# Surf Rescue - word problem - pythagoras

Angela works at the local beach as a part of a surf patrol unit. During her patrol, she notices that an old lady is having difficulty in the surf. She estimates that the distance is approxiamtely 220 m along the beach and then 75 m out (in water) to the woman. Angela can run along the beach at a speed of 180 metres per minute and can swim at a speed of 80 metres per minute.

1. How long will it take her to reach the woman if she run the full 220 m on the beach and swim 75 m?

2. In hindsight, Angela thinks that it may be quicker to swim directly out to the woman in trouble. How long would it have taken her if she had swum the entire distance?

3. Find the quickest time (down to the nearest metre) for Angela to reach the woman.

Could you please show all working out and explain it? Thank you.

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"Show all working out and explain it" --- so you can just copy it and put your name on it and hand it in? Better idea: you show us what you've been able to do with the problem and where you get stuck, and we'll help you along and give you hints. –  Gerry Myerson Jul 30 '13 at 8:51
I used to give this very same problem to my Calculus student some 26-28 years ago! A classic :) –  Andrea Mori Jul 30 '13 at 11:41
Still do, but to supply power from a substation to an island, with different costs for land and sea cables... –  DJohnM Jul 30 '13 at 16:13

Since it’s homework, I prefer not to show everything, but I’ll do my best to get you pointed in the right direction. Begin by drawing a sketch. Here’s a really crude one that should be adequate:

                                                     L old lady
|
| 75 m
|
----------------------------------------------- shoreline
A<----------------- 220 m ------------>P
Angela

1. In this version Angela runs $220$ m from $A$ to $P$ and then swims $75$ m from $P$ to $L$. You know her speeds in metres per minute running and swimming, so just calculate separately how long it takes her to run $220$ m and to swim $75$ m and add the times.

2. For this version use the Pythagorean theorem $|AL|^2=|AP|^2+|PL|^2$ to calculate the distance $|AL|$, and then use Angela’s known swimming speed to calculate how long it would have taken her to swim directly from $A$ to $L$.

3. For this version imagine that at some point $X$ along the stretch of shoreline from $A$ to $P$ Angela stops running and swims directly from $X$ to $L$. Let $x$ be the distance in metres that she runs from $A$ to $X$; clearly $|XP|=220-x$. How long does it take her to run $x$ metres, expressed as a function of $x$? How long does it take her to swim from $X$ to $L$, expressed as a function of $x$? You’ll need to apply the Pythagorean theorem again, this time to the triangle $\triangle XPL$. Adding these two times will give you her total time as a function of $x$, say $t=f(x)$. Now use the standard calculus procedure to find where this function has a minimum. Start by finding $f\,'(x)$; what do you do next?

                                                 L old lady
|
| 75 m
|
------------------|----------------------------- shoreline
A<-----------X----- 220 m ------------>P
Angela

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This isn't my homework, it's my younger brother's. I posted it up here so i can get an answer and explain it to him. –  NomNom Jul 30 '13 at 9:28
The algebra is a bit simpler if you make the variable x represent the distance from X to P –  DJohnM Jul 30 '13 at 16:12
@Brian sir, if it interests you, when I read this question, I first ran left and then right, while you first ran right and then left. I find it very interesting that all of us think differently even in case of direction. I would be curious to know if there are people who first ran right and then again right. Or, if somebody first ran left and then again left. I wonder if it has anything to do with our overall psychology or how we approach real-life problems. (By run, I mean imagine Angela to run, but of course that is obvious.) I realise it is not a correct portal to discuss this,still posting –  Ramit Jul 31 '13 at 14:56
@Ramit: I actually have her running only to the right; the way I drew the diagram, she never runs to the left. If you’re curious, my original impulse was to draw the diagram the other way around, so that she ran to the left. I switched it in order to have her running in the direction of the positive $x$-axis. –  Brian M. Scott Jul 31 '13 at 21:06
Point P is to the right of point A. And L is in the left of P. (My mistake, I should have mentioned swim, rather than run for the 2nd part) Btw, it's interesting and sensible to draw it corresponding to +ve x and y axis. Thanks for sharing. –  Ramit Aug 1 '13 at 6:15