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Baseball players A baseball diamond is a square 90 ft on a

side. A player runs from first base to second at a rate of 16 ft>sec.
a. At what rate is the player’s distance from third base changing
when the player is 30 ft from first base?
b. At what rates are angles ∅1 and ∅2 (see the figure) changing at
that time?
c. The player slides into second base at the rate of 15 ft>sec.
At what rates are angles ∅1 and ∅2 changing as the player
touches base?

Baseball players A baseball diamond is a square 90 ft on a side. A player runs from-example-1

1 Answer

5 votes

To answer these questions, we can use the concept of related rates. We need to consider the geometry of the baseball diamond to calculate the rates of change of distances and angles. The baseball diamond is a square with sides of 90 feet each, and the player is running from first base to second base.

a. To find the rate at which the player's distance from third base is changing when the player is 30 feet from first base, we can use the Pythagorean theorem. Let x be the distance from first base to the player, and y be the distance from third base to the player. The Pythagorean theorem relates these two distances:

x^2 + y^2 = 90^2

Now, differentiate both sides of the equation with respect to time t (in seconds):

2x(dx/dt) + 2y(dy/dt) = 0

We are given dx/dt = 16 ft/sec, and we want to find dy/dt when x = 30 ft. Substituting these values into the equation:

2(30)(16) + 2y(dy/dt) = 0

2(480) + 2y(dy/dt) = 0

960 + 2y(dy/dt) = 0

2y(dy/dt) = -960

dy/dt = -960 / (2y)

Now, we can find y when x = 30 ft:

30^2 + y^2 = 90^2

900 + y^2 = 8100

y^2 = 8100 - 900

y^2 = 7200

y = √7200

Now, substitute y back into the equation for dy/dt:

dy/dt = -960 / (2√7200)

dy/dt ≈ -6.67 ft/sec

So, when the player is 30 feet from first base, the rate at which the player's distance from third base is changing is approximately -6.67 ft/sec.

b. To find the rates at which angles ∅1 and ∅2 are changing at that time, we can use trigonometric relationships. We can consider the right triangle formed by the player, third base, and second base.

tan(∅1) = y / x

Now, differentiate both sides with respect to time t:

sec^2(∅1) * d∅1/dt = (dy/dt * x - y * dx/dt) / (x^2)

We have already calculated dy/dt = -6.67 ft/sec and dx/dt = 16 ft/sec, and we know x and y for this moment. Substituting these values, we can find d∅1/dt.

Similarly, you can find d∅2/dt by considering the right triangle formed by the player, first base, and second base. Note that ∅2 = 90° - ∅1.

c. When the player slides into second base at the rate of 15 ft/sec, we can use a similar approach as in part (b) to find the rates at which angles ∅1 and ∅2 are changing as the player touches base. You'll have the player's distance from first base and third base changing at different rates, and you can use these values to calculate the rates of change for ∅1 and ∅2.

User Shawn Esterman
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