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**Unformatted text preview: **Arc Length (Section 6.3)
Let C be a curve deﬁned by the parametric equations x = f (t) and y = g (t) for
a ≤ t ≤ b, where f and g have continuous ﬁrst derivatives. Then the length of C is:
b L=
a dx
dt 2 + dy
dt 2 dt L is how long the curve is if you imagine it stretched out in a straight line segment.
Example 0.1. Find the length of x = cos t, y = t + sin t on 0 ≤ t ≤ π . What if we have y = f (x) on an interval a ≤ x ≤ b with f continuously differentiable? This is a special case of the above formula where x = t and y = f (t).
Then
2
b
dy
dx
1+
L=
dx
a
Example 0.2. Find the length of y = x3/2 from x = 0 to x = 4. Example 0.3. Find the length of y = x
2 2/3 from x = 0 to x = 2. How do we overcome this problem? We can integrate with respect to y instead!
If x = g (y ) is continuously diﬀerentiable for c ≤ y ≤ d, then the length of x = g (y )
is
d 2 1+ L= dx
dy x
2 from x = 0 to x = 2. c dy Let’s try again:
Example 0.4. Find the length of y = 2/3 ...

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