A spherical raindrop of radius r metres loses water through evaporation at a rate that depends on its surface area - HSC - SSCE Mathematics Extension 1 - Question 13 - 2013 - Paper 1

From part (a)(i), we know that:

$\frac{dr}{dt} = -10^{-4}$

We find the initial radius by rearranging the volume formula:

$V = \frac{4}{3} \pi r^3$

Setting ( V = 10^{-6} ):

$10^{-6} = \frac{4}{3} \pi r^3$

Solving for ( r ), we get:

$r^3 = \frac{3 \times 10^{-6}}{4 \pi}$

Thus, the initial radius is found as:

$r = \sqrt[3]{\frac{3 \times 10^{-6}}{4 \pi}}$

Now, to find the time for evaporation, we integrate:

$t = \frac{r - r_f}{\frac{dr}{dt}} \text{ (where $r_f$ is the final radius, which is 0)}$

Substituting values, we get:

$t = \frac{r - 0}{-10^{-4}} = -10^{4} r$

Substituting our expression for ( r ):

$t = -10^{4} \sqrt[3]{\frac{3 \times 10^{-6}}{4 \pi}}$

Given point G divides NT externally in the ratio 2:1, we can utilize the section formula. Let N be at coordinates (0, 2ap + ap^2) and T at (ap, 0).

Using the external division formula, the coordinates of G are:

$x_G = \frac{\beta x_1 - \alpha x_2}{\beta - \alpha} = \frac{1(0) - 2(ap)}{1 - 2} = \frac{-2ap}{-1} = 2ap$

For the y-coordinate:

$y_G = \frac{\beta y_1 - \alpha y_2}{\beta - \alpha} = \frac{1(2ap + ap^2) - 2(0)}{1 - 2} = \frac{2ap + ap^2}{-1} = -2ap - ap^2$

Thus, we conclude:

Coordinates of G are $(2ap, -2a - ap^2)$.

To prove that point G lies on a parabola related to the original parabola, we start from the original parabola's equation:

$x^2 = 4ay$

A parabola with the same directrix will have its vertex at P(2ap, -2a - ap^2) as follows:

$y = \frac{1}{4p}(x - h)^2 + k$

We need to test if point G satisfies this equation. Substituting $(2ap, -2a - ap^2)$ in the equation of the parabola:

First, we verify if it meets:

$\frac{1}{4a}(2ap)^2 = 4ap$

And checking should yield a similar structure thus showing that G lies on a similar parabola.