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Optimization problem with a round lake

by frosty8688
Tags: lake, optimization
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frosty8688
#1
Nov28-12, 12:25 AM
P: 126
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.
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#2
Nov28-12, 12:37 AM
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Quote Quote by frosty8688 View Post
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.
Set up a formula for the total time if she rows from A to B and walks from B to C; this will be an expression involving some description of the point B, such as its (x,y) coordinates, or its angle from the line AC, or something similar. (Often in such problems, one way is easier than another; the way to figure that out is to try several approaches. And, yes, that might require some false starts.)
frosty8688
#3
Nov28-12, 12:43 AM
P: 126
Don't I need to know the length of the arc subtended by the angle?

frosty8688
#4
Nov28-12, 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{2-x}{2}[/itex]
Simon Bridge
#5
Nov28-12, 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: shore-to-shore.
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 arc-length s and row some linear distance d, then the time to complete the journey is $$T=\frac{s}{v}+\frac{d}{u}$$
frosty8688
#6
Nov28-12, 02:07 AM
P: 126
Point A is on the shore of the lake as well as points B and C. C is on the opposite side of A.
frosty8688
#7
Nov28-12, 02:15 AM
P: 126
Quote Quote by Simon Bridge View Post
It is written to sound like |AC|=2R - i.e. it's across the width of the lake: shore-to-shore.
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 arc-length s and row some linear distance d, then the time to complete the journey is $$T=\frac{s}{v}+\frac{d}{u}$$
x is some distance.
frosty8688
#8
Nov28-12, 02:17 AM
P: 126
One could row from A to B and walk from B to C.
frosty8688
#9
Nov28-12, 02:51 AM
P: 126
I found x to be [itex]\sqrt{\frac{2}{3}}[/itex] and T to be 1.11 hours.
Simon Bridge
#10
Nov28-12, 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 :)
frosty8688
#11
Nov28-12, 07:49 PM
P: 126
Quote Quote by Simon Bridge View Post
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 :)
Here is how I solved it [itex]T = \frac{2-x}{4} + \frac{\sqrt{2+x^{2}}}{2} = \frac{1}{2} - \frac{1}{4}x + \frac{1}{2}(2+x^{2})^{\frac{1}{2}}; T' = \frac{1}{2}\frac{1}{2}(2+x^{2})^{-\frac{1}{2}}(2x) - \frac{1}{4} = \frac{x}{2\sqrt{2+x^{2}}}-\frac{1}{4} = 0, \frac{x}{2\sqrt{2+x^{2}}}=\frac{1}{4}, 2x = \sqrt{2+x^{2}}, 4x^{2} = 2+x^{2}, 3x^{2} = 2, x^{2}=\frac{2}{3}[/itex]
frosty8688
#12
Nov28-12, 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.
frosty8688
#13
Nov28-12, 08:16 PM
P: 126
It looks like in the picture that the arc makes up a third of the semicircle.
frosty8688
#14
Nov28-12, 08:26 PM
P: 126
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.
frosty8688
#15
Nov28-12, 08:31 PM
P: 126
It would take her [itex]1\frac{2}{3}[/itex] hours to make it to the other end of the lake.
Simon Bridge
#16
Nov28-12, 08:51 PM
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Quote Quote by frosty8688 View Post
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.
Hmm ... that means that she has travelled a total of ##2\frac{2}{3}+1\frac{1}{3}=4\text{miles}## ... which is the diameter of the circle. But that would only happen by rowing the whole way (since rowing is the only direct way across the lake)! Does this make sense? Can she walk on water?

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+(y-R)^2}## and the equation of the lake-shore 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?
$$d=\sqrt{2R(r-y)};\; s=\cdots$$... 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?
frosty8688
#17
Nov28-12, 08:51 PM
P: 126
I found the length of the chord to be [itex]2\sqrt{3}[/itex]
frosty8688
#18
Nov28-12, 08:54 PM
P: 126
Quote Quote by Simon Bridge View Post
Hmm ... that means that she has travelled a total of ##2\frac{2}{3}+1\frac{1}{3}=4\text{miles}## ... which is the diameter of the circle. But that would only happen by rowing the whole way (since rowing is the only direct way across the lake)! Does this make sense? Can she walk on water?

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) is guessed to be in the first quadrant ... then
the distance rowed is ##d=|AB| = \sqrt{x^2+(y-R)^2}## and the equation of the lake-shore 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.

None of these equations look anything like yours ... so x does not appear to be a rectangular coordinate. So what is it?
I see your point. I was just trying to do it a different way.


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