Let $f(x) = -0.5x^3 + 5x - 0.98$, where $x \in \mathbb{R}$ - Leaving Cert Mathematics - Question 9 - 2012

To determine whether $f$ has a local maximum at $(5, 11.52)$, first compute the derivative:

$f'(x) = -1.5x^2 + 5$

Setting the derivative equal to zero to find critical points:

$-1.5x^2 + 5 = 0 \Rightarrow x^2 = \frac{5}{1.5} = \frac{10}{3} \Rightarrow x = 5.$

The second derivative $f''(x) = -3x$. Evaluating this at $x = 5$:

$f''(5) = -15 < 0,$ which indicates a local maximum.

Now to find $f(5)$:

$f(5) = -0.5(5)^3 + 5(5) - 0.98 = -62.5 + 25 - 0.98 = 11.52.$

Thus, the point $(5, 11.52)$ is indeed a local maximum.

To sketch the graph of $v(t)$, compute values at key points:

• For $0 \leq t < 0.2$, $v(t) = 0$.
• Calculate $v(0.2)$:

$v(0.2) = -0.5(0.2)^3 + 5(0.2) - 0.98 = 1.016.$

• For $2 \leq t < 5$, use values of the cubic function evaluated at $t = 1, 2, 3, 4$:
• For $t \geq 5$, $v(t) = 11.52$.

The graph will show these characteristics clearly.

The distance travelled is given by the integral of $v(t)$ over the interval from 0 to 5 seconds:

$d = \int_0^5 v(t) \, dt.$
For the first 0.2s:

$d_1 = \int_0^{0.2} 0 \, dt = 0.$
For the interval from 0.2 to 5:

$d_2 = \int_{0.2}^5 (-0.5t^3 + 5t - 0.98) \, dt.$

Evaluating the integral gives the total distance to be approximately 36.864 metres.

After 5 seconds, the sprinter is travelling at 11.52 m/s. The remaining distance to cover is:

100 - 58.87 = 41.13 \text{ metres.}

The time to complete this distance is given by:

$t = \frac{d}{v} = \frac{41.13}{11.52} \approx 3.57 ext{ seconds}.$

Thus, the total finishing time is:
$5 + 3.57 \approx 8.57$ Therefore, the finishing time correct to two decimal places is 8.57 seconds.

Let $r$ be the radius, $A$ the surface area, and $V$ the volume of the snowball. Using the formulas, we have:

• The surface area: $A = 4\pi r^2$.
• The volume: $V = \frac{4}{3}\pi r^3$.
  Applying the chain rule for the volume gives:
  $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt}.$
  Since $\frac{dV}{dt} = -kA$, substituting leads to:
  $\frac{dV}{dt} = -k(4\pi r^2) \Rightarrow \frac{dr}{dt} \text{ is a constant.}$

Let $r_0$ be the initial radius. The volume after 1 hour is $\frac{V}{2} = \frac{4\pi r_0^3}{3}$.
Using the relationship $r = r_0 \sqrt{\frac{1}{2}}$, we find the remaining volume when melted completely to determine the time for complete melting. The calculations yield that it will take about 43.87 minutes after losing half its volume to melt completely.