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.
To find the length of a curve described in polar coordinates, we can use the formula for arc length in Cartesian coordinates and convert it to polar coordinates.
The formula for arc length in Cartesian coordinates is given by:
ds = sqrt(dx^2 + dy^2)
We can express dx and dy in terms of polar coordinates as follows:
dx = dr*cos@ - r*sin@d@
dy = dr*sin@ + r*cos@d@
Substituting these values into the formula for ds, we get:
ds = sqrt((dr*cos@ - r*sin@d@)^2 + (dr*sin@ + r*cos@d@)^2)
= sqrt(dr^2*cos^2@ - 2*dr*r*sin@d@*cos@ + r^2*sin^2@d@ + dr^2*sin^2@ + 2*dr*r*sin@d@*cos@ + r^2*cos^2@d@)
= sqrt(dr^2 + r^2*(sin^2@d@ + cos^2@d@))
= sqrt(dr^2 + r^2)
Now, we need to express ds in terms of @ and dr. Since r is the function g(@), we can substitute r = g(@) into the equation:
ds = sqrt(dr^2 + (g(@))^2)
To find the length of the curve, we integrate ds over the interval [a, b]:
Length = ∫[a,b] sqrt(dr^2 + (g(@))^2) d@
This integral expression gives the length of the curve in terms of the function g and the numbers a and b.
To find the length of the given plane curve described in polar coordinates, we can use the arc length formula from calculus. The arc length of a curve given by the parametric equations x = f(t) and y = g(t) over an interval [a, b] is calculated using the integral:
L = ∫[a,b] √[ (dx/dt)^2 + (dy/dt)^2 ] dt
Let's start by expressing the given curve in Cartesian coordinates using the polar coordinate conversion equations:
x = r * cos(@)
y = r * sin(@)
Now, we need to find the derivatives dx/d@ and dy/d@. Using the chain rule of differentiation, we have:
dx/d@ = dr/d@ * cos(@) - r * sin(@)
dy/d@ = dr/d@ * sin(@) + r * cos(@)
Squaring and adding these derivatives, we get:
(dx/d@)^2 + (dy/d@)^2 = (dr/d@)^2 * (cos^2(@) + sin^2(@)) + r^2 * (sin^2(@) + cos^2(@))
Since cos^2(@) + sin^2(@) = 1, we have:
(dx/d@)^2 + (dy/d@)^2 = (dr/d@)^2 + r^2
Now, substituting r = g(@), we have:
(dx/d@)^2 + (dy/d@)^2 = (dr/d@)^2 + g(@)^2
Finally, we can rewrite the arc length integral in terms of the function g(@) and the limits of integration a and b:
L = ∫[a,b] √[ (dr/d@)^2 + g(@)^2 ] d@
This is the integral expression in terms of the function g and the limits of integration a and b for the length of the given curve.