A rod AB,13 text{ feet} long,moves with its e | vedclass

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(C) Note that $	riangle OAB$ is a right-angled triangle. Let $OA = x 	ext{ ft}$ and $OB = y 	ext{ ft}$.By Pythagoras theorem,$x^2 + y^2 = 13^2 = 16

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$\frac{5}{4} 	ext{ ft/sec}$ downwards.

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(C) Note that $	riangle OAB$ is a right-angled triangle. Let $OA = x 	ext{ ft}$ and $OB = y 	ext{ ft}$.By Pythagoras theorem,$x^2 + y^2 = 13^2 = 169$.Therefore,$y^2 = 169 - x^2$.Differentiating with respect to time $t$,we get $2y \frac{dy}{dt} = -2x \frac{dx}{dt}$,which simplifies to $y \frac{dy}{dt} = -x \frac{dx}{dt}$.Given that at $x = 5 	ext{ ft}$,$\frac{dx}{dt} = 3 	ext{ ft/sec}$.When $x = 5$,$y = \sqrt{169 - 5^2} = \sqrt{144} = 12 	ext{ ft}$.Substituting these values: $12 \frac{dy}{dt} = -5(3) = -15$.Thus,$\frac{dy}{dt} = -\frac{15}{12} = -\frac{5}{4} 	ext{ ft/sec}$.The negative sign indicates that $B$ is moving downwards towards $O$.Therefore,$B$ is moving at the rate of $\frac{5}{4} 	ext{ ft/sec}$ downwards.

$A$ rod $AB$,$13 \text{ feet}$ long,moves with its ends $A$ and $B$ on two perpendicular lines $OX$ and $OY$ respectively. When $A$ is $5 \text{ feet}$ from $O$,it is moving away at the rate of $3 \text{ feet/sec}$. At this instant,$B$ is moving at the rate of:

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