A ladder of length 10 m on horizontal ground rests against a vertical wall. The bottom of the ladder is moved away from the wall at a constant speed of $0.5{\text{ m}}{{\text{s}}^{ - 1}}$. Calculate the speed of descent of the top of the ladder when the bottom of the ladder is 4 m away from the wall.

let x, y (m) denote respectively the distance of the bottom of the ladder from the wall and the distance of the top of the ladder from the ground

then,

${x^2} + {y^2} = 100$     M1A1

$2x\frac{{{\text{d}}x}}{{{\text{d}}t}} + 2y\frac{{{\text{d}}y}}{{{\text{d}}t}} = 0$     M1A1

when $x = 4,{\text{ }}y = \sqrt {84}$ and $\frac{{{\text{d}}x}}{{{\text{d}}t}} = 0.5$     A1

substituting, $2 \times 4 \times 0.5 + 2\sqrt {84} \frac{{{\text{d}}y}}{{{\text{d}}t}} = 0$     A1

$\frac{{{\text{d}}y}}{{{\text{d}}t}} = - 0.218{\text{ m}}{{\text{s}}^{ - 1}}$     A1

(speed of descent is $0.218{\text{ m}}{{\text{s}}^{ - 1}}$)

[7 marks]