(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).