Calculus- Early Transcendentals, 2021a

414 Polar Coordinates, Parametric Equations
Using some properties of derivatives, including the chain rule, we can convert this to use parametric
equations x = f (t), y = g(t):
∫ b
a
√
1 +
( ) dy 2
dx =
dx
=
=
∫ b
a
∫ v
u
∫ v
u
√ (dx ) 2 ( dx
+
dt dt
√ (dx ) 2 ( ) dy 2
+ dt
dt dt
√
( f ′ (t)) 2 +(g ′ (t)) 2 dt.
Here u and v are the t limits corresponding to the x limits a and b.
Example 11.17: Length of Cycloid Arch
Find the length of one arch of the cycloid.
) 2 ( ) dy 2
dt
dx dx dx
Solution. From x = t − sint, y = 1 − cost, we get the derivatives f ′ = 1 − cost and g ′ = sint, so the length
is
∫ 2π
√
∫ 2π
(1 − cost) 2 + sin 2 √
tdt= 2 − 2costdt.
0
Nowweusetheformulasin 2 (t/2)=(1 − cos(t))/2or4sin 2 (t/2)=2 − 2cost to get
∫ 2π
0
0
√
4sin 2 (t/2) dt.
Since 0 ≤ t ≤ 2π, sin(t/2) ≥ 0, so we can rewrite this as
∫ 2π
0
2sin(t/2) dt = 8.
♣
Exercises for 11.5
Exercise 11.5.1 Consider the curve of 11.4.6 in section 11.4. Find all values of t for which the curve has
a horizontal tangent line.
Exercise 11.5.2 Consider the curve of 11.4.6 in section 11.4. Find the area under one arch of the curve.
Exercise 11.5.3 Consider the curve of 11.4.6 in section 11.4. Set up an integral for the length of one arch
of the curve.