Example: Circular Arc y= � 1 − x2 describes the graph of a semicircle. We’ll ﬁnd the arc length of the piece of this semicircle above the interval 0 ≤ x ≤ a. (See Figure 1.) a Figure 1: Arc length of y = √ 1 − x2 over 0 ≤ x ≤ a. We’ll use the variable α to denote the arc length along the circle. We could calculate the exact value of α using trigonometry, but we’ll ﬁrst ﬁnd it using calculus. We start by ﬁnding y � . −x √ 1 − x2 � ds = 1 + (y � )2 dx � � �2 −x = 1+ √ dx 1 − x2 � �2 −x Yuck. Let’s simplify 1 + √ in a separate calculation: 1 − x2 � �2 −x � 2 1 + (y ) = 1 + √ 1 − x2 2 x = 1+ 1 − x2 1 − x2 + x2 = 1 − x2 1 1 + (y � )2 = . 1 − x2 y� = Plugging this in and using our formula for arc length, we get: � a dx √ α = 1 − x2 0 = �a sin−1 x�0 = sin−1 a. = a.) α (So sin α a 1 α a 1 Figure 2: The arc length equals α. This is a little deeper than it looks; we went a distance α along the arc of a circle that has radius 1 and ended up at a point whose x-coordinate was a. Previously, if we had an angle whose measure was α radians, we’d say: sin α = a. You may have been told that radians measured the arc length along the curve of the circle, but this is the ﬁrst time you’ve been able to derive it. Remember that our ﬁrst deﬁnition of the exponential function ex involved the slope of its graph, but later we were able to deﬁne the natural log function as an integral. The same sort of thing is happening here; if you want to know what radians are you have to calculate this arc length. This gives you a new deﬁnition of the arcsine function, which gives you a new deﬁnition of the sine function, which leads to an improved deﬁnition and understanding of trig functions.