Science:Math Exam Resources/Courses/MATH100/December 2014/Question 07 (b)/Solution 2

So we need to minimise the distance $d$ for $t \in [0, 3]$ . We do this using the closed interval method:

$\begin{aligned} d (0) = \sqrt{36} = 6 & d (3) = \sqrt{45 - 72 + 36} = \sqrt{9} = 3 \end{aligned}$

The critical numbers are then given by $\frac{d}{d t} (d) = \frac{1}{2} (5 t^{2} - 24 t + 36)^{- 1 / 2} (10 t - 24) = \frac{10 t - 24}{2 \sqrt{5 t^{2} - 24 t + 36}} = 0$ So the only critical number is when $10 t - 24 = 0$ . This is at $t = 12 / 5$ . Then

$\begin{aligned} d (12 / 5) & = \sqrt{\frac{144}{5} - 24 \cdot \frac{12}{5} + 36} \\ = \sqrt{\frac{144 - 2 \times 144 + 180}{5}} \\ = \sqrt{\frac{36}{5}} < \sqrt{\frac{36}{4}} = 3 \end{aligned}$

By the closed interval method the minimum distance is reached when $t = \frac{12}{5}$ seconds.