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

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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:

rh=1624=23\frac{r}{h} = \frac{16}{24} = \frac{2}{3}

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

r=23hr = \frac{2}{3}h

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

V=13πr2h=13π(23h)2hV = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left( \frac{2}{3}h \right)^2 h

Now expanding this:

V=13π(49h2)h=427πh3V = \frac{1}{3} \pi \left( \frac{4}{9}h^2 \right) h = \frac{4}{27} \pi h^3

This confirms that V=427πh3V = \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:

dVdt=8\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:

dVdt=dVdhdhdt\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}

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

dVdh=ddh(427πh3)=12πh227=4πh29\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=4πh29dhdt8 = \frac{4\pi h^2}{9} \cdot \frac{dh}{dt}

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

8=4π(12)29dhdt8 = \frac{4\pi (12)^2}{9} \cdot \frac{dh}{dt}

Calculating ( 12^2 = 144 ):

8=4π(144)9dhdt8 = \frac{4\pi (144)}{9} \cdot \frac{dh}{dt}

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

dhdt=894π144=72576π=18π\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.

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