
#1
Nov2812, 12:25 AM

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1. A person from point A wants to get to point C diammetrically across a round lake. This person is on the shore and can walk at a rate of 4 mi/hr and row at a rate of 2 mi/hr. Which method should she use?
2. radius = 2 mi, triangle with angle θ has the points ABC 3. I started out by calculating the area of the semi circle which is about 6.6 mi. I found that she can walk there in less time than it takes to row. How do I prove this using differentiation. 



#2
Nov2812, 12:37 AM

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#3
Nov2812, 12:43 AM

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Don't I need to know the length of the arc subtended by the angle?




#4
Nov2812, 12:52 AM

P: 126

Optimization problem with a round lake
One equation for walking would be [itex]\frac{\sqrt{2+x^{2}}}{4}[/itex] and the equation for rowing would be [itex]\frac{2x}{2}[/itex]




#5
Nov2812, 01:44 AM

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It is written to sound like AC=2R  i.e. it's across the width of the lake: shoretoshore.
But the question makes more sense if point C or point A is on the lake itself ... I'm betting point A since the alternative would involve dragging a boat at 4mi/hr. And yes  you'll want to know the length of arc subtended by some angle. Do the algebra first and then put the numbers in ... let the walking speed be v and the rowing speed u, derive the formula. What does "x" stand for in what you have written? Note: if you walk some arclength s and row some linear distance d, then the time to complete the journey is $$T=\frac{s}{v}+\frac{d}{u}$$ 



#6
Nov2812, 02:07 AM

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Point A is on the shore of the lake as well as points B and C. C is on the opposite side of A.




#7
Nov2812, 02:15 AM

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#8
Nov2812, 02:17 AM

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One could row from A to B and walk from B to C.




#9
Nov2812, 02:51 AM

P: 126

I found x to be [itex]\sqrt{\frac{2}{3}}[/itex] and T to be 1.11 hours.




#10
Nov2812, 03:19 AM

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Can x be any distance? A distance around the shore? Or the distance betwen point A and point B in some coordinate system? You calculations are meaningless without context
Like when I described distances I said what they were for. So in my description you'd row from A to B and d=AB, then walk around to C  which is a distance s. ... distance should at least have units. Technically  you only save time if the arclength AB is more than twice the chord (since you row the chord at half speed). But you have to use calculus :) 



#11
Nov2812, 07:49 PM

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#12
Nov2812, 07:50 PM

P: 126

The answer could be wrong, since I forgot to take in account the length of the arc. I did this as if it were a right triangle.




#13
Nov2812, 08:16 PM

P: 126

It looks like in the picture that the arc makes up a third of the semicircle.




#14
Nov2812, 08:26 PM

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She would be rowing for [itex]\frac{2}{3}[/itex] of the way or [itex] 2\frac{2}{3}[/itex] miles and would walk for [itex]\frac{1}{3}[/itex] of the way or [itex]1\frac{1}{3}[/itex] miles.




#15
Nov2812, 08:31 PM

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It would take her [itex]1\frac{2}{3}[/itex] hours to make it to the other end of the lake.




#16
Nov2812, 08:51 PM

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I cannot tell where you made the mistake because you have not answered my question: what distance does x represent? For example  if we make the center of the lake the origin O of a Cartesian coordinate system, so that A=(0,R), C=(0,R), and B=(x,y) ... then the distance rowed is ##d=AB = \sqrt{x^2+(yR)^2}## and the equation of the lakeshore is ##x^2+y^2=R^2##. If ##\angle AOB = \theta## then the distance walked is given by ##s=(\pi\theta)R## I'm not sure how you can get the arclength s easily in terms of rectangular coordinates, but you can get the rowing distance d in terms of ##\theta## using the cosine rule. Lessee: looks like it is easier to use y than x  but that's just a question of labels right? $$d=\sqrt{2R(ry)};\; s=\cdots$$... something ... to do dT/dy=0 I'd have to look up the derivative of an arctangent or an arccosine or something like that... None of these equations look anything like yours ... so x does not appear to be a rectangular coordinate. So what is it? 



#17
Nov2812, 08:51 PM

P: 126

I found the length of the chord to be [itex]2\sqrt{3}[/itex]




#18
Nov2812, 08:54 PM

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