C4 January 2010 Q7
7.
Figure 2 shows a sketch of the curve \(C\) with parametric equations \[x = 5t^2 - 4, \qquad y = t(9 - t^2)\]
The curve \(C\) cuts the \(x\)-axis at the points \(A\) and \(B\).
(a) Find the \(x\)-coordinate at the point \(A\) and the \(x\)-coordinate at the point \(B\). (3)
The region \(R\), as shown shaded in Figure 2, is enclosed by the loop of the curve.
(b) Use integration to find the area of \(R\). (6)
| Scheme | Marks |
|---|---|
| \(y = 0 \Rightarrow t(9 - t^2) = t(3 - t)(3 + t) = 0\) | |
| \(t = 0, 3, -3\) Any one correct value | B1 |
| At \(t = 0\), \(x = 5(0)^2 - 4 = -4\) Method for finding one value of \(x\) | M1 |
| At \(t = 3\), \(x = 5(3)^2 - 4 = 41\) | |
| \(\left(\text{At } t = -3,\ x = 5(-3)^2 - 4 = 41\right)\) | |
| At \(A\), \(x = -4\); at \(B\), \(x = 41\) Both | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 10t\) Seen or implied | B1 |
| \(\displaystyle\int y\,\mathrm{d}x = \int y\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t = \int t(9 - t^2)10t\,\mathrm{d}t\) | M1 A1 |
| \(\displaystyle = \int \left(90t^2 - 10t^4\right)\mathrm{d}t\) | |
| \(= \dfrac{90t^3}{3} - \dfrac{10t^5}{5}\quad (+C)\qquad \left(= 30t^3 - 2t^5\quad (+C)\right)\) | A1 |
| \(\left[\dfrac{90t^3}{3} - \dfrac{10t^5}{5}\right]_0^3 = 30 \times 3^3 - 2 \times 3^5\quad (= 324)\) | M1 |
| \(A = 2\displaystyle\int y\,\mathrm{d}x = 648\quad \left(\text{units}^2\right)\) | A1 |
| (6) | |
| (9 marks) |