Answer: (b) \(\displaystyle \frac{\sqrt3}{2}\) m min\(^{-1}\), or \(0.866\) m min\(^{-1}\).

Answer: (b) \(\displaystyle \frac{\sqrt3}{2}\) m min\(^{-1}\), or \(0.866\) m min\(^{-1}\).

(a) In an equilateral triangle, the height is \(\frac{\sqrt3}{2}\) times the side length. For the water cross-section,

\(h=\frac{\sqrt3}{2}x\).

Hence

\(x=\frac{2h}{\sqrt3}\).

The area of the triangular cross-section of water is

\(\frac12 xh\).

Substitute \(x=\frac{2h}{\sqrt3}\):

The container has length 5 m, so the volume is

(b) Differentiate \(V=\frac{5\sqrt3}{3}h^2\) with respect to \(h\):

\(\displaystyle \frac{dV}{dh}=\frac{10\sqrt3}{3}h\).

Using the chain rule,

When \(h=0.1\),

Given \(\frac{dV}{dt}=0.5\),

\(\displaystyle 0.5=\frac{\sqrt3}{3}\frac{dh}{dt}\).

Therefore

So the depth is increasing at \(\frac{\sqrt3}{2}\) m min\(^{-1}\), or approximately \(0.866\) m min\(^{-1}\).