just use the good old Pythagorean Theorem. If B is at (0,0) at noon, then at time t hours,
A is at (-100,-35t)
B is at (0,25t)
The distance d is thus
d^2 = 100^2 + (35t+25t)^2
= 100^2 + 3600t^2
at 4:00 pm, t=4, so
d = 260
now, we get
2d dd/dt = 7200t
dd/dt = 7200*4/(2*260) = 720/13
Or, you can consider the distance x and y, traveled by the ships A and B:
d^2 = 100^2 + (x+y)^2
2d dd/dt = 2(x+y)(dx/dt + dy/dt)
2*260 dd/dt = 2(140+100)(35+25)
dd/dt = 720/13
Related Rates
I am having trouble finding the equation to use to fit this all together.
At noon, ship A is 100km west of ship B. Ship A is sailing south at 35 km/s and ship B is sailing north at 25 km/s. How fast is the distance between the ships changing at 4:00pm.
1 answer