Hello, jsu03!

I hope you had a good diagram.

1. A baseball diamond is a square with sides 90 ft. At the moment a batter hits the ball,

a runner from first base starts running to second base with a speed of 22 ft/sec

and a runner from second base starts running to third base with a speed of 28 ft/sec.

At what rate is the distance between these two runners changing 1 second later?

Is this distance increasing or decreasing? Code:

C
y *
D * * 90-x
o * B
* o
* * x
* *
* * A
* *
* *
* *
* *
* *
*

At time $\displaystyle t$, the runner on 1st has run $\displaystyle x$ feet from $\displaystyle A\text{ to }B,$

. . where $\displaystyle \frac{dx}{dt} = 22$ ft/sec. .Note that: $\displaystyle BC \:=\:90-x$

At the same time, the runner on 2nd has run $\displaystyle y$ feet from $\displaystyle C\text{ to }D,$

. . where $\displaystyle \frac{dy}{dt} = 28$ ft/sec.

Draw line segment $\displaystyle BD$, and let $\displaystyle z = BD.$

Since $\displaystyle BCD$ is a right triangle: .$\displaystyle z^2 \:=\:(90-x)^2 + y^2$

Differentiate with respect to time: .$\displaystyle 2z\frac{dz}{dt} \:=\:-2(90-x)\frac{dx}{dt} + 2y\frac{dy}{dt} $

. . and we have: .$\displaystyle \frac{dz}{dt} \;=\;\frac{1}{z}\left[(x-90)\frac{dx}{dt} + y\frac{dy}{dt}\right] $ .[1]

When $\displaystyle t = 1\!\;\;x = 22,\;y = 28,\;\frac{dx}{dt} = 22,\;\frac{dy}{dt} = 28 $

. . and: .$\displaystyle z \;=\;\sqrt{22^2+28^2} \:=\:\sqrt{1269} \;=\;3\sqrt{141} $

Substitute into [1]: .$\displaystyle \frac{dz}{dt} \;=\;\frac{1}{3\sqrt{141}}\bigg[(22-90)(22) + (28)(28)\bigg] \;=\;\frac{-712}{3\sqrt{141}}$

Therefore: .$\displaystyle \frac{dz}{dt} \;=\;-19.98707226 \;\approx\;\boxed{-20\text{ ft/sec}}\quad\hdots\quad\boxed{\text{decreasing}}$