Answer to Question #329755 in Calculus for hieuo

Question

A baseball player is running from the second base to the third base at 20ft/sec. At what rate is his

distance from the home plate changing when he is 30ft from the third base. The baseball diamond

is a square 90ft on aside.

Answer 1

The picture represents a baseball diamond and a player.

The player is located at the point $E$. He is between the third $(C)$ and the second base ($B$ ). The arrow shows the velocity vector. $|CE|=30ft$. $|AE|$ is the distance between the home plate and the player. Using the Pythagorean theorem, we get: $|AE|^2=|AB|^2+|EB|^2=90^2+60^2={11700}$. $|AF|$ denotes the distance between the home plate and the player, when he will pass a small distance $|EF|$. $|EF|=v\Delta t$, where $\Delta t$ is a small period of time. $|AF|^2=|AB|^2+(|FE|+|BE|)^2=90^2+(v\Delta t+60)^2$. It is enough to find the derivative of $|AF|-|AE|$ with respect to $\Delta t$ and substitute $\Delta t=0.$ It will show how fast does the distance change. $|AF|'-|AE|'=|AF|'=\frac{1}{|AF|}v(v\Delta t+60)$. After setting $\Delta t =0$ we get: $60v\frac{1}{|AE|}=\frac{1}{\sqrt{11700}}60\cdot20\approx11.09$. Thus, the distance between the home plate and the player changes with the velocity $11.09ft/sec$. This is the velocity, when the player is at the point $E$ . It changes with time.