FP3 June 2014 Q7
7. A circle \(C\) with centre \(O\) and radius \(r\) has cartesian equation \(x^2 + y^2 = r^2\) where \(r\) is a constant.
(a) Show that \(1 + \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = \dfrac{r^2}{r^2 - x^2}\) (3)
(b) Show that the surface area of the sphere generated by rotating \(C\) through \(\pi\) radians about the \(x\)-axis is \(4\pi r^2\). (5)
(c) Write down the length of the arc of the curve \(y = \sqrt{(1 - x^2)}\) from \(x = 0\) to \(x = 1\) (1)
| Scheme | Marks |
|---|---|
| \(y = \sqrt{r^2 - x^2} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -x(r^2 - x^2)^{-\frac{1}{2}}\) Or \(2x + 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = px(r^2 - x^2)^{-\frac{1}{2}}\) Or \(px + qy\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) Attempts to differentiate explicitly or implicitly to give one of the given forms | M1 |
| \(1 + \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 1 + \dfrac{x^2}{r^2 - x^2}\) or \(1 + \dfrac{x^2}{y^2}\) Substitutes their derivative into \(1 + \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2\) | M1 |
| \(= \dfrac{r^2 - x^2 + x^2}{r^2 - x^2} = \dfrac{r^2}{r^2 - x^2}\ ^*\) cso | A1* |
| (3) |
Notes
This is cso and so there must be no errors e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{x}{y}\) could give the correct answer but loses the A1 but allow to show equivalence of lhs and rhs
| Scheme | Marks |
|---|---|
| \(S = (2\pi)\displaystyle\int y\sqrt{\frac{r^2}{r^2 - x^2}}\,\mathrm{d}x\) M1: Use of \(\displaystyle\int y\sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\,\mathrm{d}x\) using their answer to part (a) (must be \(y\) and not \(y^2\)) \(2\pi\) not required here A1: Correct expression including \(2\pi\) (may be implied by later work but must appear before any integration) | M1A1 |
| \(= (2\pi)\displaystyle\int \sqrt{r^2 - x^2}\sqrt{\frac{r^2}{r^2 - x^2}}\,\mathrm{d}x\) Substitutes for \(y\) in terms of \(x\). Dependent on first M. | dM1 |
| \(= \left[2\pi rx\right]_{-r}^{r}\) or \(\left[2\pi rx\right]_0^r\) Substitutes the limits \(r\) and \(-r\) or 0 and \(r\) into an expression of the form \(\underline{k\pi rx}\) and subtracts. The use of the 0 limit can be taken on trust if omitted. Dependent on both previous method marks. | ddM1 |
| If they reach \(2\pi r^2\) correctly then double, then some justification is needed e.g. some mention of symmetry | |
| \(= 4\pi r^2\ ^*\) cso | A1 |
| (5) |
Notes
Note that \(S = 2 \times 2\pi\displaystyle\int_0^r y\sqrt{\frac{r^2}{r^2 - x^2}}\,\mathrm{d}x\) followed by correct work could score full marks as could the correct use of \(S = (2\pi)\displaystyle\int y\sqrt{\tfrac{r^2}{y^2}}\,\mathrm{d}x\)
| Scheme | Marks |
|---|---|
| arc length \(= \tfrac{\pi}{2}\) Ignore any working | B1 |
| (1) | |
| (9 marks) |