Expert: Scotto - 6/17/2010

Question
A baseball diamond is a square with side 90ft.  A batter hits the ball and runs towards first base with a speed of 24 ft/s.

a) At what rate is his distance from 2nd decreasing when he is halfway to first?  (I already got this answer from a post on this website http://en.allexperts.com/q/Calculus-2063/2008/3/Related-Rates-35.htm...however I am having difficulty with the next part as I really don't understand related rates)

b) At what rate is his distance from 3rd increasing at the same time?

Any and all help will be greatly appreciated.

Thanks,

Gabe.

Answer
When a diamond is drawn (a tipped square), put the man at x units from home at spot A,
so the distance from home to A is x.  This forms a right triangle with sides as
from home to x and from home to 3rd.  The hypotenuse is the distance.

Let the length of the hypotenuse be y.  The equation would be x² + 90² = y².
Since 9²=81, 90²=8100, so we have x² + 8100 = y².

Now when you think about it, both x and y will vary with time, so this is really
x²(t) + 8100 = y²(t).  Taking the derivative (with the chain rule) gives 2x(t)x'(t) = 2y(t)y'(t).
Note we can divide by 2, and just remember that both are functions of t, giving xx' = yy'.

Now looking back at what we've done, x is the distance from home, x' is the speed of the runner,
y is the distance to 3rd, and y' is the change in the distance to 3rd.  Next, x is the only variable, since x' is said to be 24ft/sc, y is defined by the pythagorean theorm to be
y = √(x²+8100).  Putting these in gives us y' and x as the only variables.  This equation
can be made into y' in terms of x.