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0, 1
0, 1
0, 1 the
such that for every x
0, 1 the vertical xsection S x of S has measure zero and for every y
horizontal ysection S y contains almost every number in 0, 1 . This example is quite easy to produce if you
have studied enough set theory and it can be found here. Unless you have read the set theory, must assume
the existence of the Sierpinski example for the purposes of this exercise.
Using Sierpinski’s example, prove that the statement of Fichtenholz’s theorem becomes false if we widen
it to include functions that are almost Riemann integrable.
Exercises 5 and 6 are offered as a resource for a special project. 18 Calculus of Several Variables
Exercises on Velocity, Speed and Curve Length
1. Point at the equation
t t cos t, t sin t, t 2 and assign it as a definition in Scientific Notebook. Then point at the integral Þ t dt and click on
0
Evaluate and then point at the result and then click on Simplify to show that the length of the curve is 401 5 2 1 1 5 ln
10 1
2 5 5 2 1 2. Prove that every piecewise smooth curve in R k is rectifiable.
This assertion follows at once from the relationship between speed and curve length and the fact
that a function that is piecewise continuous on an interval must be itegrable.
3. Given that
if 0 t 1
t t, t sin t if t 0 0, 0 , prove that is not rectifiable.
1. Solution: We observe first that for each t
t t and therefore, whenever 0 t 1
4 t 0, we have
2 1 1 t sin 1 cos 1
t
t
t2
we have
1 t sin 1 cos 1
1
t
t
t
t 1 t sin 1
t
t cos 1
t cos 1
t 1
4 Now given any positive integer n, if
4 2n 2n
4 1
t then
cos 1
t cos 1
t In other words, if n is a positive integer and if
4
8n 1
then 4 t cos 1
t 1
t t 1.
2 8n 1
4 1 1
t 1
2 1
4 and so Þ 4
8n 1
4
8n1 Þ t dt 4
8n 1
4
8n1 1
t 1
4 1
2 dt 1
2 1
4 log 1
2 1
4 log 8n 1
8n 1 4
8n log 1 1
2 Now since
lim
nÝ log 1 2
8n 1 1
4 1
n it follows from the comparison test that the series
log 1 2
8n 1 is divergent and we deduce that the series Þ 402 4
8n 1
4
8n1 t dt 4
8n 1
1
4 log 1 2
8n 1 is divergent. Therefore
Þ t dt Ý.
0 lim 4. Prove that if f is an increasing continuous function from an interval a, b into R and if
t t, f t
for all t
a, b then is rectifiable.
1. Solution: Suppose that t 0 , t 1 , t 2 , , t n is a partition of the interval a, b . We have n n tj tj 1 j1 tj f tj tj 1, f tj 1 j1
n t j t j 1  f t j f tj 1  f tj 1 j1
n tj tj 1 f tj j1 b af b fa. 5. Suppose that is a rectifiable curve with domain a, b in a metric space X and that, for each t
a, b we
have defined t to be the length of the restriction of the curve to the interval a, t . Prove that the function
must be continuous on the interval a, b .
1. Solution: We shall explain the continuity of at each number t in the open interval a, b and leave
as an exercise the task of proving the one sided continuity of at the numbers a and b. Suppose that
a ...
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 Fall '08
 STAFF
 Math, Calculus

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