Integral calculus
     Applications of the definite integral
      Length of plane curve arc length
         Arc length of a parametric curve
         Arc length of a curve in polar coordinates
Length of plane curve, arc length
The length of an element dl of a curve approximates to
 
The limit of the sum of such elements, as their number increases to infinity and each increment dx tends to zero, represents the length L of the curve.
Thus, if  f  is continuously differentiable on the closed interval [a, b], the length of the curve  y = f (x)  from   x = a  to  x = b is
Example:   Find the arc length of the upper semicircle of the circle x2 + y2 = r2.
Solution:  Let first calculate the derivative of the upper semicircle
then, using the above formula for the arc length
Therefore, the circumference of a circle is 2rp.
Arc length of a parametric curve
If a curve is given by the parametric equations x = f (t) and  y = g (t) such that the derivatives,  f ' and  g'  are continuous on the closed interval [t1, t2] from  f (t1) = a  to  f (t2) = b, so that
then,
Example:   Find the arc length of the common cycloid  x = r (t - sin t)  and  y = r (1 - cos t)  inside the interval   0 < t < 2p, as is shown in the below figure.
Solution:  The common cycloid is the curve described by a fixed point on the circumference of a circle with the radius r, as the circle rolls without slipping on a straight line.
The circle rolls without slipping, then  arc PA = OA = r t  therefore
P (x, y) where,    x = r t - r cos (t - p/2) = r (t - sin t)   and    y = r + r sin (t - p/2) = r (1 - cos t).
Since,
then, using the above formula for the arc length of a parametric curve
Arc length of a curve in polar coordinates
Suppose given a curve in polar coordinates by r = r (q ) where q changes inside the interval  a < q  < b while a point passes along the arc from a to b.
To find the arc length of a curve in polar coordinates we use the equations that relate Cartesian and polar coordinates 
x = r cos q   and    y = r sin q,   where r denotes  r(q )
and calculate differentials dx and  dy  applying the chain rule
dx = cosq dr - r sinq dq   and    dy = sinq dr + r cosq dq
so that,
therefore,
Example:   Find the length of the cardioid  r = a (1 - cosq ) shown in the below figure.
Solution:  The cardioid is a heart-shaped curve generated by a fixed point on a circle as it rolls round another circle of the same radius. The diameter of circles is a, and q  is the polar angle.
Since   r = a(1 - cosq )   then   r' = a sinq
so that,      r 2 + r' 2  =  4 a2 sin 2(q /2).
Using the above formula for the length of a curve in 
polar coordinates
Therefore, the length of the cardioid equals eight diameters of the circles.
Functions contents G
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