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| Integral
calculus |
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Applications
of the definite integral
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Length
of plane curve, arc length
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Arc length
of a parametric curve
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Arc length of a curve in polar coordinates
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Length
of plane curve, arc length
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The
length of an element dl
of a curve approximates to
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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.
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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
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Example:
Find the arc length of
the upper semicircle of the circle x2
+ y2 = r2.
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| Solution: Let
first calculate the derivative of the upper semicircle |
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| then, using the above
formula for the arc length |
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Therefore,
the circumference of a circle is 2rp.
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Arc length of a parametric curve
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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
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then,
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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.
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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.
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The circle rolls without slipping,
then arc PA
= OA = r
t therefore
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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).
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Since,
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then, using the above
formula for the arc length of a parametric curve
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Arc length of a curve in polar coordinates
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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.
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To find the arc length of a curve in
polar coordinates we use the equations that relate Cartesian and
polar coordinates
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x = r cos q
and y
= r sin q,
where
r
denotes r(q
)
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and calculate differentials dx
and dy
applying the chain rule
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dx
= cosq
dr -
r sinq
dq
and dy
= sinq
dr + r cosq
dq
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so that,
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therefore,
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Example:
Find the length of
the cardioid r
= a (1 -
cosq
) shown in the below figure.
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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.
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| Since
r
= a(1 -
cosq
) then r'
= a sinq |
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that, r
2 + r'
2 = 4 a2 sin
2(q
/2).
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| Using the above formula
for the length of a curve in |
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polar coordinates |
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Therefore, the length of the cardioid
equals eight diameters of the circles.
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| Calculus contents
F |
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