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$A$ 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
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The correct answer is:
$\frac{5}{4} \mathrm{ft} / \mathrm{sec}$ downwards.
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.
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