<h2>
Answer:</h2>
<em><u>Rate of change of distance at that instant = 0.066 ft/sec </u></em>
<h2>
Step-by-step explanation:</h2>
In the question,
The baseball field is in the square diamond shape.
Side length of square = 9090 ft.
Now,
The player runs from the 2nd base to the 3rd base.
Speed of running of player, v = 30 ft/ sec.
Distance of the player from third base = 20 ft.
Now,
Let us say the distance of the player from the Home base is = l
So,
In the triangle using the Pythagoras theorem, we get,
![l^{2}=x^{2}+s^{2}](https://tex.z-dn.net/?f=l%5E%7B2%7D%3Dx%5E%7B2%7D%2Bs%5E%7B2%7D)
where, 'x' is the distance of the player from the third base and 's' is the side of the square field base.
So,
![l^{2}=x^{2}+s^{2}\\On\,differentiating\,we\,get,\\2l\frac{dl}{dt}=2x\frac{dx}{dt}\\l\frac{dl}{dt}=x\frac{dx}{dt}\\](https://tex.z-dn.net/?f=l%5E%7B2%7D%3Dx%5E%7B2%7D%2Bs%5E%7B2%7D%5C%5COn%5C%2Cdifferentiating%5C%2Cwe%5C%2Cget%2C%5C%5C2l%5Cfrac%7Bdl%7D%7Bdt%7D%3D2x%5Cfrac%7Bdx%7D%7Bdt%7D%5C%5Cl%5Cfrac%7Bdl%7D%7Bdt%7D%3Dx%5Cfrac%7Bdx%7D%7Bdt%7D%5C%5C)
Now,
Also, at the moment when, x = 20,
Length from the Home base is,
![l^{2}=20^{2}+9090^{2}\\l=9090.022\,ft.](https://tex.z-dn.net/?f=l%5E%7B2%7D%3D20%5E%7B2%7D%2B9090%5E%7B2%7D%5C%5Cl%3D9090.022%5C%2Cft.)
Now,
On putting we get,
![l\frac{dl}{dt}=x\frac{dx}{dt}\\(9090.022)\frac{dl}{dt}=20(30)\\\frac{dl}{dt}=0.066\,ft./sec.](https://tex.z-dn.net/?f=l%5Cfrac%7Bdl%7D%7Bdt%7D%3Dx%5Cfrac%7Bdx%7D%7Bdt%7D%5C%5C%289090.022%29%5Cfrac%7Bdl%7D%7Bdt%7D%3D20%2830%29%5C%5C%5Cfrac%7Bdl%7D%7Bdt%7D%3D0.066%5C%2Cft.%2Fsec.)
<em><u>Therefore, the rate of change of distance from the home plate is 0.066 ft/s</u></em>