How Do Polar Coordinates Relate to Integral Equations in Homework Problems?

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Homework Help Overview

The discussion revolves around the relationship between polar coordinates and integral equations, particularly focusing on a parametrized curve C in polar coordinates and the function F defined in terms of x and y. Participants are exploring how to express and differentiate these functions in the context of the problem.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to convert the function F into polar coordinates and are discussing the implications of differentiating a function that depends on other functions. Questions arise about the differentiation process and the roles of r(t) and theta(t) in this context.

Discussion Status

There is an ongoing exploration of how to differentiate the vector function R(t) with respect to t, with some participants expressing confusion about the presence of multiple functions and the differentiation process. Guidance has been offered regarding the use of the product rule and the need to express R(t) in terms of its components.

Contextual Notes

Participants are working under the constraints of not knowing the explicit forms of r(t) and theta(t), which adds to the complexity of the differentiation task. There is an acknowledgment of the continuous derivatives of r and theta as part of the problem setup.

Denver Dang
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Homework Statement


Let the curve C be paramatized into polar coordinates given by:

[tex]\[r\left( t \right)=\left( r\left( t \right)\cos \theta \left( t \right),\,\,\,\,\,r\left( t \right)\sin \theta \left( t \right) \right),\,\,\,\,\,a\le t\le b\][/tex]
where r and theta is continuous derivatives and r(t) > 0.

Show that:

[tex]\[F\left( r\left( t \right) \right)\cdot r'\left( t \right)=\theta '\left( t \right)\,\,\,\,\,and\,\,\,\,\,\int_{C}{F\cdot dr=\theta \left( b \right)}-\theta \left( a \right)\][/tex]


Homework Equations



F is given by:

[tex]\[F\left( x,y \right)=\frac{-y\,i+x\,j}{{{x}^{2}}+{{y}^{2}}}\][/tex]


The Attempt at a Solution



No idea... I get stupid when it changes into polar coords.
So anyone with a little hint maybe ?


Regards
 
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Start by calculating what F is in polar coordinates: x = r cos theta, y = r sin theta.
 
Then I get:

[tex]F\left( r,\theta \right)=\frac{\cos \left( \theta \right)j-\sin \left( \theta \right)i}{r}[/tex]
 
Last edited:
Denver Dang said:
Then I get:

[tex]F\left( r,\theta \right)=\frac{\cos \left( \theta \right)j-\sin \left( \theta \right)i}{r}[/tex]

So far, so good. Now what's R'(t) where R(t)=r(t)cos(theta(t))i+r(t)sin(theta(t))j? Try not to confuse the scalar r(t) with the vector R(t). It's R'(t) you want to dot with F.
 
Well, that's what confuses me :)
How do I differentiate a function with functions in it ? And with respect to what ? r or theta ? If I do it with respect to both, then I get to different R'(t), don't I ?
 
Differentiate with respect to t. You'll need to use the product rule.
 
Hmmm...
But there is no t in R(t) ? :S
 
Denver Dang said:
Hmmm...
But there is no t in R(t) ? :S

There are two functions r(t) and theta(t)!
 
Ehhh, I feel pretty stupid right now :S
Do I know the functions r(t) and theta(t) ? :S
 
  • #10
Denver Dang said:
Ehhh, I feel pretty stupid right now :S
Do I know the functions r(t) and theta(t) ? :S

No, but you know dr(t)/dt=r'(t) and dtheta(t)/dt=theta'(t). Just write out R(t)' in terms of them.
 

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