A baseball player is running from the at 20 ft/sec. At what rate is his distance from the home plate changing when he is 30 ft from the third base. The baseball diamond is a square 90 ft on a side.
draw a diagram. I assume he is running from 2nd base to 3rd base, other wise it is trivial.
The distance z when he is x ft from 3rd base is
z^2 = x^2 + 90^2
2z dz/dt = 2x dx/dt
To find the rate at which the distance of the baseball player from the home plate is changing, we can use the concept of related rates. We know the baseball diamond is a square with sides of 90 ft. Let's denote the home plate as point A, the third base as point B, and the baseball player's current position as point C.
We are given that the baseball player is running at a constant speed of 20 ft/sec. This means that the speed, or the rate of change, of the distance of the baseball player from the home plate is 20 ft/sec.
We can begin by drawing a diagram illustrating the situation:
A --------------- B
| |
| |
| C |
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We want to find the rate at which the distance of point C to point A is changing. This can be calculated using the Pythagorean theorem. Let's denote the distance between point C and point A as x.
Using the Pythagorean theorem, we have:
x^2 = (90^2) + (30^2)
Simplifying this equation, we get:
x^2 = 8100 + 900
x^2 = 9000
Taking the square root of both sides, we find:
x ≈ 94.868 ft
Now, to find the rate at which x is changing, we need to differentiate both sides of the equation with respect to time (t):
d/dt (x^2) = d/dt (9000)
The left side can be rewritten as:
2x(dx/dt) = 0
Substituting the known values, we have:
2(94.868)(dx/dt) = 0
Simplifying and solving for dx/dt, we find:
dx/dt ≈ 0 ft/sec
Therefore, when the baseball player is 30 ft from the third base (point B), the rate at which his distance from the home plate (point A) is changing is approximately 0 ft/sec. This implies that the player's position with respect to the home plate is not changing at that moment.