# Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/158

Jump to: navigation, search
This page needs to be proofread.

90. Derivative of the arc in rectangular coördinates. Let s be the length[1] of the arc AP measured from a fixed point A on the curve.

Denote the increment of s (= arc PQ) by $\Delta s$. The definition of the length of arc depends on the assumption that, as Q approaches P,

$\lim \left ( \frac{\mbox{chord} PQ}{\mbox{arc} PQ} \right ) = 1.$

If we now apply the theorem in §89 to this, we get

 (G) In the limit of the ratio of chord PQ and a second infinitesimal, chord PQ may be replaced by arc PQ (= $\Delta s$).

From the above figure

 (H) $(\mbox{chord} PQ)^2 = (\Delta x)^2 + (\Delta y)^2,$

Dividing through by $(\Delta x)^2$, we get

 (I) $\left ( \frac{\mbox{chord} PQ}{\Delta x} \right )^2 = 1 + \left ( \frac{\Delta y}{\Delta x} \right )^2$.

Now let Q approach P as a limiting position; then $\Delta x \dot= 0$ and we have

$\left ( \frac{ds}{dx} \right )^2 = 1 + \left ( \frac{dy}{dx} \right )^2$.
[Since $\lim_{\Delta x \to 0} \left ( \tfrac{chord PQ}{\Delta x} \right )= \lim_{\Delta x \to 0} \left ( \tfrac{\Delta s}{\Delta x} \right ) = \tfrac{ds}{dx}$, (G).]
 (24) ∴ $\frac{ds}{dx} = \sqrt{1 + \left ( \frac{dy}{dx} \right )^2}.$

Similarly, if we divide (H) by $(\Delta y)^2$ and pass to the limit, we get

 (25) $\frac{ds}{dy} = \sqrt{\left ( \frac{dx}{dy} \right )^2 + 1}.$

Also, from the above figure,

$\cos \theta = \frac{\Delta x}{\mbox{chord} PQ}$ $\sin \theta = \frac{\Delta y}{\mbox{chord} PQ}.$

Now as Q approaches P as a limiting position $\theta \dot= \tau$, and we get

 (26) $\cos \tau = \frac{dx}{ds}$, $\sin \tau = \frac{dy}{ds}.$
[Since from (G) $\lim \tfrac{\Delta x}{\mbox{chord} PQ} = \lim \tfrac{\Delta x}{\Delta x} = \tfrac{dx}{ds}$, and $\lim \tfrac{\Delta y}{\mbox{chord} PQ} = \lim \tfrac{\Delta y}{\Delta s} = \tfrac{dy}{ds}$.]
1. Defined in § 209.