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Subject: Math/ Trig problem- an interesting one rss

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Personally, I've given up on The Riddler's math and probability questions because I'm too thick to understand the solutions let alone solve them. shake
I still check the column on Fridays to see if I can manage to solve The Riddle Express puzzle, though.
Today's was an interesting one, I thought.

You’re a lifeguard standing on the beach, right at the edge of the water, and gazing out over the ocean.
You see someone drowning 100 meters to the right of you and 100 meters away from shore.
You can run 100 meters in 15 seconds and swim 100 meters in 75 seconds. (The beach drops off steeply, meaning that you can’t run in the water.)

What’s the fastest you can get to the victim?


The solution is some combination of run and swim, no doubt; I don't think that jumping in and swimming right from where you stand is going to be better than running to your right for a while before you begin the swim.

The original link is here; you can even submit an answer for a virtual prize and some recognition at that site.
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David Jones
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a) I don't think trig is required to solve this problem. (minor spoiler)
Spoiler (click to reveal)
You only need the Pythagorean Theorem.

b) My first instinct was to say that calculus is required, but if you remember how parabolas work, you can actually solve the problem with basic algebra.
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The amazing thing about this problem is that it is analogous to the refraction of light passing through different media.
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MABBY wrote:
Personally, I've given up on The Riddler's math and probability questions because I'm too thick to understand the solutions let alone solve them. shake
I still check the column on Fridays to see if I can manage to solve The Riddle Express puzzle, though.
Today's was an interesting one, I thought.

You’re a lifeguard standing on the beach, right at the edge of the water, and gazing out over the ocean.
You see someone drowning 100 meters to the right of you and 100 meters away from shore.
You can run 100 meters in 15 seconds and swim 100 meters in 75 seconds. (The beach drops off steeply, meaning that you can’t run in the water.)

What’s the fastest you can get to the victim?


The solution is some combination of run and swim, no doubt; I don't think that jumping in and swimming right from where you stand is going to be better than running to your right for a while before you begin the swim.

The original link is here; you can even submit an answer for a virtual prize and some recognition at that site.


This is an Algebra 1 problem. (source: I write math textbooks for middleschool and highschool). It's an algebra 1 problem that requires many steps, but it is an algebra 1 problem. I'd have to double check, but it might even be an 8th grade math problem since it doesn't require function use or quadratics. edit-> (I overstated here: the initial framing of the question requires no functions. obviously you'll need a function to look at the efficiencies of line AB as you run along it.)

Spoiler (click to reveal)

Let's start by sketching out what we know.

We know the victim is 100 meters to the right and 100 meters out to sea. So, we need a sketch of that. I'm going to take the time to also draw a triangle on the diagram, which will keep track of the distances between me and the victim. We'll label the three vertices of the triangle A, B, and C just for ease of referring to them.



We know we can swim 100 meters/75 seconds
We know we can run 100 meters/15 seconds

We want to find out the fastest way to get to the victim. I'm going to make some observations here:

* The shortest distance between two points is a straight line.
* But, we are not interested in distance: we are interested in time.
* So really, we want to compare the time it takes to swim across the line AC with the time it takes to run along AB, then swim along BC.

Since the solution we want is in terms of time, but our diagram is in terms of distance, the diagram is not that useful right now. Let's re-write the sides of our triangle so they record time instead of distance.

The line AB is 100 meters on land: that will translate to 15 seconds.
The line BC is 100 meters on water: that will translate to 75 seconds.
The line AC is a little more complicated. We know from the Pythagorean theorem that a right triangle's longest side can be calculated using a^2+b^2=c^2. So the distance of AC is...

100^2+100^2=c^2
20,000 = c^2
c =~ 141.42 meters

So we know the distance of AC (141.42 meters). To calculate the time it would take our guy to cross that distance, we use ratios: (this method is taught in middleschool)

(100 meters/75 seconds) = (141.42 meters)/(AC seconds)
100m AC = 141.42m * 75s (cross multiply)
AC = 1.4142 * 75s (divide both sides by 100m)
AC = 106.065 seconds. (simplify the right side of the equation)

OK, so now we know our triangle sides in terms of seconds it takes to traverse them:

AC = 106.065 seconds.
AB = 15 seconds
BC = 75 seconds

This would mean the shortest time between AC and AB+BC is 75+15 = 90 seconds, which means you should run along the beach and then swim straight to the victim.

Followup questions:

* What if you started swimming but didn't follow AC? Why can we discount all such answers?
* What if you ran along shore, but started swimming before reaching point B? How would you set up the calculations for such a problem? (hint: this is the nuts and bolts of the true solution, since you'll have to set up a function)
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The last answer is however wrong, because it doesn't consider running a distance 100-x along the shore then swimming in the water sqrt(100+x^2). It assumes the ideal is x=0, but actually it isn't.

Before I go off and solve that, this is actually directly related to what happens when light is refracted. That's usually presented in elementary work as Snell's law, about the ratio of sines. But it turns out that minimizes time (hence you could solve this problem using Snell's law, here the grazing case). And that turn out to have an explanation in quantum theory. Feynman's short volume QED is about that.

Now to actually solve for x I need paper and a pen, trying to do it on this phone (especially with getting painfully cold fingers) is not going to work. If anyone wants to beat me to it, please do.
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Spoiler (click to reveal)
We need to find the point at which running along the beach moves the same speed towards the person that swimming directly towards them does.

If the lifeguard runs (1-x) meters down the beach, then he swims the hypotenuse of a triangle with sides x and 100.

At any point along the beach path there is an angle between the beach path and the straight line towards the swimmer; call it theta. Then as the lifeguard runs along the beach, he is moving towards the swimmer at the rate (100/15)cos(theta) m/sec, while if he swims directly at the swimmer, he moves at (100/75) m/sec.

So we have: (100/15)cos(theta) = 100/75, or cos(theta) = 1/5.

cos(theta) = x/sqrt(x^2+100^2), so we get 5x = sqrt(x^2+100^2), or, squaring both sides and collecting like terms: 24x^2 = 100^2, which gives us x = 50/sqrt(6) ~ 20.41.

The lifeguard runs almost 80 meters down the beach, then swims directly towards the drowning swimmer, which has him running for a little less than 12 seconds and swimming for a little more than 76.5 seconds, for a total of ~88.48 seconds.
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I believe it does take calculus, you have to establish a minimum from a quadratic equation. To get a minimum (or maximum) you differentiate or integrate and set resulting equation = to 0 and solve.

If my calculus wasn't so fuzzy I'd have a number solution. But ...

Spoiler (click to reveal)
you know that what is necessary is time, Speed of running = 100M/15Sec = 6.666 m/s, speed of swimming = 100m/75s=1.333m/s
So set the distance run = R, the distance swum = S
Time to complete = R/6.66m/s + S/1.333m/s = T

Distance swum (hypotenuse of triangle) = S = [(100^2 + (100-R)^2) ]^.5

Substitue this formula for S into T equation
T = R/6.66 + {[(100^2+(100-R)^2)]^.5}/1.33
To find a minimum result you must take the Integral (or derivative, I can't remember which) and set it equal to 0 (zero) then solve for R.

I tried looking through my CRC handbook for the integral of this quadratic but can't find it. Maybe someone with a fancy calcualtor could crank out the number.

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Verdigris97 wrote:
Spoiler (click to reveal)
We need to find the point at which running along the beach moves the same speed towards the person that swimming directly towards them does.

If the lifeguard runs (1-x) meters down the beach, then he swims the hypotenuse of a triangle with sides x and 100.

At any point along the beach path there is an angle between the beach path and the straight line towards the swimmer; call it theta. Then as the lifeguard runs along the beach, he is moving towards the swimmer at the rate (100/15)cos(theta) m/sec, while if he swims directly at the swimmer, he moves at (100/75) m/sec.

So we have: (100/15)cos(theta) = 100/75, or cos(theta) = 1/5.

cos(theta) = x/sqrt(x^2+100^2), so we get 5x = sqrt(x^2+100^2), or, squaring both sides and collecting like terms: 24x^2 = 100^2, which gives us x = 50/sqrt(6) ~ 20.41.

The lifeguard runs almost 80 meters down the beach, then swims directly towards the drowning swimmer, which has him running for a little less than 12 seconds and swimming for a little more than 76.5 seconds, for a total of ~88.48 seconds.

I like this solution, but i think you have an error early on, Lifeguard runs (1-x)?, isn't it he runs X so the distance remaining is 100-X (side of triangle)? Also think my solution is possible.
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klbush wrote:
I believe it does take calculus, you have to establish a minimum from a quadratic equation.


Except we already know that the min/max of a quadratic occurs at the vertex of the parabola, which is x=-b/2a. No calculus required. Otherwise I agree with your solution.

(Also, you use derivatives to find minima, not integrals.)

Edit: I thought I had a way to get the square root out of the equation, but maybe not. I shouldn't be trying to do this in the middle of work
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klbush wrote:
I believe it does take calculus, you have to establish a minimum from a quadratic equation. To get a minimum (or maximum) you differentiate or integrate and set resulting equation = to 0 and solve.

If my calculus wasn't so fuzzy I'd have a number solution. But ...

Spoiler (click to reveal)
you know that what is necessary is time, Speed of running = 100M/15Sec = 6.666 m/s, speed of swimming = 100m/75s=1.333m/s
So set the distance run = R, the distance swum = S
Time to complete = R/6.66m/s + S/1.333m/s = T

Distance swum (hypotenuse of triangle) = S = [(100^2 + (100-R)^2) ]^.5

Substitue this formula for S into T equation
T = R/6.66 + {[(100^2+(100-R)^2)]^.5}/1.33
To find a minimum result you must take the Integral (or derivative, I can't remember which) and set it equal to 0 (zero) then solve for R.

I tried looking through my CRC handbook for the integral of this quadratic but can't find it. Maybe someone with a fancy calcualtor could crank out the number.



Solution:

http://www.wolframalpha.com/input/?i=d%2Fdx(15%2F100*(x)%2B7...
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I wonder if this is an application for The Brachistochrone.

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The Unbeliever wrote:
klbush wrote:
I believe it does take calculus, you have to establish a minimum from a quadratic equation. To get a minimum (or maximum) you differentiate or integrate and set resulting equation = to 0 and solve.

If my calculus wasn't so fuzzy I'd have a number solution. But ...

Spoiler (click to reveal)
you know that what is necessary is time, Speed of running = 100M/15Sec = 6.666 m/s, speed of swimming = 100m/75s=1.333m/s
So set the distance run = R, the distance swum = S
Time to complete = R/6.66m/s + S/1.333m/s = T

Distance swum (hypotenuse of triangle) = S = [(100^2 + (100-R)^2) ]^.5

Substitue this formula for S into T equation
T = R/6.66 + {[(100^2+(100-R)^2)]^.5}/1.33
To find a minimum result you must take the Integral (or derivative, I can't remember which) and set it equal to 0 (zero) then solve for R.

I tried looking through my CRC handbook for the integral of this quadratic but can't find it. Maybe someone with a fancy calcualtor could crank out the number.



Solution:

http://www.wolframalpha.com/input/?i=d%2Fdx(15%2F100*(x)%2B7...

That's pretty cool, thanks
So distance run = 79.58m, time to rescue = ~88.4 sec
This seems to pass the smell test, 90 seconds is maximum run distance + minimum swim distance time.
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galad2003 wrote:
All I know is that by the time you figure out the math the person will have drowned.


Not if you have your dog figure it out.
https://www.sciencenews.org/article/calculating-dogs
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Just don't ask a Scottish teenager.
http://www.iflscience.com/editors-blog/math-question-baffled...
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MABBY wrote:


What’s the fastest you can get to the victim?



299,792,458 m/s
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clever answer, but you can't travel that fast. the fastest speed for a human so far is 2,020.616 m/s, and that is in space, on earth it is 339.750 m/s
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klbush wrote:

I like this solution, but i think you have an error early on, Lifeguard runs (1-x)?, isn't it he runs X so the distance remaining is 100-X (side of triangle)? Also think my solution is possible.


Nope. I said in my second sentence that he runs (1-x) so that the triangle would have a side of x. That makes the algebra easier, but still gets the right answer.
 
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I can still remember when the internets was fun. gulp
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Orangemoose wrote:
I can still remember when the internets was fun. gulp


I can still remember when I was posting to sci.math before web existed. Perhaps you've forgotten what the internet was originally for.
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Lol math, the thing my brain completely dumped once I finished school.
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The same problem is presented in this episode of School of Hard Sums.
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My knowlegde of mathematics is lacking, so I am very curious to see how the really clever guys did it.

However to come up with something I have decided to turn to religion. A religion that not only takes, but one that gives. And my god is Bill Gates and his gift to mankind is MS-Excel.

So I designed a clever Excelsheet in an hour - which was well spent time - and wrote some VBA-code. I determined distance and time in distance steps of 10. So I start with 0, 10, .... 90 and 100 meter. Then I determine the two shortest times and drill deeper to 70, 71 ... 79 and 80 meters. And so on behind the decimals. However I quickly got stuck due to Excel-limitations. No god is perfect and I thought this is enough.

This are my results:
Spoiler (click to reveal)
Distance: 79.58758 meters
Time: 88.4846922834953 seconds

I hope this is accurate enough.
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anemaat wrote:
My knowlegde of mathematics is lacking, so I am very curious to see how the really clever guys did it.

However to come up with something I have decided to turn to religion. A religion that not only takes, but one that gives. And my god is Bill Gates and his gift to mankind is MS-Excel.

So I designed a clever Excelsheet in an hour - which was well spent time - and wrote some VBA-code. I determined distance and time in distance steps of 10. So I start with 0, 10, .... 90 and 100 meter. Then I determine the two shortest times and drill deeper to 70, 71 ... 79 and 80 meters. And so on behind the decimals. However I quickly got stuck due to Excel-limitations. No god is perfect and I thought this is enough.

This are my results:
Spoiler (click to reveal)
Distance: 79.58758 meters
Time: 88.4846922834953 seconds

I hope this is accurate enough.


Ah, the old Trial & error method that my algebra teacher was determined to stop me from using. Quite accurate, at least for a life guard.
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titanticore wrote:
MABBY wrote:


What’s the fastest you can get to the victim?



299,792,458 m/s

Oh, I c.
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