A Baseball Diamond has the shape of a square with sides 90 feet long. a player is running from 1st to 2nd base at a speed of 25 feet per second. find the rate at which his distance s is from home plate is changing when the player is 20 feet from 2nd base.
To find the rate at which the player's distance from home plate is changing, we can use the concept of related rates.
Let's define the following variables:
- s = the distance of the player from home plate
- t = time (in seconds)
- x = the distance of the player from 1st base
- y = the distance of the player from 2nd base
Since the baseball diamond is a square, we have x = y. The given information tells us that each side of the square is 90 feet long. Therefore, x = y = 90 ft.
We are given that dx/dt = 25 ft/s, which represents the rate at which the player is moving from 1st base towards 2nd base.
To find ds/dt, we need to determine the relationship between s, x, y, and t.
Using the Pythagorean theorem, we can write an equation relating s, x, and y:
s^2 = x^2 + y^2
Differentiating implicitly with respect to t, we get:
2s * ds/dt = 2x * dx/dt + 2y * dy/dt
Since x = y = 90 ft, we can substitute those values in the equation:
2s * ds/dt = 2(90) * 25 + 2(90) * dy/dt
Simplifying the equation further:
2s * ds/dt = 4500 + 180 * dy/dt
Now, we need to find dy/dt when s = 20 ft.
To do this, we can use the fact that x^2 + y^2 = s^2 and differentiate implicitly with respect to t:
2x * dx/dt + 2y * dy/dt = 2s * ds/dt
Substituting the known values:
2(90) * 25 + 2(90) * dy/dt = 2(20) * ds/dt
Simplifying the equation further:
4500 + 180 * dy/dt = 40 * ds/dt
We can now solve for dy/dt:
180 * dy/dt = 40 * ds/dt - 4500
dy/dt = (40 * ds/dt - 4500) / 180
Now, plug in the given values: ds/dt = 25 ft/s and s = 20 ft:
dy/dt = (40 * 25 - 4500) / 180
dy/dt = (1000 - 4500) / 180
dy/dt = -3500 / 180
dy/dt ≈ -19.44 ft/s
Therefore, when the player is 20 feet away from 2nd base, their distance from home plate is changing at a rate of approximately -19.44 feet per second.