dan815
  • dan815
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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dan815
  • dan815
|dw:1371338084197:dw|
dan815
  • dan815
the light will be corrected to keep shining on the boat at all times as he is trying to escape
dan815
  • dan815
|dw:1371338223398:dw|

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More answers

Loser66
  • Loser66
I know where do you have this question. MIT, right?
Jhannybean
  • Jhannybean
coast guards using LED lights? :)
anonymous
  • anonymous
So the light travels around the entire time?
anonymous
  • anonymous
Sounds like a polar equation to me.
Jhannybean
  • Jhannybean
Who tries escaping in that path? LOL
anonymous
  • anonymous
and that wouldn't be exactly a function.
anonymous
  • anonymous
reason why i said that is because the term "function" in the question creates ambiguity.
anonymous
  • anonymous
\[\bf r=\theta\]Try that. Looks good.
anonymous
  • anonymous
well that's the only way it's going to be done for the kind of graph ur asking.
primeralph
  • primeralph
I don't fully see the scope of the question but I can say the boatman is dumb.
primeralph
  • 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
  • anonymous
the light is rotating...
Jhannybean
  • Jhannybean
Then the coastguard is partially blind.
primeralph
  • primeralph
|dw:1371364529936:dw|
anonymous
  • anonymous
trust me dan, r = theta is correct. It feels right in my heart.
anonymous
  • anonymous
it actually is right, probably lol.
anonymous
  • anonymous
u know the answer?
anonymous
  • anonymous
The question is ambiguous...
dan815
  • dan815
noo dont tell em answer experiment lol
anonymous
  • anonymous
I suppose you could write a differential equation here since what you're describing are tangent lines
experimentX
  • experimentX
ok ok
dan815
  • dan815
yeah thats the right step oldrin
dan815
  • dan815
u know 1 thing, which is the slope between the boat movement and the light is always the same
anonymous
  • anonymous
right
Jhannybean
  • Jhannybean
NO DAN I WAS READING THAT! lol
dan815
  • dan815
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
  • dan815
i put it on the main part lol
Jhannybean
  • Jhannybean
This is like a movie scene, lmao..
dan815
  • 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
  • dan815
|dw:1371365756120:dw|
dan815
  • dan815
|dw:1371365864968:dw|
anonymous
  • anonymous
@ganeshie8 close but it's an exponential spiral
anonymous
  • anonymous
right that's logical
anonymous
  • anonymous
|dw:1371366116380:dw|
anonymous
  • 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
  • anonymous
It's clear that \(\tan\phi=\dfrac{y}x\) so we get:$$\frac{dy}{dx}=\frac{1+y/x}{1-y/x}$$
anonymous
  • 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
  • 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
  • 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
  • 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
  • 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 φ between the tangent and radial line at the point (r, φ) is constant
anonymous
  • anonymous
@dan815 I presumed the boat started 1 unit away horizontally... IDK about it starting at the origin though
anonymous
  • anonymous
Awesome question @dan815 more more more
dan815
  • dan815
lol ill ask if i see another good qusetions while reviwing xD
anonymous
  • 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
  • anonymous
\(\tan\phi=0/0\) is undefined! :-)
experimentX
  • experimentX
there is a nice way to do this on polar coordinate ... just note the geometry |dw:1371367456337:dw|
experimentX
  • experimentX
|dw:1371367785336:dw|
experimentX
  • experimentX
yes yes
anonymous
  • anonymous
shouldn't \(dr\) be the opposite side?
experimentX
  • experimentX
dr = r2 - r1
anonymous
  • anonymous
|dw:1371368245342:dw|
anonymous
  • anonymous
the shorter radius to the drug boat is \(r\), the longer is \(r+dr\)
anonymous
  • anonymous
the angle between them is \(d\theta\)
experimentX
  • experimentX
no .. lol, think interms of position vector
anonymous
  • anonymous
\(r\) is not a position vector... what you drew is not the infinitesimal change in \(r\)
anonymous
  • anonymous
What you labelled is akin to a piece of the arclength
dan815
  • 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
  • dan815
hypotenuse length*
anonymous
  • anonymous
|dw:1371368419112:dw|
experimentX
  • experimentX
take r to be position vector. that should resolve it.
anonymous
  • anonymous
\(r+dr\) is the whole side, \(dr\) is the infinitesimal change in \(r\)
dan815
  • dan815
|dw:1371368587413:dw|
anonymous
  • anonymous
well then it's not polar is it? :-p that's just parametric
anonymous
  • anonymous
@dan815 that's no right angle!
anonymous
  • anonymous
I guess for very small \(d\theta\) it can be treated as one though
dan815
  • dan815
yes
dan815
  • 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
  • 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
  • anonymous
I suppose it's just like how they "prove" the Jacobian for polar coordinates works... :-)|dw:1371368981350:dw|
anonymous
  • anonymous
@experimentX that is not how polar works though... your approach is just a parametric one 'like' polar.
anonymous
  • anonymous
\(dr\) in polar would be the change in distance from the center, not the displacemnet
experimentX
  • experimentX
it depends how you have undersood it ... dr and |dr| are two different things for me.
anonymous
  • 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
  • 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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