Calculus of One and Several Variables

8.1.1 Lengths

Consider how to define the length of the graph of a function y = f

. Consider the following picture.

which depicts a small right triangle attached as shown to the graph of a function y = f

for x ∈. If the triangle is small enough, this shows the length of the curve joined by the hypotenuse of the right triangle is essentially equal to the length of the hypotenuse. Thus, ² = ² + ² and dividing by ² yields

∘ ---------- dl ( dy)2 ∘ --------2 dx-= 1 + dx- = 1+ f ′(x) , l(a) = 0 (8.1)

(8.1)

as an initial value problem for the function l

which gives the length of this curve on . Thus the length of the curve is given by the definite integral,

∫ b∘ --------- 1 + f′(x)2dx a

Alternatively, you could consider a partition of

,a = x₀ < x₁ < < x_n = b and observe that the length of the curve between the two points and is approximately the length of the line joining these two points and that the length of the curve would be close to the sum of these as suggested in the following picture.

Thus the length of the curve would be approximately the sum of the lengths of the little straight lines in the above picture and this equals

n∑ ∘ ---------------2-----------2- (f (xi)− f (xi−1)) + (xi − xi−1) i=1

which is equal to

∑n ∘ ------------------------------ (f′(zi)(xi − xi−1))2 + (xi − xi−1)2 i=1

by the mean value theorem. Then this reduces to

∑n ∘ -′---2---- f (zi) + 1 (xi − xi−1) i=1

which is a Riemann sum for the integral ∫ _a^b

dx. One would imagine that this approximation would in the limit, as the norm of the partition converges to 0, have what should be defined as the length of the curve as the limit.

This definition gives the right answer for the length of a straight line. To see this, consider a straight line through the points

and where a < c. Then the right answer is given by the Pythagorean theorem or distance formula and is . What is obtained from the above initial value problem? The equation of the line is f = b + and so f^′ = . Therefore, by the new procedure, the length is

∘ ------------ ∘ ------------ ∫ c ( d− b)2 ( d− b)2 ∘ ---------------- 1+ ----- dx = (c− a) 1+ ----- = (a− c)2 + (d− b)2 a c− a c− a

as hoped. Thus the new procedure gives the right answer in the familiar cases but it also can be used to find lengths for more general curves than straight lines. Summarizing,

Here is another familiar example.

Here the function is f

= and so f^′ = −x∕. Therefore, the length is

∘ ------------------ ∘ ---------- ∫ π∕4 ( ∘ ------)2 ∫ π∕4 ( 1 ) 1+ − x∕ r2 − x2 dx = r r2-−-x2-dx 0 0

Using a trig substitution x = r sinθ, it follows dx = r cos

dθ and so

∫ ∫ ∫ √--r---dx = ∘----1----r cos(θ)dθ = r dθ = rθ+ C r2 − x2 1 − sin2θ

Hence changing back to the variable x it follows an antiderivative is l

= r arcsin. Then the length is

( √- ) ( r) 1 -2- π- π- π- rarcsin r − rarcsin r 2 r = r2 − r 4 = r 4.

Note this gives the length of one eighth of the circle and so from this the length of the whole circle should be 2rπ. Here is another example

Here f^′

= 2x and so the initial value problem to be solved is

∘------- dl-= 1 + 4x2,l(0) = 0. dx

Thus, in terms of the definite integral, the length of this curve is

∫ 1∘ ------- 1√ - 1 ( √-) 1√ - 1 (√ - ) 1 +4x2dx = 2 5− 4 ln − 2 + 5 = 2 5+ 4 ln 5+ 2 0

To find an antiderivative, you use the trig. substitution 2x = tanu so dx =

du. Then you find the antiderivative in terms of u and change back to x by using an appropriate triangle as described earlier.