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: $a = 4\pi r^2$ Differentiating with respect to time gives: $\frac{da}{dt} = 8\pi r \frac{dr}{dt}$ Substituting $r = 20$ cm and using $\frac{dr}{dt} = \frac{5}{32\pi}$: $\frac{da}{dt} = 8\pi (20) \cdot \frac{5}{32\pi}$ Calculating: $\frac{da}{dt} = 8 \cdot 20 \cdot \frac{5}{32}$ $\frac{da}{dt} = \frac{800}{32} = 25 \text{ cm}^2/s$

To find the values of $x$ when the ball is on the ground, we set the height $f(x)$ to zero: $-x^2 + 10x = 0$ Factoring gives: $x(x - 10) = 0$ Thus, the solutions are: $x = 0 \text{ or } x = 10$

The average height above the ground can be calculated using the formula: $\text{Average height} = \frac{1}{b - a} \int_a^b f(x) \, dx$ In this case, $a = 0$ and $b = 10$: $\text{Average height} = \frac{1}{10 - 0} \int_0^{10} (-x^2 + 10x) \, dx$ Calculating the integral: $= \frac{1}{10} \left[ -\frac{x^3}{3} + 5x^2 \right]_0^{10}$ Evaluating between 0 and 10: $= \frac{1}{10} \left[ -\frac{10^3}{3} + 5(10^2) \right]$ $= \frac{1}{10} \left[ -\frac{1000}{3} + 500 \right]$ Combining gives: $= \frac{1}{10} \left[ \frac{-1000 + 1500}{3} \right] = \frac{1}{10} \cdot \frac{500}{3} = \frac{50}{3} ext{ m}$