A baseball diamond is a square with sides of length 90 ft. A batter hits the ball and runs toward first base with a speed of 22 ft/s.

Question

At what rate is his distance from second base changing when he is halfway to first base?

Also, at what rate is his distance from third base changing at the same moment?

Answer 1

If you sketch this out, the distance between the runner forms a right triangle. Let x be the distance between the runner and first base and s be the distance between the runner and second base. So, dx/dt = -22ft/s since the distance between the runner and first base is decreasing.

Second base distance rate:

s² = 90²+x²

2sds/dt = 2xdx/dt

ds/dt =(x/s) dx/dt

When x = 45 ft, s=√(90²+45²) ≈ 100.62 ft

ds/dt = (45/100.62)(-22)

Third base distance rate:

This will be similar, except let x be the distance between home and the runner. So dx/dt = +22 ft/s since the distance between home and the runner is increasing.

ds/dt = (45/100.62)(22)