Differential geometry notes 3

8/9/2019 Differential geometry notes 3

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the parabola

R t

t, t2

) R2

,and the cubic curve

R t

t, t2, t3) R3.

Exercise 1. Sketch the cubic curve (Hint: First draw each of the projectionsinto the xy, yz, and zx planes).

Exercise 2. Find a formula for the curve which is traced by the motion of afixed point on a wheel of radius r rolling with constant speed on a flat surface(Hint: Add the formula for a circle to the formula for a line generated bythe motion of the center of the wheel. You only need to make sure that the

speed of the line correctly matches the speed of the circle).

Exercise 3. Let : I Rn, and : J Rn be a pair of differentiablecurves. Show that

(t), (t)

=

(t), (t)

+

(t), (t)

and

(t)

=

(t), (t)

(t)

.

(Hint: The first identity follows immediately from the definition of the inner-product, together with the ordinary product rule for derivatives. The secondidentity follows from the first once we recall that := , 1/2).

Exercise 4. Show that if has unit speed, i.e., (t) = 1, then its velocityand acceleration are orthogonal, i.e., (t), (t) = 0.

Exercise 5. Show that if the position vector and velocity of a planar curve : I R2 are always perpendicular, i.e., (t), (t) = 0, for all t I, then(I) lies on a circle centered at the origin of R2.

Exercise 6. Use the fundamental theorem of Calculus for real valued func-tions to show:

(b) (a) =

ba

(t) dt.

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Exercise 7. Prove that

(b) (a)

b

a

(t) dt.

(Hint: Use the fundamental theorem of calculus and the Cauchy-Schwartsinequality to show that for any unit vector u Rn,

(b) (a), u

=

ba

(t), udt

ba

(t)dt.

Then set u := ((b) (a))/(b) (a).

The previous exercise immediately yields:

Theorem 8 (Mean Value Theorem for curves). If : I Rn be a C1

curve, then for every t, s I,

(t) (s) sup[t,s]

|t s|.

1.2 Reparametrization

We say that : J R

n

is areparametrization

of : I R

n

provided thatthere exists a smooth bijection : I J such that (t) =

(t)

. In otherwords, the following diagram commutes:

For instance (t) = (cot(2t), sin(2t)), 0 t , is a reparametrization(t) = (sin(t), cos(t)), 0 t 2, with : [0, 2] [0, ] given by (t) =t/2.

The geometric quantities associated to a curve do not change underreparametrization. These include length and curvature as we define below.

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1.3 Length and Arclength

By a partitionP of an interval [a, b] we mean a collection of points {t0, . . . , tn}of [a, b] such that

a = t0 < t1 < < tn = b.

The approximation of the length of with respect to P is defined as

length[, P] :=n

i=1

(ti) (ti1),

and if Partition[a, b] denotes the set of all partitions of [a, b], then the lengthof is given by

length[] := sup

length[, P] | P Partition[a, b]

,

where sup denotes the supremum or the least upper bound.

Exercise 9. Show that the shortest curve between any pairs of points in Rn

is the straight line segment joining them. (Hint: Use the triangle inequality).

We say that a curve is rectifiable if it has finite length.

Exercise* 10 (Nonrectifiable curves). Show that there exists a curve : [0, 1] R2 which is not rectifiable (Hint: One such curve, known as

the Koch curve (Figure 1), may be obtained as the limit of a sequence ofcurves i : [0, 1] R defined as follows. Let 0 trace the line segment [0, 1].Consider an equilateral triangle of sides 1/3 whose base rests on the middlethird of [0, 1]. Deleting this middle third from the interval and the triangleyields the curve traced by 1.

Figure 1:

Repeating this procedure on each of the 4 subsegments of 1 yields 2.Similarly i+1 is obtained from i. You need to show that i converge to a(continuos) curve, which may be done using the Arzela-Ascoli theorem. Itis easy to see that this limit has infinite length, because the length of i is

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(4/3)i. Another example of a nonrectifiable curve : [0, 1] R2 is given by

(t) := (t, t sin(/t)), when t = 0, and (t) := (0, 0) otherwise. The difficultyhere is to show that the length is infinite.)

If a curve is C1, then its length may be computed as the following theoremshows. Note also that the following theorem shows that a C1 curve over acompact domain is rectifiable. First we need the following fact:

Theorem 11 (Length of C1 curves). Show that if : I Rn is a C1

curve, then

length[] =

I

(t) dt.

Proof. It suffices to show that (i) length[, P] is not greater than the aboveintegral, for any P Partition[a, b], and (ii) there exists a sequence PN ofpartitions such that limN length[, PN] is equal to the integral. The firstpart follows quickly from Exercise 7. To prove the second part, let PN bea partition given by ti := a + i(b a)/N. Recall that, by the definition ofintegral, for any > 0, we may choose N large enough so that

I

(t) dt Ni=1

(ti)b a

N

2. (1)

Next note that the mean value theorem for curves (Theorem 8), yields that

length[, PN] Ni=1

(ti)b a

N

Ni=1

( supsi[ti1,ti]

(si) (ti))

b a

N

.

But, by the triangle inequality,

supsi[ti1,ti]

(si) (ti) sup

si[ti1,ti]

(si) (ti).

Finally since is continuos on the closed interval [a, b], we may suppose that

N is so large that

supsi[ti1,ti]

(si) (ti)

2(b a).

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The last three inequalities yield that

length[, PN] Ni=1

(ti)b a

N

2(2)

Inequalities (1) and (2), together with triangle inequality yield that,

I

(t) dt length[, PN]

which completes the proof.

Exercise 12. Compute the length of a circle of radius r, and the length ofone cycle of the curve traced by a point on a circle of radius r rolling on astraight line.

Exercise 13 (Invariance of length under reparametrization). Showthat if is a reparametrization of a C1 curve , then length[] = length[],i.e., length is invariant under reparametrization (Hint: you only need torecall the chain rule together with the integration by substitution.)

Let L := length[]. The arclength function of is a mapping s : [a, b] [0, L] given by

s(t) :=ta

(u) du.

Thus s(t) is the length of the subsegment of which stretches from the initialtime a to time t.

Exercise 14 (Regular curves). Show that if is a regular curve, i.e.,(t) = 0 for all t I, then s(t) is an invertible function, i.e., it is one-to-one (Hint: compute s(t)).

Exercise 15 (Reparametrization by arclength). Show that every regu-lar curve : [a, b] Rn, may be reparametrized by arclength ( Hint: Define: [0, L] Rn by (t) := (s1(t)), and use the chain rule to show that = 1; you also need to recall that since f(f1(t)) = t, then, again bychain rule, we have (f1)(t) = 1/f(f1(t)) for any smooth function f withnonvanishing derivative.)

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1.4 Cauchys integral formula and

curves of constant widthLet : R2 be a curve and u() := (cos(), sin()) be a unit vector. Theprojection of into the line passing through the origin and parallel to u isgiven by u(t) := (t), uu.

Exercise 16 (Cauchys integral formula). Show that if : I R2 haslength L, then the average length of the projections u, over all directions,is 2L/, i.e.,

1

2

20

length[u()] d =2L

.

(Hint: First prove this fact for the case when traces a line segment. Thena limiting argument settles the general case, once you recall the definition oflength.)

As an application of the above formula we may obtain a sharp inequalityinvolving width of closed curves. The width of a set X R2 is the distancebetween the closest pairs of parallel lines which contain X in between them.For instance the width of a circle of radius r is 2r. A curve : [a, b] R2 issaid to be closed provided that (a) = (b). We should also mention that is a Ck closed curve provided that the (one-sided) derivatives of match upat a and b.

Exercise 17 (Width and length). Show that if : [a, b] R2 is a closedcurve with width w and length L, then

w L

.

Note that the above inequality is sharp, since for circles w = L/. Arethere other curves satisfying this property? The answer may surprise you.For any unit vector u(), the width of a set X R2 in the direction u, wu,is defined as the distance between the closest pairs of lines which contain Xin between them. We say that a closed curve in the plane has constant widthprovided that wu is constant in all directions.

Exercise 18. Show that if the equality in Exercise 17 holds then is a curveof constant width.

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The last exercise would have been insignificant if circles were the only

curves of constant width, but that is not the case:Exercise 19 (Reuleaux triangle). Consider three disks of radius r whosecenters are on an equilateral triangle of sides r, see Figure 2. Show thatthe curve which bounds the intersection of these disks has constant width.Also show that similar constructions for any regular polygon yield curves ofconstant width.

Figure 2:

It can be shown that of all curves of constant width w, Reuleaux trianglehas the least are. This is known as the Blaschke-Lebesque theorem. A recentproof of this result has been obtained by Evans Harrell.

Note that the Reuleaux triangle is not a C1 regular curve for it has sharpcorners. To obtain a C1 example of a curve of constant width, we may take acurve which is a constant distance away from the Reuleaux triangle. Further,a C example may be constructed by taking an evolute of a deltoid, see Grayp. 177.

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