A 2.6m pipe leans against a wall. Its bottom end is pulled away from the wall at 0.2 m/s. How fast is the top of the pipe sliding down the wall when its bottom is 0.6m from the wall?

Let $x$ be the distance of the bottom end of the pipe from the wall, and $y$ be the height of the top end of the pipe on the wall. The length of the pipe is $L = 2.6$ m. These form a right-angled triangle, so by the Pythagorean theorem:

$$x^2 + y^2 = L^2$$ $$x^2 + y^2 = (2.6)^2$$

Differentiating both sides with respect to $t$:

$$\frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(2.6^2)$$ $$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$$ $$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$$

Given $x = 0.6$ m and $L = 2.6$ m:

$$(0.6)^2 + y^2 = (2.6)^2$$ $$0.36 + y^2 = 6.76$$ $$y^2 = 6.76 - 0.36$$ $$y^2 = 6.4$$ $$y = \sqrt{6.4} \text{ m}$$

We are given $\frac{dx}{dt} = 0.2$ m/s. We found $x = 0.6$ m and $y = \sqrt{6.4}$ m.
Substitute these into the differentiated equation:

$$(0.6 \text{ m})(0.2 \text{ m/s}) + (\sqrt{6.4} \text{ m}) \frac{dy}{dt} = 0$$ $$0.12 \text{ m}^2/\text{s} + \sqrt{6.4} \text{ m} \frac{dy}{dt} = 0$$ $$\sqrt{6.4} \frac{dy}{dt} = -0.12$$ $$\frac{dy}{dt} = -\frac{0.12}{\sqrt{6.4}}$$ $$\frac{dy}{dt} \approx -\frac{0.12}{2.5298}$$ $$\frac{dy}{dt} \approx -0.04743 \text{ m/s}$$

The negative sign indicates that the top end of the pipe is descending.

The top end of the pipe is descending at approximately $\text{0.0474 m/s}$.