The lifeguard problem: find the direction of sprint.

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SUMMARY

The lifeguard problem involves optimizing the path of a lifeguard with soil and water velocities to rescue a person located at coordinates (x,y) from the origin, with the beach ending at position l. The solution requires applying Snell's Law to determine the optimal angle of approach, denoted as point c on the waterline. The calculations for time taken in both mediums yield complex expressions, indicating that a simplified direct expression for c in terms of other distances is unlikely. The discussion highlights the challenges of differentiation in this context, particularly in deriving the time equations T1 and T2.

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  • Understanding of Snell's Law in physics
  • Familiarity with calculus, specifically differentiation
  • Knowledge of basic kinematics involving velocities in different mediums
  • Ability to solve equations involving multiple variables
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  • Learn advanced differentiation techniques for complex expressions
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This discussion is beneficial for physics students, mathematicians, and anyone involved in optimization problems related to motion in different mediums, particularly in rescue operations or similar scenarios.

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Homework Statement



This i the usual lifegard problem: we have a lifeguar that has a soil and water velocity, and wants to rescue a person at position (x,y). The lifeguard is ate the origin and th beach ends at position l, he wants to reach destination aa fast as possible.

We all know that this is accomplished by respecting the snell-descart law but what matters to the lifeguard is the angle he is going to run or the point in the waterline he is aiming, call it c.

2013-03-25 15.47.03.jpg



Homework Equations



Snell-Descarte law

The Attempt at a Solution



I made the calculations but the expressian became really ugly, can this expression simplify?

2013-03-25 15.46.46.jpg

 
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yeah, to be honest, I think a direct expression of C in terms of the other distances is going to be ugly, I don't think it will simplify into a nice expression.

I am trying to check your answer. So is vp the velocity of the man on ground? And vm the velocity of the man in water? I tried to work back from your answer, and I get to
[tex]\frac{x-c}{y-l} = \frac{y-l}{x-l} \frac{l}{c}[/tex]
This doesn't make sense, so maybe it is worth checking through your working again?
 
Just a lot of yucky differentiation, right?

T1 = √[l2 + c2]1/2/v1
T2 = √[(y-l)2 + (x-c)2]/v2
d(T1+T2)/dc = 0 etc.
Yuck!
 

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