Riemann surface
matlab

Worked Example 1
Find the wind chill temperature when
(a) $t=2^{\circ} \mathrm{C}, \quad v=20 \mathrm{mph}$;
(b) $t=-10^{\circ} \mathrm{C}, \quad v=5 \mathrm{mph}$;
(c) $t=0^{\circ} \mathrm{C}, \quad v=40 \mathrm{mph}$.

.

Solution
(a) When $t=2, v=20$,
\begin{aligned} T & =33+(0.45+0.29 \sqrt{20}-0.02 \times 20)(-31) \ & =-8.8^{\circ} \mathrm{C} \end{aligned}
(b) When $t=-10, v=5$,
\begin{aligned} T & =33+(0.45+0.29 \sqrt{5}-0.02 \times 5)(-43) \ & =-9.9^{\circ} \mathrm{C} \end{aligned}

(c) When $t=0 \mathrm{KC}, v=40 \mathrm{mph}$,
\begin{aligned} T & =33+(0.45+0.29 \sqrt{40}-0.02 \times 40)(-33) \ & =-16.0^{\circ} \mathrm{C} \end{aligned}

Worked Example 2
What points are scored for a time of 124.2 seconds, and what time would give a point score of 1000 ?

Solution
For $t=124.2$,
\begin{aligned} P & =0.11193(254-124.2)^{1.88} \ \Rightarrow \quad P & =1051 \end{aligned}
(Scores are always rounded down to the nearest whole number.)
Now, to score 1000 points requires a time of $m$ seconds where
\begin{aligned} & 1000=0.11193(254-t)^{1.88} \ \Rightarrow \quad & (254-t)^{1.88}=8934.15 \ \Rightarrow \quad & 254-t=(8934.15)^{\frac{1}{1.58}} \ \Rightarrow \quad & t=254-126.364 \ & t=127.64 \end{aligned}
giving
$$t=127.64$$
All track events use a points scoring system of the form
$$P=a(b-t)^c$$
where $t$ is the time taken, with suitable constants $a, b$ and $c$.

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