Can you help me with another calculus problem?

Question

A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 24 ft/s.
(a) At what rate is his distance from second base decreasing when he is halfway to first base?
(b) At what rate is his distance from third base increasing at the same moment?

Answer 1

It is not clear what you mean by distance to second base. Distance along the base lines? That is simply 180-v*t, and the rate is simply v.

I think you mean distance from the point of the runner directly to second base. Draw a line from the runner to second base. The length of that line is the hypotenuse of a right triange, one leg being the 1st to 2nd base line, the other being the distance remaining to get to first base. Thus leg 1 = 90ft, leg 2 = 90 - v*t ft

The distance to 2nd base is then the sqrt of the sum of the squares of the two legs, or

L(t) = √[90^2 + (90 - v*t)^2].

The rate of decrease of L is d(L(t))/dt; differentiate that expression and put in a time = 45ft/v (the time it takes for the runner to reach halfway to first base.)

For the distance to third base, draw the line connecting the runner to third. You get another right triangle, but the legs are 3rd-to-home and v*t, the distance covered from home to first in time t. You actually get a simpler expression, but the method is the same.

Answer 2

uneducated guess.
a). 7.5 seconds
b). 11.25seconds