Hello, lmao!

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? Code:

* B
* |
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 .**[1]**

At 7 PM (t = 7): .x² .= .949·7² + 2500·7 + 2500 .= .66,501

. . Hence: .x .= .√(66501)

Substitute into [1]: .dx/dt .= .(1898·7 + 2500)/(2√66501) .≈ .30.6 mph

Now you can change it back to "knots" . . .