different related rates problem.

[galactus]

Here is a related rates problem I saw somewhere but I can't remember where though.

Anyway, it is slightly different from the run-of-the-mill, cliche problems we see over and over again.

"A man is standing at the topmost edge of a circle of radius R. He walks toward the center at a rate we can call 'v' ft/min.

A light is on the leftmost edge of the circle and is shining toward the center.

How fast is the man's shadow moving across the rightmost edge of the circle when he is 2/3rds the way to the center?".

This is from memory, so I believe I worded it correctly.

 

[soroban]

Hello, galactus!

you're right . . . This is certainly different!

A man is standing at the topmost edge of a circle of radius \(\displaystyle R.\)

He walks toward the center at a rate of \(\displaystyle v\) ft/min.

A light is on the leftmost edge of the circle and is shining toward the center.

How fast is the man's shadow moving across the rightmost edge of the circle when he is 2/3 the way to the center?".

               A|
              * o *
          *     |     *   P
        *     vt|       o
       *        |   * *  *
              M o   *      x 
      *    * @  | * 2@    *
    L o - - - - * - - - - o B
      *    R    O    R    *

       *                 *
        *               *
          *           *

The man starts at point \(\displaystyle A.\)

In \(\displaystyle t\) minutes, he has walked \(\displaystyle vt\) ft to point \(\displaystyle M.\)
. . \(\displaystyle MO \:=\:R - vt\)

Draw line \(\displaystyle LM\), extended to the circle at point \(\displaystyle P.\)
Draw radius \(\displaystyle OP.\)
Let \(\displaystyle \angle PLB = \theta,\;\text{ then: }\:\angle POB = 2\theta\)
Let \(\displaystyle x \:=\:\text{arc}\,\overline{PB}\)

\(\displaystyle \text{We have: }\;x \:=\:R(2\theta) \:=\:2R\theta\) .[1]


\(\displaystyle \text{In right triangle MOL,}\:\text{ we have: }\;\tan\theta \;=\;\frac{MO}{LO} \;=\;\frac{R-vt}{R} \;=\;1 - \frac{v}{R}t\)

. . \(\displaystyle \text{Hence: }\;\theta \;=\;\arctan\left(1-\frac{v}{R}t\right)\)


\(\displaystyle \text{Substitute into [1]: }\;x \;=\;2R\arctan\left(1 - \frac{v}{R}t\right)\)


\(\displaystyle \text{Differentiate: }\;\frac{dx}{dt} \;=\;2R\cdot\frac{1}{1 + (1-\frac{v}{R}t)^2}\cdot\left(-\frac{v}{R}\right)\)

. . . . . . . . . . \(\displaystyle \frac {dx}{dt}\;=\;\frac{-2v}{1+(1-\frac{v}{R}t)^2}\) .[2]


\(\displaystyle \text{When the man is 2/3 of the way to the center: }\;vt \:=\:\frac{2}{3}R \quad\Rightarrow\quad t \:=\:\frac{2R}{3v}\)


\(\displaystyle \text{Substitute into [2]: }\;\frac{dx}{dt} \;=\;\frac{-2v}{1 + \left(1 - \frac{v}{R}\cdot\frac{2R}{3v}\right)^2} \;=\;\frac{-2v}{1 + (1 - \frac{2}{3})^2} \;=\;\frac{-2v}{\frac{10}{9}}\;=\;-\frac{9}{5}v\)


\(\displaystyle \text{The man's shadow is moving along the circle at }\tfrac{9}{5}\text{ his walking speed.}\)
. . \(\displaystyle \text{(And, of course, the arc length is decreasing.})\)

 

[galactus]

Alright, Soroban! (Insert 'thumbs up' emoticon here).

I like related rates problems, and I was glad to see something different besides the same ol', same ol', worn out problems.