2022 HCI Promo Q12

2022 HCI Promo Q12

13 marks
Promo

[It is given that volume of a trapezoid\(=\)cross-sectional area\(\times \)length].

In Figure 1, a polyvinyl chloride (PVC) sheet of length \(2\)m and breadth \(0.6\)m with negligible thickness is used to construct a rain gutter in the shape of a trapezoid. The PVC sheet is bent along its length such that its breadth is divided into \(3\) equal parts as shown in Figure 2. The two walls of the rain gutter are such that they are at an angle \(\theta \) with the horizontal. The cross-section of the rain gutter is shown in Figure 3.

  1. Show that the value of the total volume of the rain gutter is given by \(0.04\left( 2\sin \theta +\sin 2\theta \right)\).[2]
  2. Use differentiation to determine the exact value of \(\theta \) for the rain gutter to hold the maximum amount of rainwater.[4]

For the rest of the question, let \(\theta =\frac{\pi }{4}\).

On a particular day with heavy downpour, it is found that a constant amount of \(0.002\)m\(^{3}\) of rainwater is collected by the rain gutter every second. The cross-section of the rain gutter on a day with light rain is shown in Figure 4.

  1. By taking \(p\) as the depth of rainwater in the rain gutter at time \(t\), show that when the depth of rainwater in the rain gutter is \(0.1\)m, the rate of change of depth of rainwater in the rain gutter is given by \(0.0025\)ms\(^{-1}\).[4]
  2. Hence find the rate of change of the total surface area of the rain gutter in contact with the rainwater at the instant when the depth of rainwater in the rain gutter is \(0.1\)m.[3]

Video Solution:

Video Solution

Video solution locked

A free account saves your progress; this video needs tuition enrolment.

Solution:

Solution locked

680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs tuition enrolment.

Create a free account to practise similar questions.

Answer:(i) \(V=0.04(2\sin\theta+\sin2\theta)\text{ m}^3\) (ii) \(\displaystyle \theta=\frac{\pi}{3}\) (iii) \(\displaystyle \frac{\mathrm{d}p}{\mathrm{d}t}=0.0025\text{ m s}^{-1}\) (iv) \(\displaystyle \frac{\mathrm{d}A}{\mathrm{d}t}=\frac{\sqrt{2}}{100}\text{ m}^2\text{ s}^{-1}\)

Need help? Join our JC Math tuition classes.

Learn more