What it exactly asks i this question?

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Homework Statement


a heat seeking particle has the property at any point (x,y) in the plane it moves in the direction of maximum temperature increase. If the temparature at (x,y) is T(x,y)=-(e^-2y)*(cosx), find an equation y=f(x) for the path o a heat seeking particle at the point (pi/4,0)

The Attempt at a Solution



I did not completely understood the problem. What is y=f(x) exactly? Is it a linearizaton at (pi/4,0) or such a thing?
 

Answers and Replies

  • #2
quasar987
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It's the y coordinate of the particle's path. They are saying that you can find a function f(x) such that when the particle has horizontal coordinate x, then its vertical coordinate is y=f(x).

First, you should answer this. Given a fucntion f(x,y), what is the direction of maximum increase? And if we know that a particle is always moving in the direction of max increase, how does this relate to the path of the particle?
 
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  • #3
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It's the y coordinate of the particle's path. They are saying that you can find a function f(x) such that when the particle has horizontal coordinate x, then its vertical coordinate is y=f(x).

First, you should answer this. Given a fucntion f(x,y), what is the direction of maximum increase? And if we know that a particle is always moving in the direction of max increase, how does this relate to the path of the particle?

I know that it maximum increases at the direction of
((e^-2y)*sinx) i + (e^-2y)*2cosx)j (the direction of gradient vector). However, how can ı relate it with a path?
 
  • #4
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The gradient will actually be (-(e^-2y)*sinx) i - (e^-2y)*2cosx)j .
 
  • #5
quasar987
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The gradient will be at all points tangent to the path.
 
  • #6
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The gradient will be at all points tangent to the path.

I think these sentence mean something. But did not understand :cry: :cry: :cry:
 
  • #7
AKG
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Let's say the path is parametrized by time, so at time t it's located at (x(t), y(t)) = x(t)i + y(t)j. The direction it's going in is it's tangent, i.e. it's derivative, x'(t)i + y'(t)j, and you want this to equal e-2y(t)sin(x(t))i + 2e-2y(t)cos(x(t))j. So you can solve for x'(t) and y'(t) in terms of x(t) and y(t). Do some clever calculus and you'll be able to solve for y in terms of x (plus some constant, whose value you determine using the given initial conditions), which is what the problem's asking for.
 
  • #8
AKG
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The gradient will actually be (-(e^-2y)*sinx) i - (e^-2y)*2cosx)j .
I don't think so. I missed it the first time I read it too, but there's a minus sign just after the "=" when he's defining T(x,y).
 
  • #9
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Let's say the path is parametrized by time, so at time t it's located at (x(t), y(t)) = x(t)i + y(t)j. The direction it's going in is it's tangent, i.e. it's derivative, x'(t)i + y'(t)j, and you want this to equal e-2y(t)sin(x(t))i + 2e-2y(t)cos(x(t))j. So you can solve for x'(t) and y'(t) in terms of x(t) and y(t). Do some clever calculus and you'll be able to solve for y in terms of x (plus some constant, whose value you determine using the given initial conditions), which is what the problem's asking for.

I am ging nearer but still some problems. I am lost between the x'(t),y'(t),x(t),y(t)... How should I get rid of this goddamn "t" ?:grumpy:
 
  • #10
AKG
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I'm not going to give you the answer. Well, you can suppress the (t)'s, just keep in mind though that y'(t) is dy/dt, not dy/dx.
 
  • #11
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Oh yea I missed that initial minus sign.
 

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