Question

$$ \begin{array}{l}{\text { Using Tangent Fins to Find Arc Length Assume } f \text { is }} \\ {\text { smooth on }[a, b] \text { and partition the interval }[a, b] \text { in the usual }} \\ {\text { way. In each subinterval }\left[x_{k-1}, x_{k}\right] \text { construct the tangent fin at }} \\\ {\text { the point }\left(x_{k-1}, f\left(x_{k-1}\right)\right) \text { as shown in the figure. }}\end{array} $$ $$ \begin{array}{l}{\text { (a) Show that the length of the } k \text { th tangent fin over the interval }} \\ {\left[x_{k-1}, x_{k}\right] \text { equals }} \\ {\sqrt{\left(\Delta x_{k}\right)^{2}+\left(f^{\prime}\left(x_{k-1}\right) \Delta x_{k}\right)^{2}}}\end{array} $$ $$ \begin{array}{l}{\text { (b) Show that }} \\ {\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \text { (length of } k \text { th tangent fin } )=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x} \\ {\text { which is the length } L \text { of the curve } y=f(x) \text { from } x=a} \\ {\text { to } x=b .}\end{array} $$

Short answer

The length of the kth tangent fin is \(\sqrt{(\Delta x_k)^2 (1+ (f'(x_{k-1})^2)}\). The sum of the lengths of all the tangent fins, in the limit as the number of fins approaches infinity, equals the arc length of the curve, which is \(\int_{a}^{b} \sqrt{1+(f'(x))^2} dx\).

Step-by-step solution

Part (a): Length of the kth Tangent Fin

For a tangent line, with slope \(f'(x_{k-1})\), the change in the y-coordinate is \(\Delta y = f'(x_{k-1}) \Delta x_k\). The length of the kth tangent fin, from Pythagoras’ theorem, would then be \(\sqrt{(\Delta x_k)^2 + (\Delta y)^2}\), which equals \(\sqrt{(\Delta x_k)^2 + (f'(x_{k-1}) \Delta x_k)^2}\). Simplifying this, the length of the kth tangent fin is \(\sqrt{(\Delta x_k)^2 (1+ (f'(x_{k-1})^2)}\).

Part (b): Limit of the Sum of Tangent Fins

Let's sum the lengths of all the tangent fins and take the limit as the number of fins approaches infinity. This gives us \(\lim_{n → ∞} \sum_{k=1}^{n} \sqrt{(\Delta x_k)^2 (1+ (f'(x_{k-1})^2)}\). Notice that this is in the form of the definition of a definite integral: the limit of a sum as the number of terms approaches infinity. Therefore, this equation simplifies to \(\int_{a}^{b} \sqrt{1+(f'(x))^2} dx\), which is the formula for the length L of the curve y=f(x) from x=a to x=b.