Chapter 8.4, Problem 39E

a.

To determine

To show that the length of $kth$tangent fin over the interval $\left[x_{k-1},x_k\right]$ equals

  $\sqrt{{\left(\Delta x_k\right)}^2+{\left(f^{\prime }\left(x_{k-1}\right)\Delta x_k\right)}^2}$

Explanation of Solution

Given information:

The given information is that f is smooth on $\left[a,b\right]$ and partition the interval $\left(a,b\right)$

Proof:

The tangent fin is the hypotenuse of a right angle triangle with leg lengths $\Delta x_k$ and

  $\frac{df}{dx}|{}_{x=x_{k-1}}\Delta x_k=f^{\prime }\left(x_{k-1}\right)\Delta xk$

Therefore, taking the Euclidean distance between these points, the length of the $kth$ tangent is given by

  $\sqrt{{\left(\Delta x_k\right)}^2+{\left(f^{\prime }\left(x_{k-1}\right)\Delta x_k\right)}^2}$

Therefore, the length of $kth$tangent fin over the interval $\left[x_{k-1},x_k\right]$ equals

  $\sqrt{{\left(\Delta x_k\right)}^2+{\left(f^{\prime }\left(x_{k-1}\right)\Delta x_k\right)}^2}$

b.

To determine

To proof that $\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left(\text{length of }kth\text{ tangent fin}\right)={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{\left(f^{\prime }\left(x\right)\right)}^2}}dx}$ which is the length L of the curve $y=f\left(x\right)$ from $x=a\text{ to }x=b$

Explanation of Solution

Given information:

The given information is that f is smooth on $\left[a,b\right]$ and partition the interval $\left(a,b\right)$

Proof:

First write the right hand side of the equation

  $\begin{array}{l}\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left(\text{length of }kth\text{ tangent fin}\right)}\\ =\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\sqrt{{\left(\Delta x_k\right)}^2+{\left(f^{\prime }\left(x_{k1}\right)\Delta x_k\right)}^2}}\\ =\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\Delta x_k\sqrt{1+{\left(f^{\prime }\left(x_{k1}\right)\right)}^2}}\\ ={\displaystyle \underset{a}{\overset{b}{\int }}\sqrt{1+{\left(f^{\prime }\left(x_k\right)\right)}^2}}dx\end{array}$

Therefore, the equation is proved.