Final answer:
a. The player's distance from third base is changing at a rate of -13.5 ft/sec. b. The rates at which angles θ₁ and θ₂ are changing are θ₁' = arctan(30/x) * (-18 ft/sec) and θ₂' = arctan(30/(30 - x)) * (-13.5 ft/sec). c. When the player slides into second base at a rate of 15 ft/sec, the rates at which angles θ₁ and θ₂ are changing will be different.
Step-by-step explanation:
a. To find the rate at which the player's distance from third base is changing, we need to use the concept of related rates. Let x represent the player's distance from first base. We can create a right triangle with sides x, 30 ft, and the distance between third and first bases. Using the Pythagorean theorem, we can calculate the distance between third and first bases to be 40 ft. By differentiating this expression, we get dx/dt = -18 ft/sec. Now, we can use the Chain Rule to find dy/dt, where y represents the distance between the player and third base. dy/dt = (dy/dx) * (dx/dt). Since dy/dx is 30 ft/40 ft = 3/4, we can substitute dx/dt = -18 ft/sec to find dy/dt = (3/4) * (-18 ft/sec) = -13.5 ft/sec. Therefore, the player's distance from third base is changing at a rate of -13.5 ft/sec when the player is 30 ft from first base.
b. To find the rates at which angles θ₁ and θ₂ are changing, we can use trigonometric functions. Angle θ₁ is the angle between the line connecting the player to first base and the line connecting the player to third base, and angle θ₂ is the angle between the line connecting the player to second base and the line connecting the player to third base. Both angles are changing when the player is moving towards second base. We can find the rates at which these angles are changing by taking the derivative of their respective trigonometric functions. Let's call θ₁' the rate at which θ₁ is changing and θ₂' the rate at which θ₂ is changing. We have θ₁' = arctan(30/x) * (dx/dt) and θ₂' = arctan(30/y) * (dy/dt). Substituting the values we found earlier, we have θ₁' = arctan(30/x) * (-18 ft/sec) and θ₂' = arctan(30/(30 - x)) * (-13.5 ft/sec).
c. When the player slides into second base at a rate of 15 ft/sec, the rates at which angles θ₁ and θ₂ are changing will be different. Let's call θ₁'' the rate at which θ₁ is changing when the player touches second base and θ₂'' the rate at which θ₂ is changing when the player touches second base. We have θ₁'' = arctan(30/x) * (dx/dt) + arctan(15/y) * (dy/dt) and θ₂'' = arctan(30/(30 - x)) * (dx/dt) + arctan(15/(30 - y)) * (dy/dt), where x, y, dx/dt, and dy/dt are the values we calculated earlier.