**Problem Statement:**At 1:00pm, Ship A is 50 miles west of Ship B. Ship A is sailing north at 25 mph and Ship B is sailing south at 20 mph. How fast is the distance between the ships changing at 3:30pm?**Explanation:**To find how fast the distance between the two ships is changing, we can use the Pythagorean theorem. Let's set up a coordinate system where Ship B is at the origin at 1:00 pm, and Ship A is 50 miles to the west on the x-axis.Let:- \( x(t) \) be the east-west distance between the ships.- \( y(t) \) be the north-south distance between the ships.- \( z(t) \) be the straight-line distance between the ships.Initially, \( x(0) = 50 \) miles and \( y(0) = 0 \) miles.**Ships' Movement:**- Ship A moves north at 25 mph: \( y(t) = 25t \)- Ship B moves south at 20 mph: \( y(t) = 20t \)**Calculating Distance:**The distance between Ship A and Ship B at any time \( t \) is given by the Pythagorean theorem:\[ z(t) = \sqrt{x(t)^2 + y(t)^2} \]**At 3:30pm (2.5 hours later):**- \( x(t) = 50 \)- \( y(t) = 25t + 20t = 45t = 45 \times 2.5 = 112.5 \)Substitute these into the Pythagorean formula to find \( z(t) \).**Finding Rate of Change: \( \frac{dz}{dt} \):**Differentiate the distance function with respect to time to find the rate at which the distance between the ships is changing at 3:30 pm.

Construct the diagram: The triangles are similar, so 25t20t=x50-xx=2509By Pythagoras theorem, we haveD1=625t2+x2 and D2=400t2+x2-100x+2500 Then, the distance between the ships becomes D=625t2+x2+400t2+x2-100x+2500. Differentiate D with respect to t: dDdt=625t+xdxdt625t2+x2+400t+xdxdt-50dxdt400t2+x2-100x+2500 Here, at 3.00 PM t=2.5, x=2509 and dxdt=0 Then, D1=6252.52+25092≈55.9879 and D2=4002.52+25092-1002509+2500≈54.7159 Thus, the rate of change in distance becomes dDdt=6252.56252.52+25092+400t+xdxdt-50dxdt400t2+x2-100x+2500 dDdt=6252.555.9879+4002.554.7159≈46.1840 Therefore, the rate of change in distance between the two ships is 46.1840 mph at 3:00 PM.