Final answer:
When the softball player running from first to second base at 25 ft/s is 40 ft away from second base, her distance from home plate is increasing at a rate of 10 ft/s.
Step-by-step explanation:
To solve for the rate at which the player's distance from home plate is changing, we can apply the Pythagorean theorem which states that in a right triangle, the square of the length of the hypotenuse (c) is equal to the sum of the squares of the lengths of the other two sides (a and b).
In a softball diamond, which is a perfect square, the distance from home plate to second base is the hypotenuse of a right triangle where the sides are the pathways from home to first and from first to second base, both having lengths of 60 ft. Given that the player is 40 ft from second base, we can set up the following right triangle:
- Side a (first base to player) = 60 ft
- Side b (player to second base) = 40 ft
- Hypotenuse c (home plate to player) = √(60^2 + 40^2) = √(3600 + 1600) = √5200 ft
Since the player is running towards second base, we are interested in finding the rate at which the distance from home plate (hypotenuse c) is changing. Let's call this rate dc/dt. We can differentiate both sides of the Pythagorean theorem with respect to time t to obtain a relationship between the rates:
2a(da/dt) + 2b(db/dt) = 2c(dc/dt)
In our case, da/dt = 0 because the distance from the player to first base does not change as she runs from first to second. Therefore, the equation simplifies to:
2b(db/dt) = 2c(dc/dt)
Given that db/dt (speed of the player) is 25 ft/s, we can solve for dc/dt:
dc/dt = (b/c) × (db/dt) = (40/√5200) × 25 ft/s
After calculating, we find that dc/dt = 10 ft/s. Therefore, at the instant when the player is 40 ft from second base, her distance from home plate is increasing at a rate of 10 ft/s.