A rod of length 13 metres has one end P on the x-axis and the other end Q on the y-axis. If P moves on the x-axis with the speed of 12 m/sec, then the speed of the other end Q when it is 12 m from the origin is

–5 m/sec

$x^2+y^2=(13)^2=169$

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

$\Rightarrow x \frac{d x}{d t}+y \frac{d y}{d t}=0$

$\frac{d x}{d t}$ = 12 (given)

$\Rightarrow x \frac{d x}{d t}+y \frac{d y}{d t}=0$

But, y = 12 (given)    $\Rightarrow x^2+144=169 \Rightarrow x^2=25 \Rightarrow x=5$

∴  $\left.\frac{d y}{d t}\right|_{y=12}=\frac{-12(5)}{12}=-5$

∴ speed of Q = –5 m/sec.