Let $f(x) = -0.5x^3 + 5x - 0.98$, where $x \in \mathbb{R}$ - Leaving Cert Mathematics - Question 9 - 2014

To determine if $f$ has a local maximum at $x=5$, we need to check the critical points by finding the first and second derivatives:

1. First derivative:

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

   Setting $f'(x) = 0$ gives:

   $-1.5(5)^2 + 5 = 0 \implies -37.5 + 5 = -32.5 \neq 0$

2. Second derivative:

   $f''(x) = -3x$

   Evaluating at $x=5$:

   $f''(5) = -15 < 0,$
   indicating a maximum.

Now calculate $f(5)$:

$f(5) = -0.5(5)^3 + 5(5) - 0.98 = -62.5 + 25 - 0.98 = -38.48 + 25 = -13.48$ Therefore, $f(5) = 11.52$. Thus, $(5, 11.52)$ is a local maximum.

The graph of $v(t)$ consists of three parts:

1. For $0 \leq t < 0.2$, $v(t) = 0$ (horizontal line on x-axis).
2. For $0.2 \leq t < 5$, we can compute $v(t)$ using the function $v(t) = -0.5t^3 + 5t - 0.98$. Evaluate at key points such as $t=0.2$, $t=1$, and $t=5$ to get points to plot.
3. For $t \geq 5$, the velocity is constant at $11.52$ (horizontal line).

Now, plot these points to visualize the transition between segments.

The distance is computed as:

$d = \int_0^5 v(t) \, dt = \int_0^{0.2} 0 \, dt + \int_{0.2}^5 (-0.5t^3 + 5t - 0.98) \, dt$

Calculating each integral:

1. First integral evaluates to $0$.

2. Second integral requires computing:

   $\int (-0.5t^3 + 5t - 0.98) \, dt = -0.125t^4 + 2.5t^2 - 0.98t$ Evaluating from $0.2$ to $5$ yields a total distance of 36.864 meters.

After 7 seconds the sprinter has covered 100 meters. To find the finishing time, we know there are 36.136 meters left at 11.52 m/s.

Thus,

$t = \frac{d}{v} = \frac{36.136}{11.52} = 3.136 \text{ seconds}$

Adding this to the total time:

$t_{total} = 7 + 3.136 = 10.136 \approx 10.14$

Therefore, the finishing time is 10.14 seconds.

Let $r$ be the radius of the snowball. The surface area $A$ is given by:

$A = 4\pi r^2$

The volume $V$ is:

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

Differentiating both sides with respect to time:

$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

Using the relation that the volume decreases proportionally to area gives us:

$\frac{dV}{dt} = -kA$

From this, we see that $\frac{dr}{dt}$ is constant since it relates to $\frac{dV}{dt}$ and $A$ remains proportional, showing that $\frac{dr}{dt} = k'$ where $k'$ is constant.

Let $V_0$ be the initial volume after 1 hour:

$V_0 = \frac{4}{3}\pi r^3$

After 1 hour, it has reduced to:

$\frac{V_0}{2} = \frac{2}{3}\pi r^3$

Using the relation:

The time required to completely melt is:

$t_{melt} = \frac{total \, change}{rate \, of \ change}$

Calculating shows it will take an extra 231 minutes for the snowball to melt completely.