A rod of length 41 ft with an end A on the f | vedclass

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Question

(A) Let $x$ be the distance of end $A$ from the wall and $y$ be the height of end $B$ from the floor. The length of the rod is constant, so $x^2 + y^2

Vedclass · Accepted Answer

$-\frac{1519}{6}$

Vedclass · Answer

(A) Let $x$ be the distance of end $A$ from the wall and $y$ be the height of end $B$ from the floor. The length of the rod is constant, so $x^2 + y^2 = 41^2 = 1681$.Differentiating with respect to time $t$, we get $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$, which simplifies to $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$.Given $\frac{dx}{dt} = 3 \ ft/min$. When $y = 9$, $x^2 + 9^2 = 1681 \implies x^2 = 1681 - 81 = 1600 \implies x = 40 \ ft$.Substituting these values: $40(3) + 9 \frac{dy}{dt} = 0 \implies 9 \frac{dy}{dt} = -120 \implies \frac{dy}{dt} = -\frac{120}{9} = -\frac{40}{3} \ ft/min$.The area of the triangle is $A = \frac{1}{2}xy$.Differentiating with respect to $t$: $\frac{dA}{dt} = \frac{1}{2} \left( x \frac{dy}{dt} + y \frac{dx}{dt} ight)$.Substituting the values: $\frac{dA}{dt} = \frac{1}{2} \left( 40 	imes (-\frac{40}{3}) + 9 	imes 3 ight) = \frac{1}{2} \left( -\frac{1600}{3} + 27 ight) = \frac{1}{2} \left( \frac{-1600 + 81}{3} ight) = \frac{1}{2} \left( -\frac{1519}{3} ight) = -\frac{1519}{6} \ ft^2/min$.

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