The diagram shows a conical soap dispenser of radius 5 cm and height 20 cm - HSC - SSCE Mathematics Extension 1 - Question 12 - 2016 - Paper 1

We start with the volume of the soap:

[ v = \frac{1}{3}\pi r^2 h ]

Substituting for r:

[ v = \frac{1}{3}\pi \left(\frac{h}{4}\right)^2 h = \frac{1}{3}\pi \frac{h^2}{16} h = \frac{\pi}{48} h^3 ]

Now, differentiating v with respect to t:

[ \frac{dv}{dt} = \frac{\pi}{48} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{16} h^2 \frac{dh}{dt} ]

Given that the area of the top surface is decreasing at a rate of 0.04 cm²s⁻¹:

[ \frac{dA}{dt} = -0.04 ]

The area A of the top surface is:

[ A = \pi r^2 ]

From the chain rule:

[ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} ]

Setting these equal gives:

[ 2\pi r \frac{dr}{dt} = -0.04 ]

Replacing r with \frac{h}{4}:

[ \frac{dr}{dt} = \frac{dh}{dt} \cdot \frac{1}{4} ]

Thus:

[ 2\pi \left(\frac{h}{4}\right) \frac{dh}{dt} = -0.04 ]

This leads to:

[ \frac{dh}{dt} = -\frac{0.32}{rh} ]

Using the equation from (ii):

[ \frac{dv}{dt} = \frac{\pi}{16} h^2 \frac{dh}{dt} ]

Substituting h = 10:

[ \frac{dv}{dt} = \frac{\pi}{16} \cdot (10^2) \cdot \frac{dh}{dt} ]

From part (iii):

[ \frac{dh}{dt} = -\frac{0.32}{r \cdot h} ]

Let's calculate r when h = 10:

[ r = \frac{10}{4} = 2.5 ]

Thus:

[ \frac{dh}{dt} = -\frac{0.32}{2.5 \cdot 10} = -\frac{0.32}{25} = -0.0128 ]

Now substitute back:

[ \frac{dv}{dt} = \frac{\pi}{16} \cdot 100 \cdot -0.0128 ]

Calculating gives:

[ \frac{dv}{dt} = -0.8\pi \text{ cm}^3/ ext{s} ]

The total mass of X and Y is 500 g:

[ X + Y = 500 ]

Thus:

[ Y = 500 - X ]

The rate at which X is increasing:

[ \frac{dx}{dt} = kY = k(500 - x) ]

Assuming:

[ x(0) = 0\implies 0 = 500 - A \Rightarrow A = 500 ]

Differentiate:

[ \frac{dx}{dt} = 0 - (-0.004Ae^{-0.004t}) = 0.004Ae^{-0.004t} ]

Set k = 0.004:

[ \frac{dx}{dt} = 0.004(500 - x) ]

Thus, the expression matches.

At the start, setting t = 0:

[ x(0) = 500 - 500 = 0 ]