Find Length of Curve Between Two X Values

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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
 
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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]
 
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]
 
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.
 
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!).