M3 June 2016 Q2
2.
Figure 1 shows a uniform triangular lamina \(ABC\) in which \(AB = 6\) cm, \(BC = 9\) cm and angle \(ABC = 90^\circ\). The centre of mass of the lamina is \(G\). Use algebraic integration to find the distance of \(G\) from \(AB\). (6)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int xy\,\mathrm{d}x = \int\left(-\frac{2x^2}{3} + 6x\right)\mathrm{d}x\) | M1 |
| \(= \left[-\dfrac{2}{9}x^3 + 3x^2\right]_0^9\) | DM1A1 |
| \(= -162 + 243 - 0 = 81\) | A1 |
| \(\bar{x} = \dfrac{81}{27} = 3\) | M1A1cso |
| (6) | |
| (6 marks) |
Notes
M1 Attempting to obtain the correct form for \(\displaystyle\int xy\,\mathrm{d}x\), using an equation of a line. Limits not needed.
DM1 Attempting the integration. Limits not needed. Dependent on the first M mark.
A1 Correct integration and correct limits shown. This is not ft; equation of the line must be correct.
A1 Substitute the correct limits to obtain 81
M1 Divide their value from the integration by their area of the triangle
A1cso \(\bar{x} = 3\)
ALTs:
1 Using \(C\) as the origin and \(AB\) parallel to \(y\)-axis:
Equation of line must be \(y = \dfrac{2}{3}x\)
| \(\displaystyle\int xy\,\mathrm{d}x = \int\left(\frac{2x^2}{3}\right)\mathrm{d}x == \left[\frac{2}{9}x^3\right]_0^9\) | M1DM1A1 |
| \(= 162\) | A1 |
| \(\bar{x} = \dfrac{162}{27},\ (= 6)\) Dist from \(AB = 9 - 6 = 3\) | M1, A1 |
2 Using \(AB\) along the \(x\)-axis:
Must be using \(\displaystyle\int \frac{1}{2}y^2\,\mathrm{d}x\)
(i) Origin at \(B\) equation of line is \(y = -\dfrac{3}{2}x + 9\)
(ii) Origin at \(A\) equation of line is \(y = \dfrac{3}{2}x\)
NB Ignore any work for the distance from \(BC\), whether before or after distance from \(AB\)