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# Integration Over Smooth Curves in the Plane: Smooth Curves in the Plane

Module by: Lawrence Baggett. E-mail the author

Summary: Our first project is to make a satisfactory definition of a smooth curve in the plane, for there is a good bit of subtlety to such a definition. In fact, the material in this chapter is all surprisingly tricky, and the proofs are good solid analytical arguments, with lots of ϵϵ's and references to earlier theorems.

Our first project is to make a satisfactory definition of a smooth curve in the plane, for there is a good bit of subtlety to such a definition. In fact, the material in this chapter is all surprisingly tricky, and the proofs are good solid analytical arguments, with lots of ϵϵ's and references to earlier theorems.

Whatever definition we adopt for a curve, we certainly want straight lines, circles, and other natural geometric objects to be covered by our definition. Our intuition is that a curve in the plane should be a “1-dimensional” subset, whatever that may mean. At this point, we have no definition of the dimension of a general set, so this is probably not the way to think about curves. On the other hand, from the point of view of a physicist, we might well define a curve as the trajectory followed by a particle moving in the plane, whatever that may be. As it happens, we do have some notion of how to describe mathematically the trajectory of a moving particle. We suppose that a particle moving in the plane proceeds in a continuous manner relative to time. That is, the position of the particle at time tt is given by a continuous function f(t)=x(t)+iy(t)(x(t),y(t)),f(t)=x(t)+iy(t)(x(t),y(t)), as tt ranges from time aa to time b.b. A good first guess at a definition of a curve joining two points z1z1 and z2z2 might well be that it is the range CC of a continuous function ff that is defined on some closed bounded interval [a,b].[a,b]. This would be a curve that joins the two points z1=f(a)z1=f(a) and z2=f(b)z2=f(b) in the plane. Unfortunately, this is also not a satisfactory definition of a curve, because of the following surprising and bizarre mathematical example, first discovered by Guiseppe Peano in 1890.

THE PEANO CURVE The so-called “Peano curve” is a continuous function ff defined on the interval [0,1],[0,1], whose range is the entire unit square [0,1]×[0,1][0,1]×[0,1] in R2.R2.

Be careful to realize that we're talking about the “range” of ff and not its graph. The graph of a real-valued function could never be the entire square. This Peano function is a complex-valued function of a real variable. Anyway, whatever definition we settle on for a curve, we do not want the entire unit square to be a curve, so this first attempt at a definition is obviously not going to work.

Let's go back to the particle tracing out a trajectory. The physicist would probably agree that the particle should have a continuously varying velocity at all times, or at nearly all times, i.e., the function ff should be continuously differentiable. Recall that the velocity of the particle is defined to be the rate of change of the position of the particle, and that's just the derivative f'f' of f.f. We might also assume that the particle is never at rest as it traces out the curve, i.e., the derivative f'(t)f'(t) is never 0. As a final simplification, we could suppose that the curve never crosses itself, i.e., the particle is never at the same position more than once during the time interval from t=at=a to t=b.t=b. In fact, these considerations inspire the formal definition of a curve that we will adopt below.

Recall that a function ff that is continuous on a closed interval [a,b][a,b] and continuously differentiable on the open interval (a,b)(a,b) is called a smooth function on [a,b].[a,b]. And, if there exists a partition {t0<t1<...<tn}{t0<t1<...<tn} of [a,b][a,b] such that ff is smooth on each subinterval [ti-1,ti],[ti-1,ti], then ff is called piecewise smooth on [a,b].[a,b]. Although the derivative of a smooth function is only defined and continuous on the open interval (a,b),(a,b), and hence possibly is unbounded, it follows from part (d) of (Reference) that this derivative is improperly-integrable on that open interval. We recall also that just because a function is improperly-integrable on an open interval, its absolute value may not be improperly-integrable. Before giving the formal definition of a smooth curve, which apparently will be related to smooth or piecewise smooth functions, it is prudent to present an approximation theorem about smooth functions. (Reference) asserts that every continuous function on a closed bounded interval is the uniform limit of a sequence of step functions. We give next a similar, but stronger, result about smooth functions. It asserts that a smooth function can be approximated “almost uniformly” by piecewise linear functions.

## Theorem 1

Let ff be a smooth function on a closed and bounded interval [a,b],[a,b], and assume that |f'||f'| is improperly-integrable on the open interval (a,b).(a,b). Given an ϵ>0,ϵ>0, there exists a piecewise linear function pp for which

1.  |f(x)-p(x)|<ϵ|f(x)-p(x)|<ϵ for all x[a,b].x[a,b].
2.  ab|f'(x)-p'(x)|dx<ϵ.ab|f'(x)-p'(x)|dx<ϵ.

That is, the functions ff and pp are close everywhere, and their derivatives are close on average in the sense that the integral of the absolute value of the difference of the derivatives is small.

### Proof

Because ff is continuous on the compact set [a,b],[a,b], it is uniformly continuous. Hence, let δ>0δ>0 be such that if x,y[a,b],x,y[a,b], and |x-y|<δ,|x-y|<δ, then |f(x)-f(y)|<ϵ/2.|f(x)-f(y)|<ϵ/2.

Because |f'||f'| is improperly-integrable on the open interval (a,b),(a,b), we may use part (b) of (Reference) to find a δ'>0,δ'>0, which may also be chosen to be <δ,<δ, such that aa+δ'|f'|+b-δ'b|f'|<ϵ/2,aa+δ'|f'|+b-δ'b|f'|<ϵ/2, and we fix such a δ'.δ'.

Now, because f'f' is uniformly continuous on the compact set [a+δ',b-δ'],[a+δ',b-δ'], there exists an α>0α>0 such that |f'(x)-f'(y)|<ϵ/4(b-a)|f'(x)-f'(y)|<ϵ/4(b-a) if xx and yy belong to [a+δ',b-δ'][a+δ',b-δ'] and |x-y|<α.|x-y|<α. Choose a partition {x0<x1<...<xn}{x0<x1<...<xn} of [a,b][a,b] such that x0=a,x1=a+δ',xn-1=b-δ',xn=b,x0=a,x1=a+δ',xn-1=b-δ',xn=b, and xi-xi-1<min(δ,α)xi-xi-1<min(δ,α) for 2in-1.2in-1. Define pp to be the piecewise linear function on [a,b][a,b] whose graph is the polygonal line joining the n+1n+1 points (a,f(x1)),(a,f(x1)),{(xi,f(xi))}{(xi,f(xi))} for 1in-1,1in-1, and (b,f(xn-1)).(b,f(xn-1)). That is, pp is constant on the outer subintervals [a,x1][a,x1] and [xn-1,b][xn-1,b] determined by the partition, and its graph between x1x1 and xn-1xn-1 is the polygonal line joining the points {(x1,f(x1)),...,(xn-1,f(xn-1))}.{(x1,f(x1)),...,(xn-1,f(xn-1))}. For example, for 2in-1,2in-1, the function pp has the form

p ( x ) = f ( x i - 1 ) + f ( x i ) - f ( x i - 1 ) x i - x i - 1 ( x - x i - 1 ) p ( x ) = f ( x i - 1 ) + f ( x i ) - f ( x i - 1 ) x i - x i - 1 ( x - x i - 1 )
(1)

on the interval [xi-1,xi].[xi-1,xi]. So, p(x)p(x) lies between the numbers f(xi-1)f(xi-1) and f(xi)f(xi) for all i.i. Therefore,

| f ( x ) - p ( x ) | | f ( x ) - f ( x i ) | + | f ( x i ) - l ( x ) | | f ( x ) - f ( x i ) | + | f ( x i ) - f ( x i - 1 ) | < ϵ . | f ( x ) - p ( x ) | | f ( x ) - f ( x i ) | + | f ( x i ) - l ( x ) | | f ( x ) - f ( x i ) | + | f ( x i ) - f ( x i - 1 ) | < ϵ .
(2)

Since this inequality holds for all i,i, part (1) is proved.

Next, for 2in-1,2in-1, and for each x(xi-1,xi),x(xi-1,xi), we have p'(x)=(f(xi)-f(xi-1))/(xi-xi-1),p'(x)=(f(xi)-f(xi-1))/(xi-xi-1), which, by the Mean Value Theorem, is equal to f'(yi)f'(yi) for some yi(xi-1,xi).yi(xi-1,xi). So, for each such x(xi-1,xi),x(xi-1,xi), we have |f'(x)-p'(x)|=|f'(x)-f'(yi)|,|f'(x)-p'(x)|=|f'(x)-f'(yi)|, and this is less than ϵ/4(b-a),ϵ/4(b-a), because |x-yi|<α.|x-yi|<α. On the two outer intervals, p(x)p(x) is a constant, so that p'(x)=0.p'(x)=0. Hence,

a b | f ' - p ' | = i = 1 n x i - 1 x i | f ' - p ' | = a x 1 | f ' | + i = 2 n - 1 | f ' - p ' | + x n - 1 b | f ' | a a + δ ' | f ' | + b - δ ' b | f ' | + ϵ 4 ( b - a ) x 1 x n - 1 1 < ϵ . a b | f ' - p ' | = i = 1 n x i - 1 x i | f ' - p ' | = a x 1 | f ' | + i = 2 n - 1 | f ' - p ' | + x n - 1 b | f ' | a a + δ ' | f ' | + b - δ ' b | f ' | + ϵ 4 ( b - a ) x 1 x n - 1 1 < ϵ .
(3)

The proof is now complete.

REMARK It should be evident that the preceding theorem can easily be generalized to a piecewise smooth function f,f, i.e., a function that is continuous on [a,b],[a,b], continuously differentiable on each subinterval (ti-1,ti)(ti-1,ti) of a partition {t0<t1<...<tn},{t0<t1<...<tn}, and whose derivative f'f' is absolutely integrable on (a,b).(a,b). Indeed, just apply the theorem to each of the subintervals (ti-1,ti),(ti-1,ti), and then carefully piece together the piecewise linear functions on those subintervals.

Now we are ready to define what a smooth curve is.

Definition 1:

By a smooth curve from a point z1z1 to a different point z2z2 in the plane, we mean a set CCCC that is the range of a 1-1, smooth, function φ:[a,b]C,φ:[a,b]C, where [a,b][a,b] is a bounded closed interval in R,R, where z1=φ(a)z1=φ(a) and z2=φ(b),z2=φ(b), and satisfying φ'(t)0φ'(t)0 for all t(a,b).t(a,b).

More generally, if φ:[a,b]R2φ:[a,b]R2 is 1-1 and piecewise smooth on [a,b],[a,b], and if {t0<t1<...<tn}{t0<t1<...<tn} is a partition of [a,b][a,b] such that φ'(t)0φ'(t)0 for all t(ti-1,ti),t(ti-1,ti), then the range CC of φφ is called a piecewise smooth curve from z1=φ(a)z1=φ(a) to z2=φ(b).z2=φ(b).

In either of these cases, φφ is called a parameterization of the curve C.C.

Note that we do not assume that |φ'||φ'| is improperly-integrable, though the preceding theorem might have made you think we would.

REMARK Throughout this chapter we will be continually faced with the fact that a given curve can have many different parameterizations. Indeed, if φ1:[a,b]Cφ1:[a,b]C is a parameterization, and if g:[c,d][a,b]g:[c,d][a,b] is a smooth function having a nonzero derivative, then φ2(s)=φ1(g(s))φ2(s)=φ1(g(s)) is another parameterization of C.C. Since our definitions and proofs about curves often involve a parametrization, we will frequently need to prove that the results we obtain are independent of the parameterization. The next theorem will help; it shows that any two parameterizations of CC are connected exactly as above, i.e., there always is such a function gg relating φ1φ1 and φ2.φ2.

## Theorem 2

Let φ1:[a,b]Cφ1:[a,b]C and φ2:[c,d]Cφ2:[c,d]C be two parameterizations of a piecewise smooth curve CC joining z1z1 to z2.z2. Then there exists a piecewise smooth function g:[c,d][a,b]g:[c,d][a,b] such that φ2(s)=φ1(g(s))φ2(s)=φ1(g(s)) for all s[c,d].s[c,d]. Moreover, the derivative g'g' of gg is nonzero for all but a finite number of points in [c,d].[c,d].

### Proof

Because both φ1φ1 and φ2φ2 are continuous and 1-1, it follows from (Reference) that the function g=φ1-1φ2g=φ1-1φ2 is continuous and 1-1 from [c,d][c,d] onto [a,b].[a,b]. Moreover, from (Reference), it must also be that gg is strictly increasing or strictly decreasing. Write φ1(t)=u1(t)+iv1(t)(u1(t),v1(t)),φ1(t)=u1(t)+iv1(t)(u1(t),v1(t)), and φ2(s)=u2(s)+iv2(s)(u2(s),v2(s)).φ2(s)=u2(s)+iv2(s)(u2(s),v2(s)). Let {x0<x1<...<xp}{x0<x1<...<xp} be a partition of [a,b][a,b] for which φ1'φ1' is continuous and nonzero on the subintervals (xj-1,xj),(xj-1,xj), and let {y0<y1<...<yq}{y0<y1<...<yq} be a partition of [c,d][c,d] for which φ2'φ2' is continuous and nonzero on the subintervals (yk-1,yk).(yk-1,yk). Then let {s0<s1<...<sn}{s0<s1<...<sn} be the partition of [c,d][c,d] determined by the finitely many points {yk}{g-1(xj)}.{yk}{g-1(xj)}. We will show that gg is continuously differentiable at each point ss in the subintervals (si-1,si).(si-1,si).

Fix an ss in one of the intervals (si-1,si),(si-1,si), and let t=φ1-1(φ2(s))=g(s).t=φ1-1(φ2(s))=g(s). Of course this means that φ1(t)=φ2(s),φ1(t)=φ2(s), or u1(t)=u2(s)u1(t)=u2(s) and v1(t)=v2(s).v1(t)=v2(s). Then tt is in some one of the intervals (xj-1,xj),(xj-1,xj), so that we know that φ1'(t)0.φ1'(t)0. Therefore, we must have that at least one of u1'(t)u1'(t) or v1'(t)v1'(t) is nonzero. Suppose it is v1'(t)v1'(t) that is nonzero. The argument, in case it is u1'(t)u1'(t) that is nonzero, is completely analogous. Now, because v1'v1' is continuous at tt and v1'(t)0,v1'(t)0, it follows that v1v1 is strictly monotonic in some neighborhood (t-δ,t+δ)(t-δ,t+δ) of tt and therefore is 1-1 on that interval. Then v1-1v1-1 is continuous by (Reference), and is differentiable at the point v1(t)v1(t) by the Inverse Function Theorem. We will show that on this small interval g=v1-1v2,g=v1-1v2, and this will prove that gg is continuously differentiable at s.s.

Note first that if φ2(σ)=x+iyφ2(σ)=x+iy is a point on the curve C,C, then v2(φ2-1(x+iy))=y.v2(φ2-1(x+iy))=y. Then, for any τ[a,b],τ[a,b], we have

v 1 - 1 ( v 2 ( g - 1 ( τ ) ) ) = v 1 - 1 ( v 2 ( φ 2 - 1 ( φ 1 ( τ ) ) ) ) = v 1 - 1 ( v 2 ( φ 2 - 1 ( u 1 ( τ ) + i v 1 ( τ ) ) ) ) = v 1 - 1 ( v 1 ( τ ) ) = τ , v 1 - 1 ( v 2 ( g - 1 ( τ ) ) ) = v 1 - 1 ( v 2 ( φ 2 - 1 ( φ 1 ( τ ) ) ) ) = v 1 - 1 ( v 2 ( φ 2 - 1 ( u 1 ( τ ) + i v 1 ( τ ) ) ) ) = v 1 - 1 ( v 1 ( τ ) ) = τ ,
(4)

showing that v1-1v2=g-1-1=g.v1-1v2=g-1-1=g. Hence gg is continuously differentiable at every point ss in the subintervals (si-1,si).(si-1,si). Indeed g'(σ)=v1-1'(v2(σ))v2'(σ)g'(σ)=v1-1'(v2(σ))v2'(σ) for all σσ near s,s, and hence gg is piecewise smooth.

Obviously, φ2(s)=φ1(g(s))φ2(s)=φ1(g(s)) for all s,s, implying that φ2'(s)=φ1'(g(s))g'(s).φ2'(s)=φ1'(g(s))g'(s). Since φ2'(s)0φ2'(s)0 for all but a finite number of points s,s, it follows that g'(s)0g'(s)0 for all but a finite number of points, and the theorem is proved.

## Corollary 1

Let φ1φ1 and φ2φ2 be as in the theorem. Then, for all but a finite number of points z=φ1(t)=φ2(s)z=φ1(t)=φ2(s) on the curve C,C, we have

φ 1 ' ( t ) | φ 1 ' ( t ) | = φ 2 ' ( s ) | φ 2 ' ( s ) | . φ 1 ' ( t ) | φ 1 ' ( t ) | = φ 2 ' ( s ) | φ 2 ' ( s ) | .
(5)

### Proof

From the theorem we have that

φ 2 ' ( s ) = φ 1 ' ( g ( s ) ) g ' ( s ) = φ 1 ' ( t ) g ' ( s ) φ 2 ' ( s ) = φ 1 ' ( g ( s ) ) g ' ( s ) = φ 1 ' ( t ) g ' ( s )
(6)

for all but a finite number of points s(c,d).s(c,d). Also, gg is strictly increasing, so that g'(s)0g'(s)0 for all points ss where gg is differentiable. And in fact, g'(s)0g'(s)0 for all but a finite number of ss's, because g'(s)g'(s) is either (v1-1v2)'(s)(v1-1v2)'(s) or (u1-1u2)'(s),(u1-1u2)'(s), and these are nonzero except for a finite number of points. Now the corollary follows by direct substitution.

REMARK If we think of φ'(t)=(x'(t),y'(t))φ'(t)=(x'(t),y'(t)) as a vector in the plane R2,R2, then the corollary asserts that the direction of this vector is independent of the parameterization, at least at all but a finite number of points. This direction vector will come up again as the unit tangent of the curve.

The adjective “smooth” is meant to suggest that the curve is bending in some reasonable way, and specifically it should mean that the curve has a tangent, or tangential direction, at each point. We give the definition of tangential direction below, but we note that in the context of a moving particle, the tangential direction is that direction in which the particle would continue to move if the force that is keeping it on the curve were totally removed. If the derivative φ'(t)0,φ'(t)0, then this vector is the velocity vector, and its direction is exactly what we should mean by the tangential direction.

The adjective “piecewise” will allow us to consider curves that have a finite number of points where there is no tangential direction, e.g., where there are “corners.”

We are carefully orienting our curves at the moment. A curve CC from z1z1 to z2z2 is being distinguished from the same curve from z2z2 to z1,z1, even though the set CC is the same in both instances. Which way we traverse a curve will be of great importance at the end of this chapter, when we come to Green's Theorem.

Definition 2:

Let C,C, the range of φ:[a,b]C,φ:[a,b]C, be a piecewise smooth curve, and let z=(x,y)=φ(c)z=(x,y)=φ(c) be a point on the curve. We say that the curve CC has a tangential direction at z,z, relative to the parameterization φ,φ, if the following limit exists:

lim t c φ ( t ) - z | φ ( t ) - z | = lim t c φ ( t ) - φ ( c ) | φ ( t ) - φ ( c ) | . lim t c φ ( t ) - z | φ ( t ) - z | = lim t c φ ( t ) - φ ( c ) | φ ( t ) - φ ( c ) | .
(7)

If this limit exists, it is a vector of length 1 in R2,R2, and this unit vector is called the unit tangent (relative to the parameterization φφ) to CC at z.z.

The curve CC has a unit tangent at the point zz if there exists a parameterization φφ for which the unit tangent at zz relative to φφ exists.

## Exercise 1

1. Restate the definition of tangential direction and unit tangent using the R2R2 version of the plane instead of the CC version. That is, restate the definition in terms of pairs (x,y)(x,y) of real numbers instead of a complex number z.z.
2. Suppose φ:[a,b]Cφ:[a,b]C is a parameterization of a piecewise smooth curve C,C, and that t(a,b)t(a,b) is a point where φφ is differentiable with φ'(t)0.φ'(t)0. Show that the unit tangent (relative to the parameterization φφ) to CC at z=φ(t)z=φ(t) exists and equals φ'(t)/|φ'(t)|.φ'(t)/|φ'(t)|. Conclude that, except possibly for a finite number of points, the unit tangent to CC at zz is independent of the parameterization.
3. Let CC be the graph of the function f(t)=|t|f(t)=|t| for t[-1,1].t[-1,1]. Is CC a smooth curve? Is it a piecewise smooth curve? Does CC have a unit tangent at every point?
4. Let CC be the graph of the function f(t)=t2/3=(t1/3)2f(t)=t2/3=(t1/3)2 for t[-1,1].t[-1,1]. Is CC a smooth curve? Is it a piecewise smooth curve? Does CC have a unit tangent at every point?
5. Consider the set CC that is the right half of the unit circle in the plane. Let φ1:[-1,1]Cφ1:[-1,1]C be defined by
φ1(t)=(cos(tπ2),sin(tπ2)),φ1(t)=(cos(tπ2),sin(tπ2)),
(8)
and let φ2:[-1,1]Cφ2:[-1,1]C be defined by
φ2(t)=(cos(t3π2),sin(t3π2)).φ2(t)=(cos(t3π2),sin(t3π2)).
(9)
Prove that φ1φ1 and φ2φ2 are both parameterizations of C.C. Discuss the existence of a unit tangent at the point (1,0)=φ1(0)=φ2(0)(1,0)=φ1(0)=φ2(0) relative to these two parameterizations.
6. Suppose φ:[a,b]Cφ:[a,b]C is a parameterization of a curve CC from z1z1 to z2.z2. Define ψψ on [a,b][a,b] by ψ(t)=φ(a+b-t).ψ(t)=φ(a+b-t). Show that ψψ is a parameterization of a curve from z2z2 to z1.z1.

## Exercise 2

1. Suppose ff is a smooth, real-valued function defined on the closed interval [a,b],[a,b], and let CR2CR2 be the graph of f.f. Show that CC is a smooth curve, and find a “natural” parameterization φ:[a,b]Cφ:[a,b]C of C.C. What is the unit tangent to CC at the point (t,f(t))?(t,f(t))?
2. Let z1z1 and z2z2 be two distinct points in C,C, and define φ:[0,1]cφ:[0,1]c by φ(t)=(1-t)z1+tz2.φ(t)=(1-t)z1+tz2. Show that φφ is a parameterization of the straight line from the point z1z1 to the point z2.z2. Consequently, a straight line is a smooth curve. (Indeed, what is the definition of a straight line?)
3. Define a function φ:[-r,r]R2φ:[-r,r]R2 by φ(t)=(t,r2-t2).φ(t)=(t,r2-t2). Show that the range CC of φφ is a smooth curve, and that φφ is a parameterization of C.C.
4. Define φφ on [0,π/2)[0,π/2) by φ(t)=eit.φ(t)=eit. For what curve is φφ a parametrization?
5. Let z1,z2,...,znz1,z2,...,zn be nn distinct points in the plane, and suppose that the polygonal line joing these points in order never crosses itself. Construct a parameterization of that polygonal line.
6. Let SS be a piecewise smooth geometric set determined by the interval [a,b][a,b] and the two piecewise smooth bounding functions uu and l.l. Suppose z1z1 and z2z2 are two points in the interior S0S0 of S.S. Show that there exists a piecewise smooth curve CC joining z1z1 to z2,z2, i.e., a piecewise smooth function φ:[a^,b^]Cφ:[a^,b^]C with φ(a^)=z1φ(a^)=z1 and φ(b^)=z2,φ(b^)=z2, that lies entirely in S0.S0.
7. Let CC be a piecewise smooth curve, and suppose φ:[a,b]Cφ:[a,b]C is a parameterization of C.C. Let [c,d][c,d] be a subinterval of [a,b].[a,b]. Show that the range of the restriction of φφ to [c,d][c,d] is a smooth curve.

## Exercise 3

Suppose CC is a smooth curve, parameterized by φ=u+iv:[a,b]C.φ=u+iv:[a,b]C.

1. Suppose that u'(t)0u'(t)0 for all t(a,b).t(a,b). Prove that there exists a smooth, real-valued function ff on some closed interval [a',b'][a',b'] such that CC coincides with the graph of f.f. HINT: ff should be something like vu-1.vu-1.
2. What if v'(t)0v'(t)0 for all t(a,b)?t(a,b)?

## Exercise 4

Let CC be the curve that is the range of the function φ:[-1,1]C,φ:[-1,1]C, where φ(t)=t3+t6i).φ(t)=t3+t6i).

1. Is CC a piecewise smooth curve? Is it a smooth curve? What points z1z1 and z2z2 does it join?
2. Is φφ a parameterization of C?C?
3. Find a parameterization for CC by a function ψ:[3,4]C.ψ:[3,4]C.
4. Find the unit tangent to CC and the point 0+0i.0+0i.

## Exercise 5

Let CC be the curve parameterized by φ:[-π,π-ϵ]Cφ:[-π,π-ϵ]C defined by φ(t)=eit=cos(t)+isin(t).φ(t)=eit=cos(t)+isin(t).

1. What curve does φφ parameterize?
2. Find another parameterization of this curve, but base on the interval [0,1-ϵ].[0,1-ϵ].

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