes \(x = \text{"}\dfrac{41}{9}\text{"}\) into the curve and uses the logarithmic form of arcosh | M1 |
| So \(y = 80\ln 3 - 41\) Cao | A1 |
| (7 marks) |