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(B) Let the ladder be $AC = 17 \,m$. Let the wall be $AB$ of height $x$ and the ground be $BC$ of length $y$.In $	riangle ABC$, by Pythagoras theorem

Vedclass · Accepted Answer

$\frac{8}{15} \,m/sec$

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(B) Let the ladder be $AC = 17 \,m$. Let the wall be $AB$ of height $x$ and the ground be $BC$ of length $y$.In $	riangle ABC$, by Pythagoras theorem:$x^2 + y^2 = 17^2 = 289$Given that the lower end slips away at the rate of $\frac{dy}{dt} = 1 \,m/sec$.When $y = 8 \,m$, we have $x^2 + 8^2 = 289 \Rightarrow x^2 = 289 - 64 = 225 \Rightarrow x = 15 \,m$.Differentiating $x^2 + y^2 = 289$ with respect to time $t$:$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$Substituting the values $x = 15$, $y = 8$, and $\frac{dy}{dt} = 1$:$15 \frac{dx}{dt} + 8(1) = 0$$15 \frac{dx}{dt} = -8$$\frac{dx}{dt} = -\frac{8}{15} \,m/sec$.The negative sign indicates that the height $x$ is decreasing. Therefore, the upper end is coming down at the rate of $\frac{8}{15} \,m/sec$.

$A$ ladder of length $17 \,m$ rests with one end against a vertical wall and the other on the level ground. If the lower end slips away at the rate of $1 \,m/sec$, then when it is $8 \,m$ away from the wall, its upper end is coming down at the rate of

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