6.1 Smooth curves in the plane  (Page 3/4)

$\begin{array}{ccc}\hfill {\int }_{a}^{b}|{f}^{\text{'}}-{p}^{\text{'}}|& =& \sum _{i=1}^{n}{\int }_{{x}_{i-1}}^{{x}_{i}}|{f}^{\text{'}}-{p}^{\text{'}}|\hfill \\ & =& {\int }_{a}^{{x}_{1}}|{f}^{\text{'}}|+\sum _{i=2}^{n-1}|{f}^{\text{'}}-{p}^{\text{'}}|+{\int }_{{x}_{n-1}}^{b}|{f}^{\text{'}}|\hfill \\ & \le & {\int }_{a}^{a+{\delta }^{\text{'}}}|{f}^{\text{'}}|+{\int }_{b-{\delta }^{\text{'}}}^{b}|{f}^{\text{'}}|+\frac{ϵ}{4\left(b-a\right)}{\int }_{{x}_{1}}^{{x}_{n-1}}1\hfill \\ & <& ϵ.\hfill \end{array}$

The proof is now complete.

REMARK It should be evident that the preceding theorem can easily be generalized to a piecewise smooth function $f,$ i.e., a function that is continuous on $\left[a,b\right],$ continuously differentiable on each subinterval $\left({t}_{i-1},{t}_{i}\right)$ of a partition $\left\{{t}_{0}<{t}_{1}<...<{t}_{n}\right\},$ and whose derivative ${f}^{\text{'}}$ is absolutely integrable on $\left(a,b\right).$ Indeed, just apply the theorem to each of the subintervals $\left({t}_{i-1},{t}_{i}\right),$ and then carefully piece together the piecewise linear functions on those subintervals.

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

By a smooth curve from a point ${z}_{1}$ to a different point ${z}_{2}$ in the plane, we mean a set $C\subseteq C$ that is the range of a 1-1, smooth, function $\phi :\left[a,b\right]\to C,$ where $\left[a,b\right]$ is a bounded closed interval in $R,$ where ${z}_{1}=\phi \left(a\right)$ and ${z}_{2}=\phi \left(b\right),$ and satisfying ${\phi }^{\text{'}}\left(t\right)\ne 0$ for all $t\in \left(a,b\right).$

More generally, if $\phi :\left[a,b\right]\to {R}^{2}$ is 1-1 and piecewise smooth on $\left[a,b\right],$ and if $\left\{{t}_{0}<{t}_{1}<...<{t}_{n}\right\}$ is a partition of $\left[a,b\right]$ such that ${\phi }^{\text{'}}\left(t\right)\ne 0$ for all $t\in \left({t}_{i-1},{t}_{i}\right),$ then the range $C$ of $\phi$ is called a piecewise smooth curve from ${z}_{1}=\phi \left(a\right)$ to ${z}_{2}=\phi \left(b\right).$

In either of these cases, $\phi$ is called a parameterization of the curve $C.$

Note that we do not assume that $|{\phi }^{\text{'}}|$ 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 ${\phi }_{1}:\left[a,b\right]\to C$ is a parameterization, and if $g:\left[c,d\right]\to \left[a,b\right]$ is a smooth function having a nonzero derivative, then ${\phi }_{2}\left(s\right)={\phi }_{1}\left(g\left(s\right)\right)$ is another parameterization of $C.$ 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 $C$ are connected exactly as above, i.e., there always is such a function $g$ relating ${\phi }_{1}$ and ${\phi }_{2}.$

Let ${\phi }_{1}:\left[a,b\right]\to C$ and ${\phi }_{2}:\left[c,d\right]\to C$ be two parameterizations of a piecewise smooth curve $C$ joining ${z}_{1}$ to ${z}_{2}.$ Then there exists a piecewise smooth function $g:\left[c,d\right]\to \left[a,b\right]$ such that ${\phi }_{2}\left(s\right)={\phi }_{1}\left(g\left(s\right)\right)$ for all $s\in \left[c,d\right].$ Moreover, the derivative ${g}^{\text{'}}$ of $g$ is nonzero for all but a finite number of points in $\left[c,d\right].$

Because both ${\phi }_{1}$ and ${\phi }_{2}$ are continuous and 1-1, it follows from [link] that the function $g={\phi }_{1}^{-1}\circ {\phi }_{2}$ is continuous and 1-1 from $\left[c,d\right]$ onto $\left[a,b\right].$ Moreover, from [link] , it must also be that $g$ is strictly increasing or strictly decreasing. Write ${\phi }_{1}\left(t\right)={u}_{1}\left(t\right)+i{v}_{1}\left(t\right)\equiv \left({u}_{1}\left(t\right),{v}_{1}\left(t\right)\right),$ and ${\phi }_{2}\left(s\right)={u}_{2}\left(s\right)+i{v}_{2}\left(s\right)\equiv \left({u}_{2}\left(s\right),{v}_{2}\left(s\right)\right).$ Let $\left\{{x}_{0}<{x}_{1}<...<{x}_{p}\right\}$ be a partition of $\left[a,b\right]$ for which ${\phi }_{1}^{\text{'}}$ is continuous and nonzeroon the subintervals $\left({x}_{j-1},{x}_{j}\right),$ and let $\left\{{y}_{0}<{y}_{1}<...<{y}_{q}\right\}$ be a partition of $\left[c,d\right]$ for which ${\phi }_{2}^{\text{'}}$ is continuous and nonzero on the subintervals $\left({y}_{k-1},{y}_{k}\right).$ Then let $\left\{{s}_{0}<{s}_{1}<...<{s}_{n}\right\}$ be the partition of $\left[c,d\right]$ determined by the finitely many points $\left\{{y}_{k}\right\}\cup \left\{{g}^{-1}\left({x}_{j}\right)\right\}.$ We will show that $g$ is continuously differentiable at each point $s$ in the subintervals $\left({s}_{i-1},{s}_{i}\right).$

Fix an $s$ in one of the intervals $\left({s}_{i-1},{s}_{i}\right),$ and let $t={\phi }_{1}^{-1}\left({\phi }_{2}\left(s\right)\right)=g\left(s\right).$ Of course this means that ${\phi }_{1}\left(t\right)={\phi }_{2}\left(s\right),$ or ${u}_{1}\left(t\right)={u}_{2}\left(s\right)$ and ${v}_{1}\left(t\right)={v}_{2}\left(s\right).$ Then $t$ is in some one of the intervals $\left({x}_{j-1},{x}_{j}\right),$ so that we know that ${\phi }_{1}^{\text{'}}\left(t\right)\ne 0.$ Therefore, we must have that at least one of ${u}_{1}^{\text{'}}\left(t\right)$ or ${v}_{1}^{\text{'}}\left(t\right)$ is nonzero. Suppose it is ${v}_{1}^{\text{'}}\left(t\right)$ that is nonzero. The argument, in case it is ${u}_{1}^{\text{'}}\left(t\right)$ that is nonzero, is completely analogous. Now, because ${v}_{1}^{\text{'}}$ is continuous at $t$ and ${v}_{1}^{\text{'}}\left(t\right)\ne 0,$ it follows that ${v}_{1}$ is strictly monotonic in some neighborhood $\left(t-\delta ,t+\delta \right)$ of $t$ and therefore is 1-1 on that interval. Then ${v}_{1}^{-1}$ is continuous by [link] , and is differentiable at the point ${v}_{1}\left(t\right)$ by the Inverse Function Theorem. We will show that on this small interval $g={v}_{1}^{-1}\circ {v}_{2},$ and this will prove that $g$ is continuously differentiable at $s.$

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