Answer:
answer = 12.87 km/h
Step-by-step explanation:
Given
Ship A is sailing east at 25 km/h = [tex]\frac{dx}{dt}[/tex]
ship B is sailing north at 20 km/h =[tex]\frac{dy}{dt}[/tex]
here x and y are the sailing at t = 4 : 00 pm for ship A and B respectively
so we get x = 4 ×25 =100 km/h
y = 4× 20 = 80 km/h
let z is the distance between the ships, we need to find [tex]\frac{dz}{dt}[/tex] at t = 4 hr
At noon, ship A is 130 km west of ship B (12:00 pm)
so equation will be
[tex]z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we | | get \\[/tex]
[tex]z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h[/tex]
derivative first equation w . r. to t we get
[tex]2z\frac{dz}{dt} =[/tex][tex]-2(130-x)\frac{dx}{dt}[/tex][tex]+2y\frac{dy}{dt}[/tex]
[tex]\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}][/tex]
[tex]\frac{dz}{dt} = \frac{( -20\times25 + 80\times20)}{\sqrt{7300} }[/tex]
[tex]= \frac{1100}{85.44}\\ = 12.87km/h[/tex]
