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 Applications of Differentiation: 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 of the water in the cone, we use the relationship between the dimensions of the cone and the volume formula:

Start with the formula for the volume of a cone:

$V = \frac{1}{3} \pi r^2 h$ .
Since the cone's dimensions are proportionate, we can use similar triangles:

$\frac{r}{h} = \frac{16}{24} \Rightarrow r = \frac{2}{3} h$ .
Substitute this expression for r into the volume formula:

$V = \frac{1}{3} \pi \left(\frac{2}{3} h\right)^2 h$
Simplifying, we have:

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

Step 2

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

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Answer

From the previous step, we know:

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

Taking the derivative of V with respect to time t gives:

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

To find ( \frac{dV}{dh} ):

Differentiate the volume with respect to h:

$\frac{dV}{dh} = \frac{4}{27} \pi \cdot 3h^2 = \frac{12}{27} \pi h^2 = \frac{4}{9} \pi h^2$ .
At h = 12:

$\frac{dV}{dh} \Big|_{h=12} = \frac{4}{9} \pi (12)^2 = \frac{4}{9} \pi (144) = \frac{576}{9} \pi = \frac{192}{3} \pi = 64\pi$ .
We know that water flows into the container at a rate of 8 cm³ s⁻¹:

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

Using the relationship we derived:

$8 = 64 \pi \frac{dh}{dt}$ .

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

$\frac{dh}{dt} = \frac{8}{64\pi} = \frac{1}{8\pi}$ .

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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