Suppose a player is running from first to second base 20ft/s. Find rate at which distance from home plate is changing when the player is 30ft away from 2nd base?(home plate to 1st and 3rd base is 90ft)
bases are arranged in a square, 90 ft on a side
f = home to 1st , s = distance from 1st
distance to home plate (p)
... p^2 = f^2 + s^2 = 90^2 + 60^2
2 p dp/dt = 2 s ds/dt
dp/dt = (s / p) ds/dt
To find the rate at which the distance from home plate is changing when the player is 30ft away from 2nd base, we can use the Chain Rule from calculus, which states that the rate of change of a composite function is given by the product of the derivatives of the individual functions.
Let's define the variables:
- Let x represent the distance from home plate to 2nd base.
- Let y represent the distance from home plate to the player.
The distance from the player to 2nd base will be the difference between the distance from home plate to 2nd base (x) and the distance from home plate to the player (y), so we can define it as z = x - y.
Now, we need to find dz/dt, which represents the rate at which z is changing with respect to time.
Given that the player is running at a constant speed of 20ft/s from 1st to 2nd base, the rate at which the player is changing his distance from 2nd base (y) can be expressed as dy/dt = -20ft/s (negative sign indicates the player is moving away from 2nd base).
To find dz/dt, we differentiate z = x - y with respect to time (t) using the Chain Rule:
dz/dt = dx/dt - dy/dt
Since x is not changing with time (since it is the distance from home to 2nd base), dx/dt = 0.
Substituting dy/dt = -20ft/s, we have:
dz/dt = 0 - (-20ft/s)
dz/dt = 20ft/s
Therefore, the rate at which the distance from home plate is changing when the player is 30ft away from 2nd base is 20ft/s.