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

The volume of the soap is given as ( v = \frac{1}{3} \pi r^2 h ). Substituting ( r = \frac{h}{4} ) gives:

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

Differentiating with respect to time t:

[ \frac{dv}{dt} = \frac{1}{48} \pi (3h^2 \frac{dh}{dt}) = \frac{\pi}{16} h^2 \frac{dh}{dt} ]

Now, as per the problem conditions, the area of the circle formed by the top surface is decreasing, thus it will be negative. Hence:

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

We know ( \frac{dv}{dt} = -0.04 ) cm³/s. From earlier, we have:

[ -0.04 = -\frac{\pi}{16} h^2 \frac{dh}{dt} ]

Rearranging gives us:

[ \frac{dh}{dt} = \frac{0.04 \cdot 16}{\pi h^2} = \frac{0.64}{\pi h^2} ]

Using ( r = \frac{h}{4} ), we can express ( r ) in terms of ( h ) as:

[ \frac{dh}{dt} = -\frac{0.32}{rh} \text{ (as we substitute to re-arrange)} ]

To find the rate of change of volume when h = 10 cm:

Substituting into the equation derived previously:

[ \frac{dh}{dt} = -\frac{0.32}{(\frac{10}{4})(10)} \ = -\frac{0.32}{2.5 \cdot 10} \ = -\frac{0.32}{25} ]

Calculating gives us:

[ \frac{dh}{dt} = -0.0128 cm/s ]

Thus, the volume decreases at a rate of ( 0.2 cm^3/s ) when h = 10 cm.

Using the mass balance for the reaction, we know that:

[ x + y = 500 \Rightarrow y = 500 - x ]

Given the rate of formation of X:

[ \frac{dx}{dt} = k \cdot y = k(500 - x) \text{, where } k ext{ is a constant.} ]

Differentiating the expression for x:

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

At t = 0, substituting gives x = 0:

[ 0 = 500 - A \Rightarrow A = 500 ]

Now substituting back into the derived equation will yield the correct rate change at t = 0.

Therefore, A = 500.