Question

At noon, ship a is 60 km west of ship

Answer 1

DA/dt=15 km/h and dB/dt=5 km/h
base of our triangle if drawn is 60 km
to find dD/dt when t=4 hours we shall have:

(A+B)²+60²=D²
d/dt[(A+B)²+60²]=d/dt(D²)
[d(A+B)]/dt*2(A+B)(2)=dD/dt*2D
B=5km/h*4hr=20km
A=15km/hr*4=60km
20²+60²=D²
D=√4000

(15+5)*(2)*80*(2)=2*√4000*dD/dt
dD/dt=6400/√4000=101.192 km/hr

Answer 2

The two ships are sailing in opposite sides, Hence, the resultant speed is given by 15+5 = 20 km/hr.

Therefore, from the below triangle, we have

$\frac{dx}{dt} = 20 \text{ km/hr}$

Let the distance between the ships is y. On applying Pythagorous theorem, we have

$y^2=x^2+60^2\\ \text{On differentiating, we get}\\ 2y\frac{dy}{dt} =2x\frac{dx}{dt}\\\frac{dy}{dt}= \frac{x}{y} \frac{dx}{dt}$

On substituting the value of y as $y=\sqrt{60^2+x^2}$

$\frac{dy}{dt} =\frac{x}{\sqrt{60^2+x^2}} \frac{dx}{dt}$

Since, the at noon the ship is 60 km to each other. Hence, for 4 PM, i.e. t=4, we have

$x=4 \times \frac{dx}{dt} \\ x= 4 \times 20 \\x=80$

On substituting the value in above, we get

$\frac{dy}{dt} =\frac{80}{\sqrt{60^2+80^2}} (20$

$\frac{dy}{dt} = 16.0 \text{ km/hr}$

Therefore,  the distance between the ships changing at a rate of 16 km/hr at 4:00 pm