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Calculus/Related Rate of Change Question


Reaching first base after a dropped third strike, Ryan Braun, attempting to steal second base, takes off running toward second. The throw from the catcher gets through the shortstop, who was covering second, and the ball rolls into centerfield. From second base, Braun takes off for third base, running at 21.5 feet per second. Assuming the ball travels in a straight line from home plate over second base and rolls into center field at 40 feet per second, how fast is the distance between Braun and the ball changing when Braun is 70 feet from third base and the ball is 60 feet past second base into center field? (Hint: What angle does the balls path make with Brauns path?)

Sorry it took so long, but I finally got a chance to answer this.

It does not give how far it is between bases, but from standard baseball it is 90',

The main angle to look at is 135 measured at 2nd base.
The ball is 60' past 2nd travelling at 40'/s.

Since the person is 70' from 3rd, he is x-70 feet from 2nd,
where x is the distance from 2nd to 3rd.  Since x is 90',
the distance from 2nd is only 20'.  The speed away from 2nd is 21.5'/s.

The distance betwen them is given by c = a + b - 2ab*cos(135)
where a is how far the runner has moved and b is how far the ball has moved.
Taking the squareroot of both sides gives c = √[a + b - 2ab*cos(135)].

Taking the derivative of c(t) = a(t) + b(t) - 2a(t)b(t)*cos(135) with respect to t gives
2c(t)(dc/dt) = 2a(t)(da/dt) + 2b(t)(db/dt) - 2(a(t)(db/dt) + b(t)(da/dt))(-√2/2).

At whatever the time is, it is known that a(t) = 20; da/dt = 21.5; b(t) = 60; db/dt = 40; and
c is the length of the 3rd leg on the triangle with side 20, side 60, and angle 135.

Once c has been found, dc/dt can be solved for by dividing both sides by 2c.
This gives the value for dc/dt.


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