2024 JPJC Promo Q2

2024 JPJC Promo Q2

5 marks
Promo

Water is poured at a rate of \(0.1\)m\(^{3}\) per minute into a container in the form of an open cone. The semi-vertical angle of the cone is \({{30}^{{}^\circ }}\). At time \(t\) minutes after the start, the radius of the water surface is \(r\) m (see diagram). Find the rate of increase of the depth of water when the volume of the water is \(3\)m\(^{3}\).[5]

[The volume of a cone of base radius \(r\) and height \(h\) is given by \(V=\frac{1}{3}\pi {{r}^{2}}h\)]

Video Solution:

Video Solution

Video solution locked

Solution:

Solution locked

Sign in to view the step-by-step solution

Finding similar questions...
Answer: \(\frac{\text{d}h}{\text{d}t}=0.0228\) m/min (nearest to \(3\) s.f.) or \(\frac{1}{30{{\pi }^{\frac{1}{3}}}}\) m/min.

Need help? Join our JC Math tuition classes.

Learn more