A water container is made in the shape of a hollow inverted right circular cone with semi-vertical angle of 30°, as shown in Figure 1. The height of the container is 50 cm. When the depth of the water in the container is h cm, the surface of the water has radius r cm and the volume of water is V cm³. (a) Show that $V = \frac{1}{9} \pi h^3$. [You may assume the formula $V = \frac{1}{3} \pi r^2 h$ for the volume of a cone.] Given that the volume of water in the container increases at a constant rate of 200 cm³/s, (b) find the rate of change of the depth of the water, in cm s⁻¹, when $h = 15$. Give your answer in its simplest form in terms of $\pi$.

A-Level Edexcel Maths Pure Applications of Differentiation: A water container is made in the

A water container is made in the shape of a hollow inverted right circular cone with semi-vertical angle of 30°, as shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 9

Question 5

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A water container is made in the shape of a hollow inverted right circular cone with semi-vertical angle of 30°, as shown in Figure 1. The height of the container is... show full transcript

Worked Solution & Example Answer:A water container is made in the shape of a hollow inverted right circular cone with semi-vertical angle of 30°, as shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 9

Step 1

Show that $V = \frac{1}{9} \pi h^3$.

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Answer

To find the volume of water in the cone, we can use the formula for the volume of a cone:

$V = \frac{1}{3} \pi r^2 h$

Given the semi-vertical angle of the cone is $30°$ , we can relate the radius and height using trigonometry. From the triangle formed:

$\tan(30°) = \frac{r}{h}$

Thus, we can express $r$ in terms of $h$ :

$r = h \tan(30°) = h \cdot \frac{1}{\sqrt{3}} = \frac{h}{\sqrt{3}}$

Substituting this value into the volume formula:

$V = \frac{1}{3} \pi \left(\frac{h}{\sqrt{3}}\right)^2 h$

This expands to:

$V = \frac{1}{3} \pi \frac{h^2}{3} h = \frac{1}{9} \pi h^3$

Therefore, it is shown that $V = \frac{1}{9} \pi h^3$ .

Step 2

Find the rate of change of the depth when $h = 15$.

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Answer

Given that the volume increases at a rate of 200 cm³/s, we can write this as:

$\frac{dV}{dt} = 200$

Now, we can differentiate the volume with respect to time:

$\frac{dV}{dt} = \frac{d}{dt}\left(\frac{1}{9} \pi h^3\right) = \frac{1}{9} \pi \cdot 3h^2 \frac{dh}{dt}$

This simplifies to:

$\frac{dV}{dt} = \frac{\pi}{3} h^2 \frac{dh}{dt}$

Substituting $\frac{dV}{dt} = 200$ and $h = 15$ :

$200 = \frac{\pi}{3} (15^2) \frac{dh}{dt}$

Calculating $15^2 = 225$ gives us:

$200 = \frac{\pi}{3} \cdot 225 \cdot \frac{dh}{dt}$

Multiplying both sides by 3:

$600 = 225 \pi \frac{dh}{dt}$

Solving for $\frac{dh}{dt}$ :

$\frac{dh}{dt} = \frac{600}{225\pi} = \frac{8}{3\pi} \text{ cm/s}$

Thus, the rate of change of the depth of the water when $h = 15$ is $\frac{8}{3\pi}$ cm/s.

A water container is made in the shape of a hollow inverted right circular cone with semi-vertical angle of 30°, as shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 9

Worked Solution & Example Answer:A water container is made in the shape of a hollow inverted right circular cone with semi-vertical angle of 30°, as shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 9

Show that $V = \frac{1}{9} \pi h^3$.

Find the rate of change of the depth when $h = 15$.

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