Consider a plane curve which is described in polar coordinates (r, @) by r = g(@) for @ for all [a,b] (@ representing theta).
Starting from the known expression for the length of a plane curve in Cartesian coordinates and the equation x = rcos@, y = rsin@, obtain an integral expression in terms of the function g and the numbers a and b for the length of the given curve.
How would I do this question. Thanks.
Paging a math expert! Would some math expert come in please?!
To find the length of the given plane curve described in polar coordinates, we can use the arc length formula for a curve in Cartesian coordinates. Since we have the equations x = rcos@ and y = rsin@, we can express the length of the curve in terms of these variables.
The arc length formula in Cartesian coordinates is given by:
L = ∫ [a,b] √(dx/dt)^2 + (dy/dt)^2 dt
To express this formula in terms of polar coordinates, we use the chain rule to find dx/dt and dy/dt. Recall that r = g(@), so we can write:
x = rcos@
y = rsin@
Differentiating both equations with respect to @, we get:
dx/d@ = (dr/d@)cos@ - rsin@
dy/d@ = (dr/d@)sin@ + rcos@
Now, substitute these values back into the arc length formula:
L = ∫ [a,b] √[(dr/d@)^2 (cos@)^2 + (-rsin@)^2 + (dr/d@)^2 (sin@)^2 + (rcos@)^2] d@
Simplifying this expression:
L = ∫ [a,b] √[(dr/d@)^2 (cos^2@ + sin^2@) + r^2] d@
Since cos^2@ + sin^2@ = 1, we can further simplify:
L = ∫ [a,b] √[(dr/d@)^2 + r^2] d@
This is the integral expression for the length of the given curve in terms of the function g(@) and the interval [a, b].