A container is made in the shape of a hollow inverted right circular cone. The height of the container is 24 cm and the radius is 16 cm, as shown in Figure 2. Water is flowing into the container. When the height of water is h cm, the surface of the water has radius r cm and the volume of water is V cm³. (a) Show that $V = \frac{4}{27} \pi h^3$. [The volume V of a right circular cone with vertical height h and base radius r is given by the formula $V = \frac{1}{3} \pi r^2 h$] Water flows into the container at a rate of 8 cm³ s⁻¹. (b) Find, in terms of π, the rate of change of h when h = 12.

A-Level Edexcel Maths Pure Exponential & Logarithms: A container is made in the shape

A container is made in the shape of a hollow inverted right circular cone - Edexcel - A-Level Maths Pure - Question 6 - 2009 - Paper 3

Question 6

A container is made in the shape of a hollow inverted right circular cone. The height of the container is 24 cm and the radius is 16 cm, as shown in Figure 2. Water ... show full transcript

Worked Solution & Example Answer:A container is made in the shape of a hollow inverted right circular cone - Edexcel - A-Level Maths Pure - Question 6 - 2009 - Paper 3

Step 1

Show that V = \frac{4}{27} \pi h^3

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Answer

To find the volume V of the water in the conical container, we first establish the relationship between the radius r and height h using similar triangles. The full height of the cone is 24 cm, and the radius of the base is 16 cm. Thus, we have:

$\frac{r}{h} = \frac{16}{24} = \frac{2}{3}$

From this, we can express r in terms of h:

$r = \frac{2}{3}h$

Substituting this into the volume formula of a cone, we get:

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

Now expanding this:

$V = \frac{1}{3} \pi \left( \frac{4}{9}h^2 \right) h = \frac{4}{27} \pi h^3$

This confirms that $V = \frac{4}{27} \pi h^3$ .

Step 2

Find, in terms of π, the rate of change of h when h = 12.

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Answer

Given that water flows into the container at a rate of 8 cm³ s⁻¹, we have:

$\frac{dV}{dt} = 8$

We need to find the rate of change of height h with respect to time. First, we differentiate the volume formula with respect to time:

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

Now, we compute the derivative of V with respect to h:

$\frac{dV}{dh} = \frac{d}{dh} \left( \frac{4}{27} \pi h^3 \right) = \frac{12\pi h^2}{27} = \frac{4\pi h^2}{9}$

Substituting this back into the rate of change equation:

$8 = \frac{4\pi h^2}{9} \cdot \frac{dh}{dt}$

To find ( \frac{dh}{dt} ) when h = 12:

$8 = \frac{4\pi (12)^2}{9} \cdot \frac{dh}{dt}$

Calculating ( 12^2 = 144 ):

$8 = \frac{4\pi (144)}{9} \cdot \frac{dh}{dt}$

Solving for ( \frac{dh}{dt} ):

$\frac{dh}{dt} = \frac{8 \cdot 9}{4 \pi \cdot 144} = \frac{72}{576\pi} = \frac{1}{8\pi}$

Thus, the rate of change of h when h = 12 is (\frac{1}{8\pi}) cm s⁻¹.

A container is made in the shape of a hollow inverted right circular cone - Edexcel - A-Level Maths Pure - Question 6 - 2009 - Paper 3

Worked Solution & Example Answer:A container is made in the shape of a hollow inverted right circular cone - Edexcel - A-Level Maths Pure - Question 6 - 2009 - Paper 3

Show that V = \frac{4}{27} \pi h^3

Find, in terms of π, the rate of change of h when h = 12.

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