Abaseball diamond is a square with side 90 ft. a batter hits the ball and runs toward first base with a speed of 26 ft/s.
(a) at what rate is his distance from second base decreasing when he is halfway to first base?
ft/s
(b) at what rate is his distance from third base increasing at the same moment?
ft/s

(a) Distance decreases at a rate of 11.628 ft/s.

(b) Distance increases at a rate of 11.628 ft/s.

Step-by-step explanation:

Refer to the diagram attached to at the bottom of this answer. We will call a the distance between the batter and 1st base, x the distance between the batter and 2nd base and y the distance between the batter and 3rd base.

The speed of the batter toward first base is 26 ft/s, which means:

Since the diamond is a square, the angle between trajectories in each base is a square angle, so we can use Pythagoras theorem:

Finding the first derivative:

When the batter is halfway to 1st, a = 45 ft, then:

So the distance between the batter and 2nd base decreases at 11.628 ft/s

Now, using Pythagoras again to find y:

and we know that

Finding the first derivative:

at  b = 45 ft

So the distance between the batter and 3rd base increases at 11.628 ft/s

18.75 gallons

step-by-step explanation: