Chapter 5: Applications of Integration
Section 5.4: Arc Length
Example 5.4.2
For the curve defined by x=y2, obtain the length of the arc from 0,0 to 1,1.
Solution
Mathematical Solution
Since x=gy=y2, the arc length can be obtained as
s
=∫011+ⅆⅆ y y22ⅆy
=∫011+4 y2ⅆy
=125−14ln−2+5
≐1.478942857
An antiderivative for 1+4 y2 is obtained by making the substitution y=tanθ/2, which then requires knowing that the antiderivative of sec3θ is (secθtanθ+lnsecθ+tanθ/2. Methods for dealing with integrals of this complexity appear in Chapter 6. Meanwhile, Maple will be used to obtain exact and approximate numeric answers, as needed.
Maple Solution
The tutor shown in Figure 5.4.2(a) applies only to functions fx. Hence, to find arc length for x=y2, the tutor must be called with y=x.
As in Figure 5.4.1(a), the red curve is a graph of the function; the blue curve, the integrand; and the green, the arc-length function.
The keen observer will see that the forms of the exact value of the arc length given by the tutor and by direct integration in the Mathematical Solution, are not the same. But they are equivalent, as the numerical approximations suggest.
Figure 5.4.2(a) Arc Length tutor
Surprisingly, the ArcLength command admits curves in the form x=gy.
Application of the ArcLength command
Tools≻Load Package: Student Calculus 1
Loading Student:-Calculus1
Apply the ArcLength command, including the option to return the inert integral. Context Panel: Evaluate and Display Inline Context Panel: Evaluate Integral
ArcLengthy2,y=0..1,output=integral = ∫014y2+1ⅆy=125−14ln−2+5
In order to show the equivalence of 125−14ln−2+5 and 18 ln9+45+125, it suffices to show
18 ln9+45
=−14 ln−2+5
=−1412ln5−22
=18ln15−22
=18ln15−45+4
=18ln19−45
=18ln19−45 9+459+45
=18ln9+4581−80
=18ln9+45
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