F.5 APPLICATIONS OF DIFFERENTI

3.
(a)
Take km and min to be the units of distance and time respectively.


(a)
At t (min) after noon :
OA = 5 + (1/2)t (km)
OB = (√2)t (km)

Hence, the coordinates of A = (0, (5 + 0.5t))
and the coordinates of B = ((√2)tcos45°, (√2)tsin45°) = (t, t)

AB² = (t - 0)² + [t - (5 + 0.5t)]²
AB² = t² + (0.5t - 5)²
AB² = t² + 0.25t² - 5t + 25
AB² = 1.25t² - 5t + 25

d(AB²)/dt = 2.5t - 5 = 2.5(t - 2)
d(AB²)²/d²t = 2.5 > 0

When t = 2 (min), d(AB²)/dt = 0 and d(AB²)²/d²t > 0
Hence, AB² has the minimum at t = 2 (min)
Also, AB has the minimum at t = 2 (min)

Hence, at 2 min after noon,the two cars will be closest to each other.


(b)
When t = 2 (min)
min. AB² = 1.25(2)² - 5(2) + 25 = 20 (km²)
min. AB = √20 (km) = 2√5 (km)
Hence, the shortest distance between the two cars = 2√5 km