A ladder 5 m long is leaning against a wall. | vedclass

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(B) Let the ladder be $PQ$ with length $5 \ m$. Let $P$ be the bottom of the ladder at a distance $x$ from the wall and $Q$ be the top of the ladder a

Vedclass · Accepted Answer

$8/3 \ m/sec$

Vedclass · Answer

(B) Let the ladder be $PQ$ with length $5 \ m$. Let $P$ be the bottom of the ladder at a distance $x$ from the wall and $Q$ be the top of the ladder at a height $y$ from the ground.By the Pythagorean theorem,we have $x^2 + y^2 = 5^2 = 25$.Given that the bottom of the ladder is moving away from the wall at a rate of $\frac{dx}{dt} = 2 \ m/sec$.We need to find the rate at which the height $y$ is decreasing,i.e.,$-\frac{dy}{dt}$,when $x = 4 \ m$.When $x = 4$,substituting into $x^2 + y^2 = 25$ gives $4^2 + y^2 = 25$,so $y^2 = 25 - 16 = 9$,which means $y = 3 \ m$.Differentiating $x^2 + y^2 = 25$ with respect to $t$,we get:$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$Substituting the known values $x = 4$,$y = 3$,and $\frac{dx}{dt} = 2$:$2(4)(2) + 2(3) \frac{dy}{dt} = 0$$16 + 6 \frac{dy}{dt} = 0$$6 \frac{dy}{dt} = -16$$\frac{dy}{dt} = -\frac{16}{6} = -\frac{8}{3} \ m/sec$.Since the height is decreasing,the rate of decrease is $\frac{8}{3} \ m/sec$.

$A$ ladder $5 \ m$ long is leaning against a wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of $2 \ m/sec$. How fast is its height on the wall decreasing when the foot of the ladder is $4 \ m$ away from the wall?

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