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?

Question

Isaiah Allen · Accepted Answer

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: https://tutor.hix.ai

Adam Adkins · Answer

undefined 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.