A baseball diamond is a square 90 ft on a side. A player runs from first base to second base at 15 ft/sec. At what rate is the player's distance from home base increasing when he is half way from first to second base?
To find the rate at which the player's distance from home base is increasing when he is halfway from first to second base, we can use the concept of related rates from calculus.
First, let's establish some notation:
- Let's denote the distance of the player from home base as "x" ft.
- We'll denote the time variable as "t" seconds.
- The speed of the player is given as 15 ft/sec.
Next, let's determine the relationship between the variables:
We have a square with sides measuring 90 ft. The player starts at first base (one corner) and is running towards second base (opposite corner).
The distance from home base is given by the diagonal of the square, which forms a right triangle with the sides of the square. By the Pythagorean theorem, we know that the diagonal (distance from home base) equals the square root of the sum of the squares of the sides. Thus:
x = sqrt((90 ft)^2 + (90 ft)^2)
Differentiating both sides with respect to time "t", we get:
dx/dt = (1/2) * (2 * 90 ft) * (1/2) * (2 * 90 ft) * d/dt(sqrt((90 ft)^2 + (90 ft)^2))
Simplifying this, we have:
dx/dt = 90 ft * 90 ft * (dx/dt) / (sqrt((90 ft)^2 + (90 ft)^2))
Since we are interested in the rate at which the player's distance from home base is increasing when he is halfway from first to second base, we have x = 45 ft.
Plugging this into the equation, we get:
dx/dt = 90 ft * 90 ft * (dx/dt) / (sqrt((90 ft)^2 + (90 ft)^2)), where x = 45 ft
Now we have:
dx/dt = 90 ft * 90 ft * (dx/dt) / (sqrt((90 ft)^2 + (90 ft)^2)), where x = 45 ft
Simplifying further, we have:
dx/dt = 90 ft * 90 ft * dx/dt / (sqrt(2 * (90 ft)^2))
dx/dt = 90 ft * 90 ft * dx/dt / (sqrt(2 * (90 ft)^2))
dx/dt = 90 ft * dx/dt / sqrt(2)
Now we can substitute the given speed of the player (15 ft/sec) into the equation above to find the rate at which the player's distance from home base is increasing when he is halfway from first to second base.
dx/dt = 90 ft * (15 ft/sec) / sqrt(2)
Finally, we can calculate the rate:
dx/dt = (90 ft * 15 ft/sec) / sqrt(2)
Approximating this with a calculator, we find:
dx/dt ≈ 954.92 ft/sec
So, the rate at which the player's distance from home base is increasing when he is halfway from first to second base is approximately 954.92 ft/sec.