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 by:
[ v = \frac{1}{3} \pi r^2 h
] Using the relation from part (i), we can substitute ( r ) with ( \frac{h}{4} ):
[ v = \frac{1}{3} \pi \left(\frac{h}{4}\right)^2 h
] Simplifying gives:
[ v = \frac{1}{3} \pi \frac{h^2}{16} h = \frac{\pi}{48} h^3
] To find ( \frac{dv}{dt} ), we differentiate with respect to time: [ \frac{dv}{dt} = \frac{\pi}{16} h^2 \frac{dh}{dt}]

The area of the top surface of the soap is given by:
[ A = \pi r^2 ] From the problem statement, we know that ( \frac{dA}{dt} = -0.04 ) cm²/s. Thus, we can differentiate the area with respect to time: [ \frac{dA}{dt} = 2\pi r \frac{dr}{dt}] Setting these equal gives us: [ -0.04 = 2\pi r \frac{dr}{dt}] Using the relationship from part (i):
( r = \frac{h}{4} ), we can express ( r ) in terms of ( h ):
[ \frac{dr}{dt} = \frac{1}{4} \frac{dh}{dt}] Substituting this into the area equation: [ -0.04 = 2\pi \left(\frac{h}{4}\right) \left(\frac{1}{4} \frac{dh}{dt}\right) \rightarrow -0.04 = \frac{\pi}{8} h \frac{dh}{dt}] Solving for ( \frac{dh}{dt} ) results in: [ \frac{dh}{dt} = -0.32 \frac{r}{h}]

From part (iv), we know that: [ \frac{dv}{dt} = \frac{\pi}{16} h^2 \frac{dh}{dt}
] We already established that when ( h = 10 ), we substitute into the equation: [ \frac{dh}{dt} = -0.32 \frac{r}{h}] Using ( r = \frac{h}{4} = \frac{10}{4} = 2.5 ):
[ \frac{dh}{dt} = -0.32 \frac{2.5}{10} = -0.08] Now substituting back: [ \frac{dv}{dt} = \frac{\pi}{16} (10^2)(-0.08) = -0.2 \text{ cm}^3/s]

Given the total mass of compounds X and Y is 500 g, we represent the mass of Y as ( y(t) = 500 - x(t) ). The rate of increase of mass of compound X is proportional to the mass of compound Y, thus: [ \frac{dx}{dt} = k(500 - x)] where ( k ) is a proportionality constant.

We can obtain the solution of the differential equation from part (i) as: [ x = 500 - A e^{-0.004t}] By substituting initial conditions, when ( t = 0 ):
[ x(0) = 0 \implies 500 - A = 0] Thus, ( A = 500 ) which confirms the solution.