
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 x^{2}
+ y^{2} = r^{2}.

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 [t_{1},
t_{2}]
from f (t_{1})
= a to f
(t_{2}) = 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
heartshaped 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 a^{2 }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.











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