A bad date

Calculus Level 5

Here is something a little more light-hearted than my usual flux problems, in the spirit of the holidays.

Romeo and Juliet have a date at a lake whose shore line is given by $x^2+y^2=1$ , with distances measured in kilometers. Due to an unfortunate misunderstanding, they find themselves at the points $(1,0)$ and $(-1,0)$ , respectively, as they notice their error. Naturally, they wish to meet up as soon as possible, by all means necessary. They will each run along the lake for a while, northwards, towards the point $(0,1)$ , then jump into the water and swim towards each other (they have the option to swim all the way to the center of the lake or run all the way to the point $(0,1)$ , of course). They can both run at a speed of 10 km/h and swim at 5 km/h. How many meters should each of them run? Round your answer to the nearest integer.

None of the others 1047 0 524 785 1571

2 solutions

Otto Bretscher
Dec 21, 2018

Comrade Huan has submitted a fine mathematical solution. The point of this problem, besides its (small) entertainment value, is that a critical point does not always produce the desired extremum; one needs to "check."

One can also reason this problem out as follows:

Since the two lovers swim half as fast as they run, swimming will save them time only if they can cut the distance in less than half. But any arc to a semicircle is less than twice as long as the corresponding chord, so that swimming would be a waste of time. The lovers are better off running all the way to $(0,1)$ , travelling $\frac{\pi}{2}$ km or about $\boxed{1571}$ meters.

Do you know what the smallest possible value of $b$ is such that the elliptic lake described by $x^2 + \frac {y^2}{b^2} = 1$ requires some swimming? You need elliptic integrals to solve this and I'm not familiar with them.

Henry U - 2 years, 5 months ago

Gute Frage! My intuition tells me that this would happen when the circumference of the ellipse exceeds 8, in which case they would be well advised to swim all the way to the center of the lake. My trusted Swiss Logarithmentafel tells me that this happens just above $b=1.5196$ .

Otto Bretscher - 2 years, 5 months ago

And for all $b < 1.5196$ , running all the way is the best strategie, right?

Henry U - 2 years, 5 months ago

@Henry U – I think a hyperbolic ( $xy = 1$ ) lake could make this more interesting. I get that the optimal place to start swimming is for $0.9 \leq x \leq 0.905$ . (Romeo and Juliet could start anywhere on the graph where $|x| > 0.905$ )

Henry U - 2 years, 5 months ago

@Henry U – Yes, the case of a hyperbola is more interesting. I'm finding that they should swim at $x=\pm \frac{1}{\sqrt[4]{3}}$ , but I'm computing this somewhat hastily, "on the go." I will double-check. It would make a good problem for an introductory calculus class.

Otto Bretscher - 2 years, 5 months ago

Huan Bui
Dec 20, 2018

Fortunately, Romeo and Juliet's reunion might only involve sweat and tears, but not the water from the lake!

Assume, by symmetry, that the distance each person has to travel has both the running and swimming component, so $L(\phi) = \phi + \cos\phi$ , with $0 \leq \phi \leq \pi/2$ . This gives the time $t(\phi) = \frac{\phi}{10} + \frac{\cos\phi}{5}$ . We notice that $t(\phi)$ attains a minimum at $\phi = \frac{\pi}{2}$ , for $0 \leq \phi \leq \pi/2$ , in which case the distance each has to run, in meters, is $L = 1000\frac{\pi}{2} \approx \boxed{1571}$ .

After all that sweat and tears, a swim might have been refreshing... but time is of the essence. Thank you, Comrade!

Otto Bretscher - 2 years, 5 months ago

The second derivative of total time with respect to the angle is always negative within the range of the angle. Therefore the time doesn't possess a local minimum. However, it exhibits a minimum at an angle of π/2.

A Former Brilliant Member - 2 years, 4 months ago

A bad date

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