Length of a curve

  • Thread starter soandos
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166
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is there a way to find the length of a curve between two x values?
if so, what is it.
thanks
 
135
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Yes since the element of length is (assuming two dimensions)

[tex]ds=\sqrt{dx^2+dy^2}[/tex]

If you integrate you get the length of the curve [itex]s[/itex] from [itex][a,b][/itex]

[tex]s=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt=\int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}dx[/tex]
 

HallsofIvy

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In three dimensions, write x, y, z as functions of parameter t and do the same:
[tex]\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2+ \left(\frac{dz}{dt}\right)^2}dt[/tex]
 
6
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I am sorry for my broken english.

How can one find the length of a curve, if the coordinate system is not rectangular (for instance, it is spherical)?

Please if not inconvenient to you, point out my mistakes in my english.
 

HallsofIvy

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By using exactly the same formulas but converting from the other coordinate system.

For example, suppose a path is given in spherical coordinates by [itex]\rho= 1[/itex], [itex]\phi= \pi/3[/itex], [itex]\theta= t[/itex], with parameter t. In Cartesian coordinates that is [itex]x= \rho cos(\theta) sin(\phi)= (\sqrt{3}/2)cos(t)[/itex], [itex]y= \rho sin(\theta)sin(\phi)= (\sqrt{3}/2)sin(t)[/itex], [itex]z= \rho cos(t)= cos(t)[/itex].

And my only criticism is that you should stop apologizing for your English. It is excellent. Far better than my (put whatever language you wish in here!).
 

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