There is a lighthouse on an iscolated island, Imagine this Lighthouse is located on the origin, There is a drugboat caught in the light of this lighthouse house, that is being controlled by a person, this person in the lighthouse keeps readjusting the light so that he keep sight on this drugboat while the police helicopters catch him the boat driver thinks it is a good idea to drive a 45 degree from the lightpointed at him, to try and escape, what does the graph of the path of this boat driver look like, and find the function of the graph.

- dan815

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- dan815

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- dan815

the light will be corrected to keep shining on the boat at all times as he is trying to escape

- dan815

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## More answers

- Loser66

I know where do you have this question. MIT, right?

- Jhannybean

coast guards using LED lights? :)

- anonymous

So the light travels around the entire time?

- anonymous

Sounds like a polar equation to me.

- Jhannybean

Who tries escaping in that path? LOL

- anonymous

and that wouldn't be exactly a function.

- anonymous

reason why i said that is because the term "function" in the question creates ambiguity.

- anonymous

\[\bf r=\theta\]Try that. Looks good.

- anonymous

well that's the only way it's going to be done for the kind of graph ur asking.

- primeralph

I don't fully see the scope of the question but I can say the boatman is dumb.

- primeralph

The light can go from the tip of the beam to infinity so where exactly on that path is he to start with?

- anonymous

the light is rotating...

- Jhannybean

Then the coastguard is partially blind.

- primeralph

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- anonymous

trust me dan, r = theta is correct. It feels right in my heart.

- anonymous

it actually is right, probably lol.

- anonymous

u know the answer?

- anonymous

The question is ambiguous...

- dan815

noo dont tell em answer experiment lol

- anonymous

I suppose you could write a differential equation here since what you're describing are tangent lines

- experimentX

ok ok

- dan815

yeah thats the right step oldrin

- dan815

u know 1 thing, which is the slope between the boat movement and the light is always the same

- anonymous

right

- Jhannybean

NO DAN I WAS READING THAT! lol

- dan815

- dan815

i put it on the main part lol

- Jhannybean

This is like a movie scene, lmao..

- dan815

maybe i shud state the context where this is from,
this is about differential equations, u can write and expression for slope and you have to figure out what the function must be
y'=.....
solve for

- dan815

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- dan815

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- anonymous

@ganeshie8 close but it's an exponential spiral

- anonymous

right that's logical

- anonymous

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- anonymous

... so we get that the slope of our tangent vector is given by \(\tan(\phi+45^\circ)\), which we can rewrite:$$\frac{dy}{dx}=\tan(\phi+45^\circ)=\frac{\tan\phi+\tan45^\circ}{1-\tan\phi\tan45^\circ}=\frac{1+\tan\phi}{1-\tan\phi }$$

- anonymous

It's clear that \(\tan\phi=\dfrac{y}x\) so we get:$$\frac{dy}{dx}=\frac{1+y/x}{1-y/x}$$

- anonymous

Note this equation is of the form \(dy/dx=F(y/x)\) i.e. it is homogeneous and we may use a substitution of the form \(v=y/x\), yielding \(y=vx\). Observe \(dy/dx=xdv/dx+v\)

- anonymous

$$x\frac{dv}{dx}+v=\frac{1+v}{1-v}\\x\frac{dv}{dx}=\frac{1+v}{1-v}-v=\frac{1+v-v(1-v)}{1-v}=\frac{1+v^2}{1-v}\\\frac{1-v}{1+v^2}dv=\frac1xdx\\\int\left(\frac1{1+v^2}-\frac{v}{1+v^2}\right)dv=\int\frac1x dx\\\arctan v-\frac12\log(1+v^2)=\log x+C$$

- anonymous

$$\arctan v=\frac12\log x^2+\frac12\log(1+v^2)+C=\log\sqrt{x^2+x^2v^2}+C\\v=\frac{y}x\implies\arctan\frac{y}x=\log\sqrt{x^2+y^2}+C\\r=\sqrt{x^2+y^2},\theta=\arctan\frac{y}x\implies\theta+C=\log r\\r=Ce^\theta$$

- anonymous

We know when \(\theta=0\) we get \(r=1\) so we have the initial condition \(r(0)=1\) and thus it follows that \(C=1\) yielding \(r=e^\theta\) as our solution... :-)

- anonymous

I knew it'd be a logarithmic spiral ahead of time because this is in fact a special property of them :-)
http://en.wikipedia.org/wiki/Logarithmic_spiral#Definition
The spiral has the property that the angle Ï† between the tangent and radial line at the point (r, Ï†) is constant

- anonymous

@dan815 I presumed the boat started 1 unit away horizontally... IDK about it starting at the origin though

- anonymous

Awesome question @dan815 more more more

- dan815

lol ill ask if i see another good qusetions while reviwing xD

- anonymous

I don't even think it *can* start at the origin... then we wouldn't be able to talk about the tangent at said point and the entire solution so far would fail. :/

- anonymous

\(\tan\phi=0/0\) is undefined! :-)

- experimentX

there is a nice way to do this on polar coordinate ... just note the geometry
|dw:1371367456337:dw|

- experimentX

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- experimentX

yes yes

- anonymous

shouldn't \(dr\) be the opposite side?

- experimentX

dr = r2 - r1

- anonymous

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- anonymous

the shorter radius to the drug boat is \(r\), the longer is \(r+dr\)

- anonymous

the angle between them is \(d\theta\)

- experimentX

no .. lol, think interms of position vector

- anonymous

\(r\) is not a position vector... what you drew is not the infinitesimal change in \(r\)

- anonymous

What you labelled is akin to a piece of the arclength

- dan815

ya man im a little confused too no matter how small theta becomes that will still be hypotenuse can u really say that hypotenuse angle = change in radius at very small angles

- dan815

hypotenuse length*

- anonymous

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- experimentX

take r to be position vector. that should resolve it.

- anonymous

\(r+dr\) is the whole side, \(dr\) is the infinitesimal change in \(r\)

- dan815

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- anonymous

well then it's not polar is it? :-p that's just parametric

- anonymous

@dan815 that's no right angle!

- anonymous

I guess for very small \(d\theta\) it can be treated as one though

- dan815

yes

- dan815

thats wat experiment was saying since there is that symmetry, of 45,45 there it must be 45 45 90 trangle at very small lengths

- experimentX

(r, theta) gives the trajectory of the drug boat. we are considering the geometry of infinitesimal case. dr will the be the distance traveled by drug boat, not by the light. i don't see any problem.

- anonymous

I suppose it's just like how they "prove" the Jacobian for polar coordinates works... :-)|dw:1371368981350:dw|

- anonymous

@experimentX that is not how polar works though... your approach is just a parametric one 'like' polar.

- anonymous

\(dr\) in polar would be the change in distance from the center, not the displacemnet

- experimentX

it depends how you have undersood it ... dr and |dr| are two different things for me.

- anonymous

\(|dr|\) wouldn't make sense for polar since \(dr\) is just a small scalar change. You are talking about displacement parametrized with \(\theta\), where \(r(\theta)\) is a vector-valued function.

- experimentX

polar coordinate is an orthogonal coordinate system. both r and theta have unit vector associated with them. by, dr i meant \( d \vec r \) which is line element
\[ d\vec r = dr \hat r + r d \theta \hat \theta \]
if this doesn't make sense to you either my book is wrong you probably misunderstood something.

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