Question

Ship A is travelling south at the rate of 2 km/hr, at the instant that ship B, which is 32 miles south of ship A, is travelling east at rate of 4 km/hr.

a) Are they separating or approaching at the end of 2 hrs, and at what rate? b) At what time are they nearest together?

c) What is their minimum distance apart?

Answer 1

Initially distance between the ships is 32 mile = 32*1.6 km = 51.2 km

B is the initial position of ship-B

C is the position of ship-B after t hours

A is the position of ship-A after t hours

So AB = 51.2 - 2t and BC = 4t

So AC = $\sqrt{(51.2-2t)²+(4t)²}$

=> AC = $\sqrt{(51.2)²-204.8t+20t²}$

AC is the distance between the ships.

Let S = AC

As S is positive, S and S² have same type of monotonicity. So S and S² both will have same critical points.

Now,

S² = $(51.2)²-204.8t+20t²$

Differentiating with respect to t

$\frac{d(S²)}{dt} = 40t - 204.8$

=> $[\frac{d(S²)}{dt}]_{t=2} = 80 - 204.8 = -124.8$

a)

Since $\frac{dS²}{dt}$ is negative at t = 2 , after 2 hours the ships will be approaching..

Now,

$\frac{dS²}{dt} = 40t - 204.8$

=> $2S\frac{dS}{dt} = 40t - 204.8$

=> $\frac{dS}{dt} = \frac{40t - 204.8}{2S}$

=> $\frac{dS}{dt} = \frac{40t - 204.8}{2\sqrt{(51.2-2t)²+(4t)²}}$

=> $[\frac{dS}{dt} ]_{t=2}= \frac{80 - 204.8}{2\sqrt{(51.2-4)²+(8)²}}= \frac{-124.8}{47.87}=-2.61$

So the rate of approaching is 2.61 km/h

b)

For extremum value of S ,

$\frac{dS}{dt} = 0=>\frac{40t - 204.8}{2\sqrt{(51.2-2t)²+(4t)²}}=0$

=> 40t - 204.8 = 0 => t = $\frac{204.8}{40}=5.12$

Now $\frac{d²(S²)}{dt²}=40>0$

So S² is minimum at t = 5.12 and hence S is also minimum at t = 5.12.

So distance will be minimum after 5.12 hours.

c)

When t = 5.12, S = $\sqrt{(51.2-2\times 5.12)²+(4\times 5.12)²}$

= $\sqrt{(40.96)²+(20.48)²}$

= $\sqrt{2097.152}$

= 45.79

So the minimum distance is 45.79 km