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A hollow hemispherical bowl is shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 5 - 2011 - Paper 5
Question 5
A hollow hemispherical bowl is shown in Figure 1. Water is flowing into the bowl. When the depth of the water is h m, the volume V m³ is given by V = \frac{1}{12} ... show full transcript
Worked Solution & Example Answer:A hollow hemispherical bowl is shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 5 - 2011 - Paper 5
Step 1
Find, in terms of \pi, \frac{dV}{dh} when h = 0.1.
Answer
To find \frac{dV}{dh}$, we start by differentiating the volume function with respect to ( h ).
Using the given formula:
we apply the product rule:
[\frac{dV}{dh} = \frac{1}{12} \pi \left(2h(3 - 4h) + h^2(-4)\right).]
Plugging in h = 0.1: [\frac{dV}{dh} = \frac{1}{12} \pi \left(2(0.1)(3 - 4(0.1)) + (0.1)^2(-4)\right) ]
[= \frac{1}{12} \pi \left(0.2(3 - 0.4) - 0.04\right) = \frac{1}{12} \pi \left(0.2(2.6) - 0.04\right)] [= \frac{1}{12} \pi (0.52 - 0.04) = \frac{0.48}{12} \pi = 0.04 \pi.]
Therefore, ( \frac{dV}{dh} = 0.04\pi ) when ( h = 0.1 ).
Step 2
Find the rate of change of h, in ms⁻¹, when h = 0.1.
Answer
Given that water flows into the bowl at a rate of ( \frac{\pi}{800} m^3s^{-1} ), we use the chain rule to relate (\frac{dh}{dt}) and (\frac{dV}{dh}):
[\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}.]
Rearranging gives: [\frac{dh}{dt} = \frac{\frac{dV}{dt}}{\frac{dV}{dh}}.]
Substituting the values at ( h = 0.1 ): [\frac{dh}{dt} = \frac{\frac{\pi}{800}}{0.04\pi}.]
This simplifies to: [\frac{dh}{dt} = \frac{1}{800 \times 0.04} = \frac{1}{32} = 0.03125.]
Thus, the rate of change of ( h ) is approximately 0.031 ms⁻¹ when ( h = 0.1 ).