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Label the position of ship B at noon as P.
At 4:00pm ship A is 10km from P (it started 150km from P and has travelled 35km/h for 4hr or 140km.) Ship B is 100km from P (it has travelled 25km/h for 4 hours or 100km.)
Ship A's speed is 35km/h, so `(dA)/(dt)=-35` (negative since the distance from A to P is decreasing as t increases). Also ship B's speed is 25km/h so `(dB)/(dt)=25` (positive as the distance between B and P increases as t increases.)
The distance between A and B is `d=sqrt(A^2+B^2)` where A is the distance fro A to P, and B is the distance from B to P.
Then the rate at which the distance between the ships is changing is:
Substituting the known values we get:
Thus the distance between the ships is increasing at approxiamtely 21.39km/h
o - - o
The initial distance between A and B is 150 km.
Four hour has gone by, we know that A travels to east at 35km/hr. Hence, for 4hrs ship A travels -35 * 4 = - 140km. (the rate is negative since, ship A goes nearer and nearer to B).
This will affect the horizontal distance 150km, 150km - 140km = 10km. Let set that distance to be "x".
Now, we know that B travels to north at 25km/hr. Hence, for 4hrs ship B travels 25*4 = 100km. Let set that distance to be "y".
After 4 hrs, our triangle will be:
- - y = 100
A o - - o
x = 10
Let set the hypotenuse to be r. We can solve for the measure of r using Pythagorean Theorem:
x^2 + y^2 = r^2
===> 10^2 + 100^2 = r^2
===> 100 + 10000 = r^2
===> 10100 = r^2
Getting the square root of both sides we will get,
r = 100.499
Now, we differentiate implicitly the x^2 + y^2 = r^2 with respect to t (time).
===> 2xdx/dt + 2y/dt = 2rdr/dt
Dividing both sides by 2:
===> xdx/dt + ydy/dt = rdr/dt
We can now solve for the dr/dt.
Plug-in x = 10, dx/dt = -35, y = 100, dy/dt = 25 and r = 100.499.
===> (10)(-35) + (100)(25) = 100.499(dr/dt)
===> -350 + 2500 = 100.499dr/dt
===> 2150 = 100.499dr/dt
Dividing both sides by 100.499 to isolate the dr/dt on right side,
2150/100.499 = (100.499dr/dt)/100.499
===> 21.39 = dr/dt
or dr/dt = 21.39 km/hr
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