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 $\pi$, the rate of change of h when h = 12.

A-Level Edexcel Maths Pure Parametric Equations: 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 7 - 2009 - Paper 3

Question 7

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 7 - 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 will use the properties of similar triangles.

Using the height and radius of the container:

The height of the cone is 24 cm.
The radius of the cone is 16 cm.

From similar triangles: $\frac{r}{h} = \frac{16}{24} = \frac{2}{3}$ Therefore, we can express r in terms of h: $r = \frac{2}{3} h$

Substituting this into the formula for the volume of a cone: $V = \frac{1}{3} \pi r^2 h$

Substituting the value of r: $V = \frac{1}{3} \pi \left(\frac{2}{3} h\right)^2 h$ $= \frac{1}{3} \pi \left(\frac{4}{9} h^2\right) h$ $= \frac{4}{27} \pi h^3$ This shows that $V = \frac{4}{27} \pi h^3$ .

Step 2

Find, in terms of $\pi$, the rate of change of h when h = 12

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Answer

Given the volume flow rate: $\frac{dV}{dt} = 8 \text{ cm}^3/\text{s}$

From the volume formula: $V = \frac{4}{27} \pi h^3$

Differentiating both sides with respect to t using the chain rule: $\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}$

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

Substituting this into our expression: $8 = \frac{4 \pi h^2}{9} \cdot \frac{dh}{dt}$

Now substituting h = 12: $8 = \frac{4 \pi (12^2)}{9} \cdot \frac{dh}{dt}$ Calculating: $8 = \frac{4 \pi (144)}{9} \cdot \frac{dh}{dt}$ $8 = \frac{576 \pi}{9} \cdot \frac{dh}{dt}$

Now solving for $\frac{dh}{dt}$ : $\frac{dh}{dt} = \frac{8 \cdot 9}{576 \pi} = \frac{72}{576 \pi} = \frac{1}{8\pi}$ Thus the rate of change of h when h = 12 is $\frac{1}{8\pi}$ cm/s.

A container is made in the shape of a hollow inverted right circular cone - Edexcel - A-Level Maths Pure - Question 7 - 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 7 - 2009 - Paper 3

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

Find, in terms of $\pi$, the rate of change of h when h = 12

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