(a)     Given that $\frac{{{\text{d}}y}}{{{\text{d}}t}} = 0.001r$ , show that $\frac{{{\text{d}}m}}{{{\text{d}}t}} = {\left( {\frac{r}{{10\sqrt {{r^2} - {y^2}} }}} \right)^3}$.

(b)     State the geometrical meaning of $\frac{{{\text{d}}m}}{{{\text{d}}t}}$ .

(a)     $\frac{{{\text{d}}m}}{{{\text{d}}t}} = \frac{{{\text{d}}m}}{{{\text{d}}y}}\frac{{{\text{d}}y}}{{{\text{d}}t}}$     (M1)

${ = {{\sec }^2}\left( {\arcsin \frac{y}{r}} \right) \times \left( {\arcsin \frac{y}{r}} \right)' \times \frac{r}{{1000}}}$

$= \frac{1}{{{{\cos }^2}\left( {\arcsin \frac{y}{r}} \right)}} \times \frac{{\frac{1}{r}}}{{\sqrt {1 - {{\left( {\frac{y}{r}} \right)}^2}} }} \times \frac{r}{{1000}}$   (or equivalent)     A1A1A1

$= \frac{{\frac{1}{{\sqrt {{r^2} - {y^2}} }}}}{{\frac{{{r^2} - {y^2}}}{{{r^2}}}}}\frac{r}{{1000}}$     (A1)

$= \frac{{{r^3}}}{{{{10}^3}\sqrt {{{\left( {{r^2} - {y^2}} \right)}^3}} }}$   (or equivalent)     A1

$= {\left( {\frac{r}{{10\sqrt {{r^2} - {y^2}} }}} \right)^3}$     AG     N0

(b)     $\frac{{{\text{d}}m}}{{{\text{d}}t}}$ represents the rate of change of the gradient of the line OP     A1

[7 marks]

Few students were able to complete this question successfully, although many did obtain partial marks. Many students failed to recognise the difference between differentiating with respect to $t$ or with respect to $y$ . Very few were able to give a satisfactory geometrical meaning in part (b).