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.