help please!

[Guest]

Suzie walks at 4 mph and rows at 2 mph. She wishes to get from a point on the edge of a circular pond to a point directly opposite. She can row to any point and walk along the edge the rest of the way. Find the fastest route/time.

 

[soroban]

Hello, jackiemofo!

Suzie walks at 4 mph and rows at 2 mph.
She wishes to get from a point on the edge of a circular pond to a point directly opposite.
She can row to any point and walk along the edge the rest of the way.
Find the fastest route/time.

              * * *     P
          *           *
        *     x   * @/   *
       *      *     /R     *
          * @      / 2@
    A * - - - - - * - - - - * B
           R      O     R

The circular pond has center \(\displaystyle O\) and radius \(\displaystyle R.\)

She rows from \(\displaystyle A\) to \(\displaystyle P\) at 2 mph, and walk from \(\displaystyle P\) to \(\displaystyle B\) at 4 mph.
. . Let \(\displaystyle x = AP.\)

Let \(\displaystyle \angle PAO = \angle APO = \theta\)
. . Then: .\(\displaystyle \angle POB = 2\theta\)

Using trig on \(\displaystyle \Delta AOB\), we have: .\(\displaystyle x \:=\:2R\cos\theta\,\) miles.

\(\displaystyle \text{Rowing that far at 2 mph, it takes her: }\;T_1 \:=\:\frac{2R\cos\theta}{2} \:=\:R\cos\theta\,\text{ hours.}\)


\(\displaystyle \text{The length of arc }PB \text{ is: }\;R(2\theta) \,=\,2R\theta\,\text{ miles.}\)

\(\displaystyle \text{Walking that far at 4 mph, it takes her: }\;T_2 \;=\;\frac{2R\theta}{4} \:=\:\frac{1}{2}R\theta\,\text{ hours.}\)


\(\displaystyle \text{Her total time is: }\;T \;=\;r\cos\theta + \frac{1}{2}R\theta\)


And that is the function we must minimize.

 

[BigGlenntheHeavy]

\(\displaystyle Take \ any \ circular \ lake.\)

\(\displaystyle For \ sake \ of \ brevity, \ assume\ lake \ has \ a \ radius \ of \ 1 \ mile.\)

\(\displaystyle Then \ diameter \ = \ 2 \ miles \ and \ half \ the \ circumference \ = \ \pi \ miles.\)

\(\displaystyle Now, \ if \ you \ can \ row \ 2 \ mph, \ it \ will \ take \ you \ 1 \ hour \ to \ row \ straight \ accross \ the \ lake,\)

\(\displaystyle but \ if \ you \ can \ walk \ 4 \ mph. \ it \ will \ take \ you \ only \ \frac{\pi}{4} \ = \ .785 \ hours \ to \ walk\)

\(\displaystyle half \ the \ circumference \ of \ the \ lake.\)

\(\displaystyle Hence, \ the \ shortest \ time \ to \ get \ to \ the \ other \ side \ of \ the \ lake \ is \ \frac{\pi r}{4} \ hours \ and\)

\(\displaystyle the \ fastest \ route \ is \ no \ rowing \ and \ just \ walking.\)

 

[BigGlenntheHeavy]

\(\displaystyle Note: \ Observe \ that \ my \ answer \ jels \ with \ soroban's \ equation, \ to \ wit:\)

\(\displaystyle T(\theta) \ = \ rcos(\theta)+\frac{r\theta}{2}, \ \ 0 \ \le \ \theta \ \le \ \frac{\pi}{2}\)

\(\displaystyle T'(\theta) = \ -rsin(\theta)+\frac{r}{2}, \ \implies \ \theta \ = \ \frac{\pi}{6}\)

\(\displaystyle Ergo, \ T(0) \ = \ r, \ T\bigg(\frac{\pi}{6}\bigg) \ = \ \bigg(\frac{6\sqrt3+\pi}{12}\bigg)r \ = \ 1.1278r, \ and \ T\bigg(\frac{\pi}{2}\bigg) \ = \ \frac{r\pi}{4}\)

\(\displaystyle Hence, \ min. \ amt. \ of \ time \ occurs \ when \ T(\theta) \ = \ \frac{r\pi}{4} \ (Endpoint, \ walking, \ no \ rowing)\)

 

[galactus]

Here is a way I done it. It is more complicated then Glenn and Sorobans method, but works nonetheless.

Just thought I would share.

Per the diagram, the time rowed can be found with the law of cosines.

\(\displaystyle \frac{\sqrt{2r^{2}(1-cos{\theta})}}{2}\)

Then, the time walked is \(\displaystyle \frac{r({\pi}-{\theta})}{4}\)

The total time is \(\displaystyle \frac{\sqrt{2r^{2}(1-cos{\theta})}}{2}+\frac{r({\pi}-{\theta})}{4}\)

Differentiating, we get \(\displaystyle \sqrt{2}\cdot r\cdot sin{\theta}-r\sqrt{1-cos{\theta}}=0\)

This is what is to be optimized in terms of theta. Getting \(\displaystyle {\theta}=\frac{2\pi}{3}\)

Which jives with the two previous solutions. I used a different theta angle. Also, @ is theta in the diagram.