two ships leave the same port at noon ship a sales north at 10 mph and ship B sales east of 20 mph. How fast is the distance between them changing at 1 PM.
To find the rate at which the distance between the two ships is changing, we can use the Pythagorean theorem. Let the position of ship A at time t be given by the coordinates (0, 10t), and let the position of ship B be given by the coordinates (20t, 0).
The distance between the two ships at time t is given by the formula:
d(t) = sqrt((20t)^2 + (10t)^2)
To find the rate at which the distance between the two ships is changing at 1 PM (t=1), we can take the derivative of the distance function with respect to t:
d'(t) = (1/2)sqrt((20t)^2 + (10t)^2)^(-1/2)*(2*20t + 2*10t)
Now, we can plug in t=1 to find the rate at which the distance between the two ships is changing at 1 PM:
d'(1) = (1/2)sqrt((20)^2 + (10)^2)^(-1/2)*(2*20 + 2*10)
d'(1) = (1/2)sqrt(400 + 100)^(-1/2)*(40 + 20)
d'(1) = 30/sqrt(500)
d'(1) = (30/500)*(1/sqrt(5))
d'(1) = 30/(500*sqrt(5))
d'(1) = 3/(50sqrt(5))
d'(1) = 3/(50sqrt(5))
Therefore, the rate at which the distance between the two ships is changing at 1 PM is approximately 0.134 mph.