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418 Chapter 7 Applications of Definite Integrals<br />
Extending the Ideas<br />
39. Using Tangent Fins to Find Arc Length Assume f is<br />
smooth on a, b and partition the interval a, b in the usual<br />
way. In each subinterval x k1 , x k construct the tangent fin at<br />
the point x k1 , f x k1 as shown in the figure.<br />
(x k –1 , f(x k –1 ))<br />
y f(x)<br />
x k<br />
Tangent fin<br />
with slope<br />
f'(x k –1 )<br />
(a) Show that the length of the kth tangent fin over the interval<br />
x k1 , x k equals<br />
x 2<br />
k f x k1 x 2 k .<br />
(b) Show that<br />
lim<br />
n→∞<br />
n<br />
(length of kth tangent fin) b<br />
1 fx 2 dx,<br />
k1<br />
a<br />
which is the length L of the curve y f x from x a<br />
to x b.<br />
40. Is there a smooth curve y f x whose length over<br />
the interval 0 x a is always a2? Give reasons<br />
for your answer. Yes. Any curve of the form y x c, c a<br />
constant.<br />
x k –1<br />
x k<br />
x<br />
39. (a) The fin is the hypotenuse of a right triangle with leg lengths x k and<br />
df<br />
d x<br />
xxk1<br />
x k f(x k–1 ) x k .<br />
(b) lim<br />
n→∞<br />
n<br />
( x k ) 2 ( f(x x k1 )<br />
k ) 2<br />
k1<br />
lim<br />
n→∞<br />
n<br />
x k 1(x ( f) k1 ) <br />
2<br />
k1<br />
1(x)) ( f 2 dx<br />
b<br />
a
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