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(C) Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the height of the top of the ladder on the wall.By the Pythagorean theo

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$-3/2 \ m/min$

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(C) Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the height of the top of the ladder on the wall.By the Pythagorean theorem,we have $x^2 + y^2 = 10^2 = 100$.Differentiating both sides with respect to time $t$,we get $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$.Given that $x = 6 \ m$ and $\frac{dx}{dt} = 2 \ m/min$.When $x = 6$,$y = \sqrt{100 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8 \ m$.Substituting these values into the differentiated equation:$2(6)(2) + 2(8) \frac{dy}{dt} = 0$$24 + 16 \frac{dy}{dt} = 0$$16 \frac{dy}{dt} = -24$$\frac{dy}{dt} = -\frac{24}{16} = -\frac{3}{2} \ m/min$.Thus,the height is decreasing at the rate of $3/2 \ m/min$.

$A$ ladder of length $10 \ m$ is resting against a vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of $2 \ m/min$. How fast is its height on the wall decreasing when the foot of the ladder is $6 \ m$ away from the wall?

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