A baseball diamond is a square with side 90 feet. A batter hits the ball and runs toward first base with a speed of 24 f/s.
At what rate is his distance from second base changing when he is halfway to first base?
To find the rate at which the distance from the batter to second base is changing, we can use the concept of related rates.
Let's set up a coordinate system where the origin is the pitcher's mound, and the positive x-axis extends from the pitcher's mound toward home plate.
First, we need to figure out the position of the batter as a function of time. Since we know the batter's speed is 24 f/s, the function describing the batter's position can be written as x = 24t, where x is the position of the batter relative to the pitcher's mound and t is the time elapsed.
Now, let's consider the position of second base. Since the baseball diamond is a square with side length 90 feet, the coordinates of second base are (90, -90).
To find the distance between the batter and second base, we can use the distance formula:
Distance^2 = (x2 - x1)^2 + (y2 - y1)^2
Plugging in the coordinates of the batter (x, y) = (24t, 0) and second base (x, y) = (90, -90), we have:
Distance^2 = (90 - 24t)^2 + (-90 - 0)^2
Simplifying this expression, we get:
Distance^2 = (90 - 24t)^2 + 90^2
Now, to find the rate at which the distance from the batter to second base is changing, we can take the derivative of this expression with respect to time:
2 * Distance * (d(Distance) / dt) = 2 * (90 - 24t) * (-24)
Simplifying further:
Distance * (d(Distance) / dt) = (90 - 24t) * (-24)
We are interested in finding the rate at which the distance from the batter to second base is changing when the batter is halfway to first base. In this case, we can set x = 45 feet (halfway to first base) and solve for the rate.
Plugging in x = 45 and solving for (d(Distance) / dt), we get:
Distance * (d(Distance) / dt) = (90 - 24t) * (-24)
45 * (d(Distance) / dt) = (90 - 24t) * (-24)
Simplifying further:
d(Distance) / dt = [(90 - 24t) * (-24)] / 45
Now, we can plug in the value of t that corresponds to halfway to first base. Since x = 24t represents the distance traveled by the batter, when the batter is halfway to first base, x = 45 feet. Solving for t, we get:
45 = 24t
t = 45/24
t ≈ 1.875 seconds
Plugging this value of t into the expression for (d(Distance) / dt), we get:
d(Distance) / dt = [(90 - 24(1.875)) * (-24)] / 45
Simplifying further, the rate at which the distance from the batter to second base is changing when the batter is halfway to first base is approximately:
d(Distance) / dt ≈ -19.2 feet per second
To find the rate at which the batter's distance from second base is changing when he is halfway to first base, we need to use the concept of related rates.
Let's assume that the batter is running in a straight line from home plate (located at the center of the diamond) towards first base. At the halfway point between home plate and first base, the batter is 45 feet away from home plate.
To find the rate at which the batter's distance from second base is changing, we'll need to use similar triangles.
Consider a right triangle formed by the batter, second base, and a point on the perpendicular line drawn from second base to the line connecting the batter and home plate.
The hypotenuse of this right triangle represents the batter's distance from second base, which we'll call x. The base of this right triangle (the distance from home plate to the point on the perpendicular line) is 45 feet.
The side of the triangle representing the distance from second base to the point on the perpendicular line is a variable, which we'll call y.
Since the baseball diamond is a square, the distance from home plate to second base is also 90 feet. Therefore, the remaining distance from the point opposing the battter's location to second base is (90 - y).
Since we know that the distance from the batter to second base is always 45 feet (half the side of the square) and the distance from home to second base is always 90 feet (the full side of the square), we can create a ratio:
45 / (90 - y) = y / x
To solve for the rate at which the batter's distance from second base is changing, we need to differentiate both sides of this equation with respect to time.
d(45 / (90 - y)) = d(y / x)
To make this equation more manageable, let's multiply both sides by (90 - y) to remove the denominator:
45 = (y / x) * (90 - y)
Now, let's differentiate both sides with respect to time:
0 = (x(dy/dt) - y(dx/dt)) / x^2
Since the batter is running towards first base, dx/dt is equal to his speed, which is given as 24 ft/s.
Substituting this value and the known values of x and y into the equation, we get:
0 = (45(24) - 45(dy/dt)) / (45^2)
Simplifying further:
0 = 24 - (dy/dt) / 45
Now, solving for dy/dt (the rate at which the batter's distance from second base is changing):
(dy/dt) / 45 = 24
dy/dt = 24 * 45
Therefore, the rate at which the batter's distance from second base is changing when he is halfway to first base is 1,080 ft/s.