jack is running at 5 mph east and passes spot X at the same time jill, who is 8 miles south of point X, is running south at 4 mph One hour later what is the rate of change in the distance between them

Question

related rates problem

Answer 1

At time t, Jack’s position is 5t miles east of X while Jill’s position is 8+4t miles south of X.

The distance s between them is √((5t)²+(8+4t)²)=√(41t²+64t+64), by Pythagoras.

ds/dt=(1/2)(82t+64)/√(41t²+64t+64) is the rate of change (speed) of distance between them.

After an hour, t=1, this expression has a value (1/2)(146)/√169=(1/2)(146)/13=73/13=5.62mph.