I'll walk through the first one . . .
And I'm changing the units to miles and miles per hour, okay?
1. At noon, ship A is 50 miles due west of ship B.
Ship A is sailing west at 25 mph and ship B is sailing north at 18 mph.
How fast is the distance between the ships changing at 7 PM?
x * |
* | 18t
* - - - - * - - - - - - *
A 25t P 50 Q
At noon, ship A starts at point P, 50 miles west of point Q.
. . At 25 mph, in the next t hours, it moves 25t miles to point A.
At noon, ship B start at point Q, sailing north at 18 mph.
. . In the next t hours, it moves 18t miles to point B.
Let x = the distance AB.
Pythagorus says: .x² .= .(25t + 50)² + (18t)²
So we have: .x² .= .949t² + 2500t + 2500
Differentiate with respect to time: .2x(dx/dt) .= .1898t + 2500
. . and we have: .dx/dt .= .(1898t + 2500)/2x .
At 7 PM (t = 7): .x² .= .949·7² + 2500·7 + 2500 .= .66,501
. . Hence: .x .= .√(66501)
Substitute into : .dx/dt .= .(1898·7 + 2500)/(2√66501) .≈ .30.6 mph
Now you can change it back to "knots" . . .