A baseball diamond is a 90 foot square with the pitcher's mound at the square's center.
A runner races for home at 25 ft/sec.
How fast is his distance from the pitcher changing as he crosses the plate?
To find the rate at which the distance between the runner and the pitcher is changing, we need to find the derivative of the distance function with respect to time (t). Let's call the distance between the runner and the pitcher D.
Since the baseball diamond is a 90-foot square, the distance D is the hypotenuse of a right triangle formed by the runner, the pitcher, and the baseline. The length of each side of the square is 90 feet, and the runner is crossing the baseline.
To calculate this, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
So, for the triangle formed by the pitcher, runner, and baseline, we have:
D^2 = B^2 + R^2
Where B is the distance from the pitcher's mound to the baseline (45 feet), and R is the distance from the runner to the baseline. We want to find dD/dt, the rate at which D is changing with respect to time.
Differentiating both sides of the equation with respect to time, we get:
2D(dD/dt) = 0 + 2R(dR/dt)
Since the baseline distance B is constant, its derivative is 0. So, the equation becomes:
2D(dD/dt) = 2R(dR/dt)
Dividing both sides by 2D, we get:
dD/dt = (R * dR/dt) / D
Now, we need to substitute the given values into the equation:
R = 90 feet (since the runner is crossing the baseline, which is one side of the square)
dR/dt = 25 ft/sec (given)
To find D, we can use the Pythagorean theorem:
D = sqrt(B^2 + R^2)
D = sqrt((45 feet)^2 + (90 feet)^2)
D = sqrt(2025 + 8100)
D = sqrt(10125)
D ≈ 100.62 feet
Substituting these values into the equation, we have:
dD/dt = (90 feet * 25 ft/sec) / 100.62 feet
dD/dt ≈ 22.36 ft/sec
Therefore, the rate at which the distance between the runner and the pitcher is changing as he crosses the plate is approximately 22.36 ft/sec.
To find the speed at which the distance between the runner and the pitcher is changing, we can use the Pythagorean theorem. Let's assume that the pitcher is at the origin, (0,0), and the runner is at the point (x,y) on the diamond.
We know that the baseball diamond is a 90-foot square, so the distance from the pitcher's mound to each base is 90/2 = 45 feet.
Using the Pythagorean theorem, the distance d between the runner and the pitcher can be calculated as:
d = √(x^2 + y^2)
To find how fast this distance is changing, we need to differentiate the equation with respect to time (t). So, we take the derivative of both sides of the equation:
dd/dt = d/(dx/dt) + d/(dy/dt)
The velocity of the runner with respect to time can be expressed as:
dx/dt = 25 ft/s (the rate at which the runner moves along the x-axis)
dy/dt = 0 ft/s (since the runner is moving perpendicular to the y-axis)
Substituting these values into the equation:
dd/dt = √(x^2 + y^2)/(dx/dt)
We also know that the runner is crossing home plate, which is at (0, 90) on the diamond. Therefore, when the runner crosses the plate, y = 90 ft.
Now, we can substitute these values into the equation to find the rate at which the distance is changing:
dd/dt = √(x^2 + 90^2)/(25 ft/s)
To find the value of x, we can use similar triangles. Notice that the pitcher's mound is at the center of the diamond, so the distance from the runner to the x-axis is constant at 45 ft.
Therefore, we can use the following triangle to find x:
45/90 = x/(90-y)
Substituting y = 90:
45/90 = x/(90-90)
1/2 = x/0
x = 0 ft
Substituting the value of x into the equation, we get:
dd/dt = √(0^2 + 90^2)/(25 ft/s)
dd/dt = √(8100)/(25 ft/s)
dd/dt = 90/25
Therefore, the speed at which the distance between the runner and the pitcher is changing as he crosses the plate is 90/25 ft/s.