(a) (i) Air is pumped into a spherical exercise ball at the rate of 250 cm³ per second - Leaving Cert Mathematics - Question 7 - 2016

The surface area A of a sphere is given by:

$A = 4\pi r^2$

To find the rate of change of surface area, we differentiate A with respect to time:

$\frac{dA}{dt} = 8\pi r \frac{dr}{dt}$

Substituting r = 20 cm and $\frac{dr}{dt} = \frac{25}{32\pi}$:

$\frac{dA}{dt} = 8\pi (20) \left( \frac{25}{32\pi} \right)$

This simplifies to:

$\frac{dA}{dt} = 8 \cdot 20 \cdot \frac{25}{32}$

Calculating the above:

$\frac{dA}{dt} = \frac{4000}{32} = 125 ext{ cm}^2/s$.

Thus, the rate at which the surface area of the ball is increasing is: $125 ext{ cm}^2/s$.

To find when the ball is on the ground, set f(x) to 0:

$-x^2 + 10x = 0$

Factoring the equation gives:

$x(x - 10) = 0$

The solutions are:

$x = 0 \text{ or } x = 10$.

Thus, the values of x when the ball is on the ground are: 0 m and 10 m.

To find the average height, we need to calculate:

$ext{Average height} = \frac{1}{b-a} \int_a^b f(x) \, dx$

Here, a = 0 and b = 10, and we have:

$f(x) = -x^2 + 10x$

Setting up the integral:

$\int_0^{10} (-x^2 + 10x) \ dx = \left[ -\frac{x^3}{3} + 5x^2 \right]_0^{10}$

Calculating this gives:

$= \left( -\frac{10^3}{3} + 5(10^2) \right) - \left( -\frac{0^3}{3} + 5(0^2) \right) = \left( -\frac{1000}{3} + 500 \right)$

Simplifying:

$= \frac{1500}{3} - \frac{1000}{3} = \frac{500}{3}$

The interval length is:

$b - a = 10 - 0 = 10$

So, the average height becomes:

$\text{Average height} = \frac{1}{10} \cdot \frac{500}{3} = \frac{50}{3} \text{ m} \approx 16.67 ext{ m}$.