Question #281956

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?


1
Expert's answer
2021-12-23T18:05:13-0500


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 = (51.22t)2+(4t)2\sqrt{(51.2-2t)²+(4t)²}

=> AC = (51.2)2204.8t+20t2\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)2204.8t+20t2(51.2)²-204.8t+20t²

Differentiating with respect to t

d(S2)dt=40t204.8\frac{d(S²)}{dt} = 40t - 204.8

=> [d(S2)dt]t=2=80204.8=124.8[\frac{d(S²)}{dt}]_{t=2} = 80 - 204.8 = -124.8

a)

Since dS2dt\frac{dS²}{dt} is negative at t = 2 , after 2 hours the ships will be approaching..

Now,

dS2dt=40t204.8\frac{dS²}{dt} = 40t - 204.8

=> 2SdSdt=40t204.82S\frac{dS}{dt} = 40t - 204.8

=> dSdt=40t204.82S\frac{dS}{dt} = \frac{40t - 204.8}{2S}

=> dSdt=40t204.82(51.22t)2+(4t)2\frac{dS}{dt} = \frac{40t - 204.8}{2\sqrt{(51.2-2t)²+(4t)²}}

=> [dSdt]t=2=80204.82(51.24)2+(8)2=124.847.87=2.61[\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 ,

dSdt=0=>40t204.82(51.22t)2+(4t)2=0\frac{dS}{dt} = 0=>\frac{40t - 204.8}{2\sqrt{(51.2-2t)²+(4t)²}}=0

=> 40t - 204.8 = 0 => t = 204.840=5.12\frac{204.8}{40}=5.12

Now d2(S2)dt2=40>0\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 = (51.22×5.12)2+(4×5.12)2\sqrt{(51.2-2\times 5.12)²+(4\times 5.12)²}

= (40.96)2+(20.48)2\sqrt{(40.96)²+(20.48)²}

= 2097.152\sqrt{2097.152}

= 45.79

So the minimum distance is 45.79 km


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