Optimization problem with a round lake

[frosty8688]

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.

 

[Ray Vickson]

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]

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

 

[frosty8688]

One equation for walking would be $\frac{\sqrt{2+x^{2}}}{4}$ and the equation for rowing would be $\frac{2-x}{2}$

 

[Simon Bridge]

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]

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]

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]

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

 

[frosty8688]

I found x to be $\sqrt{\frac{2}{3}}$ and T to be 1.11 hours.

 

[Simon Bridge]

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]

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 $T = \frac{2-x}{4} + \frac{\sqrt{2+x^{2}}}{2} = \frac{1}{2} - \frac{1}{4}x + \frac{1}{2}(2+x^{2})^{\frac{1}{2}}; T' = \frac{1}{2}\frac{1}{2}(2+x^{2})^{-\frac{1}{2}}(2x) - \frac{1}{4} = \frac{x}{2\sqrt{2+x^{2}}}-\frac{1}{4} = 0, \frac{x}{2\sqrt{2+x^{2}}}=\frac{1}{4}, 2x = \sqrt{2+x^{2}}, 4x^{2} = 2+x^{2}, 3x^{2} = 2, x^{2}=\frac{2}{3}$

 

[frosty8688]

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]

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

 

[frosty8688]

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

 

[frosty8688]

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

 

[Simon Bridge]

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

Hmm ... that means that she has traveled 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]

I found the length of the chord to be $2\sqrt{3}$

 

[frosty8688]

Hmm ... that means that she has traveled 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.

 

[frosty8688]

The rowing distance is about 3.46.

 

[frosty8688]

The arc length is 2.09.

 

[frosty8688]

It would take her over two hours to row and walk. So she should row, which would take 2 hours.

 

[Simon Bridge]

I see your point.

Which one: I made a number of them. But I feel that if you really saw the point you would have answered the questions!

I was just trying to do it a different way.

Yes: clearly - and you still neglect to tell me which way that was!

The rowing distance is about 3.46. The arc length is 2.09.

How are you getting these values?

It would take her over two hours to row and walk. So she should row, which would take 2 hours.

Nonsense!

If you had set up your equation correctly, and rowing the whole way was the quickest route, then there would be a minima in your equation for that situation.
There wasn't - instead the minima for the equation gave a slower route that the slowest possible path (see below) - therefore, your equation is set up incorrectly.

If she rowed the whole way, then, yes: she would take $R/u=4/2$ or 2hr to get there. This is the slowest route of the ones available. For instance: if she walked the whole way, then she would take $\pi R/v = 2\pi/4 = \pi/2$ or 1.57 hr to get there ... so, of the two choices she should walk - surely?

However, there may still be some advantage in rowing some of the way.
Your calculation is simply wrong.

I am sorry - considering that you will not answer questions, will not follow advise, and will not show your reasoning, I cannot help you. Good luck.

 

[Ray Vickson]

Which one: I made a number of them. But I feel that if you really saw the point you would have answered the questions!Yes: clearly - and you still neglect to tell me which way that was!How are you getting these values?Nonsense!

If you had set up your equation correctly, and rowing the whole way was the quickest route, then there would be a minima in your equation for that situation.
There wasn't - instead the minima for the equation gave a slower route that the slowest possible path (see below) - therefore, your equation is set up incorrectly.

If she rowed the whole way, then, yes: she would take $R/u=4/2$ or 2hr to get there. This is the slowest route of the ones available. For instance: if she walked the whole way, then she would take $\pi R/v = 2\pi/4 = \pi/2$ or 1.57 hr to get there ... so, of the two choices she should walk - surely?

However, there may still be some advantage in rowing some of the way.
Your calculation is simply wrong.

I am sorry - considering that you will not answer questions, will not follow advise, and will not show your reasoning, I cannot help you. Good luck.

When I do it I find there is an even slower route (corresponding to a maximum time): row to point B, where the angle BOC is ___ radians, then walk the rest of the way, for a total time of 2.2556 hours; I am keeping the answer blank in order to leave something for the OP to do. Thus, rowing the whole way is not the slowest route. When done properly (in terms of the angle θ = BOC) the time function T(θ) has local minima at θ = 0 (row all the way) and θ = π (walk all the way) and a global maximum at an intermediate point.

RGV

 

[frosty8688]

I got the same answer as regards to rowing and walking times. I figured out that the chord AB is approximately 1 unit or 1 mile away from the point O and the radius is 2 miles. So the chord is 2$\sqrt{r^{2}-d^{2}}$ with d being the distance of the chord from point O. The answer for the chord length is 2$\sqrt{3}$ which equals about 3.46. I divided this by 2 and it came out to be 1.73 hours of rowing.

 

[frosty8688]

I then took 2 arcsin of (cL/2Ra). cL is chord length and Ra is radius and ended up with $\frac{2}{3}$∏. This is the arc length. I then divided this by 4 and it came out to be about 0.52.

 

[frosty8688]

I agree that walking is the fastest route.

 

[Simon Bridge]

I figured out that the chord AB is approximately 1 unit or 1 mile away from the point O

How? Presumably that is for the maxima?

You are right though - making x the distance from O to the center of the chord (if that is what you did) is an interesting way of doing it.
(I messed up my description before - thanks Ray.)