## The sides of a square baseball diamond are 90 feet long. When a player who is between second and third base is 60 feet from second base and

Question

The sides of a square baseball diamond are 90 feet long. When a player who is between second and third base is 60 feet from second base and heading towards third base at a speed of 22 feet per second, how fast is the distance between the player and home plate changing

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3 years 2021-08-06T17:30:57+00:00 1 Answers 29 views 0

d(L)/dt  =  6,96 ft/s

Explanation:

We have a right triangle where the hypotenuse (L) is the distance between the player and home plate, the legs are the line between third base and home plate ( 90 feet ) and distance between the player and third base (x)  (over the line between second and third base). So we can write

L²  =  (90)²  +  x²

Applying differentiation in relation to time, on both sides of the equation we have:

2*LdL/dt  =  0  + 2*x d(x)/dt     (2)

In this equation we know:

d(x)/ dt  =  22 feet/sec

x = 30 ft

We need to calculate  L when the player is at 30 feet from third base

Then

L²  =  (90)² + (30)²

L²  =  8100 + 900

L  = √9000

L =94,87 feet

Then we are in condition for calculate  d(L)/dt from the equation

2*Ld(L)/dt  =  0  + 2*x d(x)/dt

2*94,87 * d(L)/dt  =  2* 30* 22     ⇒  189,74 d(L)/dt = 1320

d(L)/dt  =  1320/ 189,74

d(L)/dt  =  6,96 ft/s