Answer:
The distance between the ships changing at 6PM is 21.29Km/h
Explanation:
Ship A is sailing east at 35Km/h and ship B is sailing West at 25Km/h
Given
dx/dt= 35
dy/dt= 25
dv/dt= ???? at t= 6PM - 2PM= 4
Therefore t=4
We know ship A travels at 150km in the x-direction and Ship A at t=4 travels at 4.35 Which is 140 also in x-direction
So, we use:
[tex] D^2 = (150 - x)^2 + y^2 [/tex];
[tex] D^2 = (150 - 140)^2 + y^2 [/tex]
But ship B travels at t=4, at 4.25 =100 in the y-direction
so, let's use the equation:
[tex] D^2 = 10^2 + 100^2 [/tex]
[tex] = D= sqrt*(10 + 100) [/tex]
Lets use 2DD' = 2xx' + 2yy'
Differentiating with respect to t we have:
D•d(D)/dt = -(10)•dx/dt + 100•dy/dt
=100.5 d(D)/dt = (-10)•35 + (100)•25
When t=4, we have x=(140-150) =10 and y=100
[tex]= D = sqrt*(10^2 + 100^2) [/tex]
=100.5
= 100.5 dD/dt = 10.35 +100.25
= dD/dt = 21.29km/h