# How do you find the rate at which a batter's distance from second base decreases when he is halfway to first base if the baseball diamond is a square with side 90 ft and a batter hits the ball and runs toward first base with a speed of 24 ft/s?

Here is an answer to the same question but with a different runner's speed.

Use this as a model, but the runner's speed is different:

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To find the rate at which the batter's distance from second base decreases when he is halfway to first base, we can use related rates and the Pythagorean theorem.

Let ( x ) be the distance of the batter from second base, and ( y ) be the distance of the batter from first base.

Using the Pythagorean theorem, we have ( x^2 + y^2 = 90^2 ).

Differentiating both sides with respect to time ( t ), we get:

[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 ]

Given that ( \frac{dx}{dt} = 24 ) ft/s (the batter's speed towards first base), and we need to find ( \frac{dy}{dt} ) when ( x = \frac{90}{2} = 45 ) ft.

Plugging in the given values, we can solve for ( \frac{dy}{dt} ):

[ 2(45)(24) + 2(45) \frac{dy}{dt} = 0 ]

[ 2(45)(24) = -2(45) \frac{dy}{dt} ]

[ \frac{dy}{dt} = -24 \text{ ft/s} ]

So, the rate at which the batter's distance from second base decreases when he is halfway to first base is 24 ft/s.

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