\(A\) rod \(A B, 13\) feet long moves with its ends \(A\) and \(B\) on two perpendicular lines \(O X\) and \(O Y\) respectively. When \(\mathrm{A}\) is 5 feet from \(\mathrm{O}\), it is moving away at the rate of \(3 \mathrm{feet} / \mathrm{sec}\). At this instant, B is moving at the rate

Note that \(\triangle \mathrm{OAB}\) is a right angled triangle. Let \(\mathrm{OA}=x \mathrm{ft}\) and \(\mathrm{OB}=y \mathrm{ft}\).
\(\therefore \quad y^2=169-x^2\)
Now, differentiating above function w.r.t. time 't', we get
\(2 y \frac{\mathrm{d} y}{\mathrm{dt}}=-2 x \frac{\mathrm{d} x}{\mathrm{dt}}\)
At \(x=5, \frac{\mathrm{d} x}{\mathrm{dt}}=3 \mathrm{ft} / \mathrm{sec}\)... [Given]
Also, at \(x=5, y=12\)
\(\therefore \quad\) (i) \(\Rightarrow 2(12) \frac{\mathrm{d} y}{\mathrm{dt}}=-2(5)(3)\)
\(\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-5}{4}\)
Negative sign indicates that B is moving downwards.
\(\therefore \quad\) B is moving at the rate \(\frac{5}{4} \mathrm{ft} / \mathrm{sec}\) downwards.