Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Lecture 19: Arclength and trigonometric functions. We begin this lecture by defining the arclength of the graph of a differentiable function y = f (x) between x = a and x = b. We motivate our definition by calculating the length of the part of the line y = mx between x = a and x = b. This is the hypotenuse of a right triangle whose legs have lengths b − a and m(b − a) respectively. √ Thus by the Pythagorean theorem, the length of the hypotenuse is given by (b − a) 1 + m2 . This motivates the following definition of arclength. (We view arclength as the limit of lengths of splines along the curve.) Definition Let f be a differentiable function on the interval [a, b]. The arclength of the graph of f is Z bp 1 + (f 0 (x))2 dx. a We immediate apply this definition to our favorite curve from plane geometry: the unit circle. The part of the unit circle in the upper half plane is the curve p 1 − x2 . y= With f (x) = √ 1 − x2 , we calculate f 0 (x) = √ −x , 1 − x2 so that p 1 + (f 0 (x))2 = √ 1 . 1 − x2 Thus the arclength A of the circle between x = 0 and x = a (actually the negative of the arclength if a is negative) is given by Z A= a √ 0 dx . 1 − x2 We have no especially nice way of computing this integral, so we just give it a name. We say Z a dx √ arcsin a = . 1 − x2 0 This entirely corresponds to our intuition from plane geometry. When studying geometry, we keep going about the arclengths of parts of circles, even before we can define what arclength actually means. We associate arcs on circles with the angles that subtend them, and the arc we are describing corresponds to the angle whose sine is a. The reason that inverse function to sin has such an odd name is that it is computing the length of an arc. For our purposes, we look at things a little differently. We have no way of describing 1 the function sin x without its inverse, because we don’t know what an angle means without the notion of arclength. However clearly arcsin is increasing as a goes from −1 to 1 (and it is odd). Thus it has an inverse. We define sin to be the inverse of arcsin. We have not yet named the domain of sin. We make the definition that π = 2 1 Z √ 0 dx . 1 − x2 This is really the usual definition for π. It is the arclength of the unit semicircle. We have π defined sin x on the interval [ −π 2 , 2 ]. On the same interval, we may define cos x by cos x = p 1 − sin2 x. It is not hard to see that for x ∈ [0, π2 ], we have that cos( π − x) = sin(x), 2 because this is just the symmetry between x and y in the definition of the unit circle. It is definitely interesting to extend sin and cos to be defined on the whole real line. We are already in a position to do this by symmetry as well, but for the moment we refrain. We will have a much clearer way of defining this extension later when we introduce complex numbers. But, for now, as long as we stay in the interval [− π2 , π2 ], we are in a position to obtain all the basic facts of calculus for trigonometric functions. Thus, for instance, x = arcsin(sinx). Differentiating in x, we get 1= p 1 d sin x), 1 − sin2 x dx ( and solving for the second factor, we obtain the famous formula that d sin x = cos x. dx Applying the symmetry cos( π − x) = sin(x), 2 we immediately obtain that d cos x = − sin x. dx 2 Using these two results, we can easily build up all the famous formulae in the calculus of trigonometric functions. sin x For instance, we define sec x = cos1 x and tan x = cos x . We readily use the quotient rule to calculate d sec x = sec x tan x, dx and d tan x = sec2 x. dx Then we are free to observe sec x sec x(sec x + tan x) sec x + tan x d (sec x + tan x) = dx sec x + tan x d = (log(sec x + tan x)). dx = In short, all the identities of calculus just come to life. The final thing I wanted to bring up today is the dual role of π. Perhaps, we all remember π as the arclength of the unit semi-circle, but we might also remember it as the area of the unit circle. The first can be a definition, but then the second should be a consequence. Here is how we see it: We calculate Z 1 p 1 − x2 dx. 0 This is just the area of one quarter of the unit circle. We will do this using the (quite natural) trigonometric substitution x = sin u. (Wasn’t x already the sin of something!?!) We obtain dx = cos udu. The integral now runs from 0 to π 2 and becomes Z π 2 cos2 udu. 0 We calculate this integral without any fancy double angle identities. We just use again the symmetry π cos( − u) = sin(u), 2 to obtain Z π Z π 2 2 cos2 udu = 0 0 3 sin2 udu. Thus Z 0 π 2 1 cos udu = 2 2 π 2 Z (cos2 u + sin2 u)du, 0 and since it is easy to integrate 1, we get Z 1 p 1 − x2 dx = 0 π . 4 How is this related to the usual Euclidean proof of the same fact? 4