Q.454. Application of differentiation.

Question 454.
At noon, ship A is 100 km west of ship B. Ship A is sailing east at 35km/hr and ship B is sailing north at 25 km/hr. How fast is the distance between the ships changing at 4.00 p.m.


Answer 454.
Let the position of ship B at noon be at the origin
=> ship A has x-coordinate = - 100 km
and B has x-coordinate = 0 km
At 4.00 p.m., ship A is at x = - 100 + 4 * 35 = 40 km
and B is at x = 0 and y = 4 * 25 = 100 km
Distance between them at 4.00 p.m.
= √[(100)^2 + (40)^2] = 20√(29) km

Distance between the ships,
s^2 = x^2 + y^2
=> 2s ds/dt = 2x dx/dt + 2y dy/dt
=> rate of change of distance between the ships, ds/dt
= (x dx/dt + y dy/dt) / s
= [40 * 35 + 100 * 25] / [20√(29)] km/hr
= (1400 + 2500) / [20√(29)] km/hr
= 195/√(29) km/hr
≈ 36.21 km/hr.

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