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(a) Use the theorem of Pythagoras to show that
s
2
D
.x/
2
C
.y.x
C
x/
y.x//
2
:
(b) Use Taylor's theorem - see (1.39) - to show that
s
2
.x/
2
.1
C
.y
0
.x//
2
/:
(c) Divide the interval Œa; b into n subintervals and let s
i
denote the length of the
i th interval. Give an argument for the approximation
n
X
x
p
1
C
.y
0
.x
i
//
2
;
l.y; a; b/
i D1
where l.y; a; b/ denotes the length of the graph of y from x
D
a to x
D
b,and
where x
i
denotes the midpoint of the i th interval.
(d) Let x
!
0 and argue by Riemann's integration theory that the limit is given by
l.y; a; b/
D
Z
b
a
p
1
C
.y
0
.x//
2
dx:
(1.51)
This is the formula we will use to compute the length of the cables. But before
we go back to the cables, we will test the accuracy of the trapezoidal rule applied
to an integral of this form.
(e) Consider a circle with unit radius. The length of the entire circle is given by 2.
In Fig.
1.12
we have graphed one-eighth of the circle. The length of this patch
is =4. Show that
D
Z
p
2=2
p
1
C
.y
0
.x//
2
dx;
4
0
where y.x/
D
p
1
x
2
:
(f) Show that
D
Z
p
2=2
4
1
p
dx:
(1.52)
1
x
2
0
(g) Use (1.52) to investigate the accuracy of the composite trapezoidal rule given
by (1.16). Use n
D
10; 20; : : : ; 100 and argue that the error is
0:1667h
2
:
(h) Now we return to the bridge and the problem of computing the length of the
cable. Suppose that the cable consists of four similar patches and suppose that
the left patch of Fig.
1.10
is given by
