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

To demonstrate this, start with the volume formula:

$v = \frac{1}{3} \pi r^2 h.$

Substituting for r using the previous result, we get:

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

Now, differentiating with respect to time t:

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

Since the area is decreasing at a rate of 0.04 cm$^2$s$^{-1}$, we use the relationship of the radius and height:

Using the formula for the area of the circle (A = \pi r^2) and differentiating:

$\frac{dA}{dt} = \frac{dA}{dh} \cdot \frac{dh}{dt} = \pi r \frac{dh}{dt}.$

Given that \frac{dA}{dt} = -0.04,

Substituting for r gives:

$-0.04 = \pi \left( \frac{h}{4} \right) \frac{dh}{dt}.$

Solving for \frac{dh}{dt}: $\frac{dh}{dt} = -\frac{0.32}{rh}$

Therefore, substituting \frac{dh}{dt} into the previous equation shows that:

$\frac{dv}{dt} = -\frac{\pi}{16} h^2.$

From the previous step, we can equate the rate of change of volume with the derivative of the area of the circle:

From our equation:

$\frac{dA}{dt} = \pi r \frac{dh}{dt}.$

And since the area decrease is given as -0.04 cm$^2$s$^{-1}$:

$-0.04 = \pi r \frac{dh}{dt}.$

Substituting in r = \frac{h}{4} yields:

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

Therefore, we can simplify and solve:

$\frac{dh}{dt} = -\frac{0.32}{rh}.$

Now substituting h = 10 cm into our previous equation:

Using \frac{dh}{dt} = -\frac{0.32}{rh} with r = \frac{h}{4} gives us:

$r = \frac{10}{4} = 2.5 ext{ cm}$

So, substituting the values:

$\frac{dh}{dt} = -\frac{0.32}{2.5 imes 10} = -0.0128 ext{ cm/s}.$

Thus, the rate of change of the volume at this height is evaluated using \frac{dv}{dt}: Substituting back into \frac{dv}{dt} = -\frac{\pi}{16} h^2 gives:

$\frac{dv}{dt} = -\frac{\pi}{16} (10)^2 = -\frac{\pi}{16} (100) = -\frac{100\pi}{16} = -\frac{25\pi}{4} ext{ cm}^3/ ext{s}.$

From the problem statement, we know that the sum of the masses of X and Y is 500 g, which gives us:

$x + y = 500.$

Thus, we express y as:

y = 500 - x.

The rate of change of x becomes proportional to y:

$\frac{dx}{dt} = k y = k(500 - x)$, where k is a constant.

This illustrates that the mass of compound X is increasing proportional to the remaining mass of compound Y.

To verify x = 500 - Ae^{-0.004t} satisfies the equation, we differentiate with respect to t:

$\frac{dx}{dt} = 0.004Ae^{-0.004t}.$

At t = 0, we find:

$x(0) = 0 \Rightarrow 0 = 500 - A \Rightarrow A = 500.$

Thus, \frac{dx}{dt} = -0.004(500 - x), demonstrating that the solution meets the required conditions.