**A baseball diamond is a square whose sides are 90 feet long. Suppose that a player is running from second to third base has a speed of 30 feet per second at the instant when he is 20 feet from third base. at what rate is the players distance from home plate changing at that instant?**

**Answer:**

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it sometimes the different value in given.

Let H be home, A be first base and B is 2nd base.

The angle HAB = 90 degrees

let x ft. be the player’s distance from first base and let y be his distance from home, then from the pythagorean theorem we have:

(1) y = sqrt(90^2 +x^2) ft = (90^2 +x^2)^0.5 ft

We are given dx/dt = 28ft per sec (a constant. Acceleration is ignored.)

Now: (2) dy/dt = (dy/dx)*(dx/dt)

dy/dx = 0.5(90^2+x^2)^ -0.5 * d/dx (90^2+x^2)

Now: d/dx (90^2+x^2) = 2x, so

dy/dx = x / (90^2 + x^2)^0.5

When the player is 30 ft from 2nd, he is 60 ft from first. So x = 60ft. We can now determine dy/dt from (2) using x = 60 ft

dy/dt = (60 /(90^2+60^2)^0.5)*28 ft/sec

= 15.53 ft/sec