Question

1. a baseball diamond is a square with side 90 ft. a batter hits the ball and runs to first base with a speed of 24 ft/s. at what rate is his distance from second base decreasing when he is halfway to first base?

Explanation:

Step1: Set up the variables

Let $x$ be the distance of the batter from home - plate. The side - length of the square baseball diamond is $a = 90$ ft. By the Pythagorean theorem, the distance $y$ from the batter to second base is $y=\sqrt{(90)^2+(90 - x)^2}=\sqrt{8100+(90 - x)^2}$.

Step2: Differentiate with respect to time $t$

Using the chain - rule, $\frac{dy}{dt}=\frac{1}{2\sqrt{8100+(90 - x)^2}}\times2(90 - x)(-1)\frac{dx}{dt}=\frac{-(90 - x)}{\sqrt{8100+(90 - x)^2}}\frac{dx}{dt}$.

Step3: Determine the values of $x$ and $\frac{dx}{dt}$

The batter runs to first base with a speed of $\frac{dx}{dt}=24$ ft/s. When the batter is halfway to first base, $x = 45$ ft.

Step4: Substitute the values into the derivative formula

Substitute $x = 45$ ft and $\frac{dx}{dt}=24$ ft/s into the formula for $\frac{dy}{dt}$.
\[

$$\begin{align*} \frac{dy}{dt}&=\frac{-(90 - 45)}{\sqrt{8100+(90 - 45)^2}}\times24\\ &=\frac{-45}{\sqrt{8100 + 2025}}\times24\\ &=\frac{-45}{\sqrt{10125}}\times24\\ &=\frac{-45}{100.62}\times24\\ &=\frac{-45\times24}{100.62}\\ &=\frac{-1080}{100.62}\approx - 10.73\text{ ft/s} \end{align*}$$

\]
The negative sign indicates that the distance $y$ is decreasing.

Answer:

The rate at which the distance from the batter to second base is decreasing is approximately $10.73$ ft/s.