Softball Related Rates Problem

[michial93]

A softball diamond is a square whose sides are 60 feet long. Suppose that a player is running from first to second base has a speed of 25 feet per second at the instant when she is 10 feet from second base. At what rate is the player's distance from third base changing at that instant?

This is my last related rates problem, I don't know if my brain is just tired now, but I can't seem to find my starting point with this problem. I tried drawing the pictures out, but the numbers are not sinking in. Can someone help me set-up and start the problem?

Thanks!

 

[srmichael]

A softball diamond is a square whose sides are 60 feet long. Suppose that a player is running from first to second base has a speed of 25 feet per second at the instant when she is 10 feet from second base. At what rate is the player's distance from third base changing at that instant?

This is my last related rates problem, I don't know if my brain is just tired now, but I can't seem to find my starting point with this problem. I tried drawing the pictures out, but the numbers are not sinking in. Can someone help me set-up and start the problem?

Thanks!

I don't have a picture to show you, but if you can picture that when the player is 10 feet from second base then the distance from that player to third base is the hypotenuse of a right triangle with one leg at 10, and the other at 60. Now, the leg that is 10 feet is actually a variable, say x, and the distance from the player to third base is also a variable, say z.

So from here can you proceed? See how far you can get and then let us know if you get stuck.

 

[soroban]

Hello, michial93!

Did you make a sketch?

A softball diamond is a square whose sides are 60 feet long.
A player is running from first to second base has a speed of 25 ft/sec at the instant
. . when she is 10 feet from second base.
At what rate is her distance from third base changing at that instant?

                B
                *  60-25t
              *   *
         60 *       o P
          *     o     *   25t
        *   o x         *
      * o                 *
  C o                       * A
      *                   *
        *               *
       60 *           * 60
            *       *
              *   *
                *
                H

The distance from 1st base $\displaystyle A$ to 2nd base $\displaystyle B$ is 60 feet.

In $\displaystyle t$ seconds, she runs to $\displaystyle P\!:\;AP = 25t$
Hence:.$\displaystyle PB \,=\,60-25t$

Her distance from third base, $\displaystyle x = CP$, is the hypotenuse of a right triangle
. . with legs $\displaystyle 60-25t$ and $\displaystyle 60.$

Hence: .$\displaystyle x^2 \;=\;(60-25t)^2 + 60^2 \;=\;625t^2 - 3000t + 7200$


Differentiate with respect to time:
. . $\displaystyle 2x\dfrac{dx}{dt} \:=\:1250t - 3000 \quad\Rightarrow\quad \dfrac{dx}{dt} \:=\:\dfrac{125(5t-12)}{x}$ .[1]



When $\displaystyle BP = 10\!:\;\;60-25t \,=\,10 \quad\Rightarrow\quad \boxed{t \,=\,2}$

And \(\displaystyle x^2 \:=\60-25\cdot2)^2 + 60^2\:=\: 3700 \quad\Rightarrow\quad \boxed{x \:=\:10\sqrt{37}}\)


Substitute into [1] . . .

 

[srmichael]

Or you could have set it up this way using the same variables I mentioned above:

Given: $\displaystyle \displaystyle \frac{dx}{dt}=-25$

Formula: $\displaystyle \displaystyle x^2+60^2=z^2$

Derivative: $\displaystyle \displaystyle 2x\frac{dx}{dt}+0=2z\frac{dz}{dt}$ (A)

You are looking for $\displaystyle \displaystyle \frac{dz}{dt}$ so you need to solve for z at the instant that $\displaystyle \displaystyle x=10$.

Pythag it out: $\displaystyle \displaystyle 10^2+60^2=z^2$

Solve for z: $\displaystyle \displaystyle z=10\sqrt{37}$. Now plug this into (A) and solve...