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 $$V = \frac{1}{12} \pi h^2 (3 - 4h), \quad 0 \leq h \leq 0.25$$ (a) Find, in terms of $\, \pi$, $\frac{dV}{dh}$ when $h = 0.1$. Water flows into the bowl at a rate of $\frac{\pi}{800} \, m^3 \, s^{-1}$. (b) Find the rate of change of $h$, in ms$^{-1}$, when $h = 0.1$.

A-Level Edexcel Maths Pure Differentiation: A hollow hemispherical bowl is s

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... 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 $h = 0.1$.

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Answer

To find $\frac{dV}{dh}$ , we need to differentiate the volume function with respect to $h$ :

$V = \frac{1}{12} \pi h^2 (3 - 4h)$

Using the product rule,

$\frac{dV}{dh} = \frac{1}{12} \pi \left(2h(3 - 4h) + h^2(-4)\right)$

At $h = 0.1$ :

$\frac{dV}{dh} = \frac{1}{12} \pi \left(2(0.1)(3 - 4(0.1)) + (0.1)^2(-4)\right)$

Calculating this gives:

$\frac{dV}{dh} = \frac{1}{12} \pi \left(0.2(3 - 0.4) - 0.04\right) = \frac{1}{12} \pi (0.2(2.6) - 0.04) = \frac{1}{12} \pi (0.52 - 0.04) = \frac{1}{12} \pi (0.48) = \frac{0.04 \pi}{1} = \frac{\pi}{25}$

Thus, $\frac{dV}{dh}$ at $h = 0.1$ is $\frac{\pi}{25}$ .

Step 2

Find the rate of change of $h$, in ms$^{-1}$, when $h = 0.1$.

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Answer

We know the rate at which the volume is increasing:

$\frac{dV}{dt} = \frac{\pi}{800} \, m^3 \, s^{-1}$

We can use the relationship:

$\frac{dh}{dt} = \frac{\frac{dV}{dt}}{\frac{dV}{dh}}$

Substituting the values:

$\frac{dh}{dt} = \frac{\frac{\pi}{800}}{\frac{\pi}{25}} = \frac{\frac{\pi}{800}}{\frac{\pi}{25}} = \frac{25}{800} = \frac{1}{32}$

At $h = 0.1$ , the rate of change of $h$ is:

$\frac{dh}{dt} = \frac{25}{800} \, ms^{-1} \approx 0.03125 \, ms^{-1}$

Rounded, this is approximately $0.031 \, ms^{-1}$ .

A hollow hemispherical bowl is shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 5 - 2011 - Paper 5

Worked Solution & Example Answer:A hollow hemispherical bowl is shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 5 - 2011 - Paper 5

Find, in terms of $\, \pi$, $\frac{dV}{dh}$ when $h = 0.1$.

Find the rate of change of $h$, in ms$^{-1}$, when $h = 0.1$.

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