At noon, ship A is 150Km west of ship B. Ship A is sailing east at 35km/h and ship B is sailing north at 25km/h. How fast is the distance between the ships changing at 4:00PM?
Let x be the distance between ship A and the initial position of ship B (B1).
Let y be the distance ship B has traveled.
Let z be the distance between the two ships.
After 4 hours:
A travels 140 km (35km/h *4 hrs). So the distance between A and B1 is 10 km. It is travelling at a speed of 35 km/h so dx/dt = - 35 km/h (note, the rate is negative since the distance is decreasing).
B travels 100 km from its initial position.
By pythagoras then, after 4 hours, z^2 = 10^2 + 100^2
=> z = sqrt(100 + 10000) = 100.50 km
Now to solve the problem, we again use Pythagoras' Theorem:
z^2 = x^2 + y^2 ..........now differentiate implicitely
=> 2z dz/dt = 2x dx/dt + 2y dy/dt
=> dz/dt = (2x dx/dt + 2y dy/dt)/2z
after 4 hours: x = 10, y = 100, dx/dt = -35, dy/dt = 25, z = 100.50
=> dz/dt = [2(10)(-35) + 2(100)(25)]/2(100.5)
=> dz/dt = 21.39 km/h ....the rate at which the distance between the two ships are changing
At noon, ship A is 150Km west of ship B.
Ship A is sailing east at 35km/h and ship B is sailing north at 25km/h.
How fast is the distance between the ships changing at 4:00pm?Code:| * B x * | * | 25t * | * - - - - * - - - - - * Q P 35t A 150-35t
Ship A starts at point P, 150 km west of point Q.
. . In t hours, it has moved 35t km east to point A.
. . Hence: .AQ = 150 - 35t
Ship B starts at point Q.
. . In t hours, it has moved 25t km north to point B.
Let x = AB
Then: .x² .= .(25t)² + (150 - 35t)²
. . and we have: .x² .= .1850t² - 10,500t + 22,500 . 
Differentiate with respect to time: .2x(dx/dt) .= .3700t - 10,500
. . and we have: .dx/dt .= .(1850t - 5250)/x . 
When t = 4, we have: . .x² .= .1850·4² - 10,500·4 + 22,500 .= .10,100
. . Hence: .x .= .√10,100 .= .10√101
Substitute into : .dx/dt .= .(1850·4 - 5350)/(10√101) .= .21.39329959
Therefore, the ships are separating at about 21.4 km/hr.