A question from life (about swimming)

[Duplicity]

A man running on a straight seaside, sees a person drowning on point A, 100 mts from the closest beach point C. At the time he sees, he is 1.000mts away from the beach point that is closest (point S)

His swimming speed is 1/x of his running speed where x>3

Find the angle (ABC) where the man needs to jump into water and swim in terms of x, in which it will be the fastest way to reach drowning person.

It is easy to approximate by using excel and trying numbers, but I could not create a limit function to demonstrate with unknown value x

For example if AC is 1000 mt, and x=5, then BC is approximately 204 meters. Thus you can calculate an angel from there.
But in formulisation I was stuck.
Thanks for any help

 

[Dr.Peterson]

Never, ever write "mts" for meters. It's 100 m.

Please show some attempt, so we can see how close you got. You aren't looking for a "limit function", but for an equation that expresses the time for the run as a function of the angle; then you'll be differentiating to find the minimum time. You could do this directly, or first work just in terms of distances (as you did in your example, perhaps using a variable for BC), and later find the angle.

 

[Duplicity]

Hi Dr. Peterson,


Thank you for the quick answer. Really appreciated. Meter as m. is well noted.



I could get to here by myself:



As in my example let length [AC] = 1000m

Let length [SC] = 1000

let length[BC] = y

Running speed = r

Swimming speed = r/x



Time for running is (1000-y)/r



Time for swimming = (SQRT (y^2+1000^2) ) / (r/x)



Total time is the sum:



ttotal = (1000-y)/r + (SQRT (y^2+1000^2) ) / (r/x)



By design we can agree that the distances [AC], [SC] and running speed will not change angel (ABC) for minimum time, considering swimming time is r/x. (Let me know if you disagree)

The only parameter that will change the angel will be x. Considering all parameters are positive numbers. (The drowning man is not on the beach and the swimming speed of rescuer is less than running speed)



Taking r = 1 simplifies to finding the min value for



(1000-y)+ (SQRT (y^2+1000^2) ) / (1/x)



Which is:



(1000-y)+ x (SQRT (y^2+1000^2) )



And it is equal to:



1000 - y + x (SQRT (y^2+1000^2) )



To find minimum value for this equation, we can eliminate the first 1000, as it will not effect for the minimum value ;



Ttotal = y + x (SQRT (y^2+1000^2) )



The angel is :



Arccos ( x/ Sqrt(x^2+1000^2)



We can consider that 1000 can be replaced with any positive value greater than zero.



I am stuck here.



Thank you for the help.

 

[Dr.Peterson]

When you wrote "Ttotal = y + x (SQRT (y^2+1000^2) )", you dropped a negative sign on the y. Until then you were fine.

But then you never differentiated to find the minimum time. You submitted this under Calculus, so I assume you know that's what you need to do. Keep in mind that your variable is y; x is constant. Don't let that confuse you. Differentiate with respect to y.

Don't worry about the angle until you've minimized. And I think the 1000 (for AC, which originally was 100) will drop out by itself at that point. Similarly, the 1000 you dropped from before y would drop out when you differentiate, so you didn't have to deliberately ignore it (but it didn't hurt).

 

[Duplicity]

When you wrote "Ttotal = y + x (SQRT (y^2+1000^2) )", you dropped a negative sign on the y. Until then you were fine.

But then you never differentiated to find the minimum time. You submitted this under Calculus, so I assume you know that's what you need to do. Keep in mind that your variable is y; x is constant. Don't let that confuse you. Differentiate with respect to y.

Don't worry about the angle until you've minimized. And I think the 1000 (for AC, which originally was 100) will drop out by itself at that point. Similarly, the 1000 you dropped from before y would drop out when you differentiate, so you didn't have to deliberately ignore it (but it didn't hurt).

Hi,
You are absolutely right. I dropped in a mistake the minus sign for y. (I did it rigt on my paper notebook, but now I am in vacation and I tried to put what I remember)
I am 45 yo and I have not done any differenciation since 1998 (the year I was graduated)

I am really interested about if the solution to this problem could be put in a way, such that Angle(ABC) = function (x).

Is there a function that you could put, and works with all values of x>1 ?

Many thanks for the replies

 

[Dr.Peterson]

I am really interested about if the solution to this problem could be put in a way, such that Angle(ABC) = function (x).

Is there a function that you could put, and works with all values of x>1 ?

Yes, indeed, there is such a function, and it's very simple.

I'll boost you over the hard part and see if you can finish (though part of me wants to insist you review differentiation).

The derivative of [MATH]1000 - y + x \sqrt{y^2+100^2}[/MATH] is [MATH]-1 + \frac{xy}{\sqrt{ y^2 + 100^2}}[/MATH]. Set that equal to 0 and solve for y; then use the fact that [MATH]\tan(\theta) = \frac{100}{y}[/MATH] to find the angle [MATH]\theta[/MATH].

 

[Duplicity]

Wow Dr. Peterson,

I am amazed. I am once again amazed with math.

Solving from there on I found the answer to be:

angel θ = \(\displaystyle tan^{-1}\left( \sqrt{x^2-1}\right)\)

Putting this to some examples I see it works. And it also automatically defines x>1.

Many thanks.

Your advise on looking into derivatives is also appreciated. I will do that. Brain needs exercise to remember.
Keep safe,

Best regards,

Alper

 

[Dr.Peterson]

Good job. I suspected I didn't need to give any more help than I did.

 

[Duplicity]

Good job. I suspected I didn't need to give any more help than I did.

Thanks. And I also understood why we need to take the derivative of the function as zero. It is a kind of parabolic graph and we are looking for the min point.

Thank you again