Applications to Geometry

10.5 Applications to Geometry

I will not define time, space, place, and motion, as being well known to all.

Sir Isaac Newton (1642–1727) Let f : [a, b]→R be a differentiable function and assume that its derivative is

continuous. Then the length of the curve y = f (x) between a and b is given by

1 +f ′ (x) 2 dx .

Assuming that f : [a, b]→[0,∞) is continuous, the volume of the solid obtained by revolving the curve y = f (x) around the x-axis is

f (x) 2 dx .

Assuming now that f : [a, b]→[0,∞) is a differentiable function with continuous derivative, the area of the surface of revolution of the graph of f around the x-axis is

10.5 Applications to Geometry 407

f (x) 1 +f ′ (x) 2 dx .

An important example in mathematical physics is played by the following sequence of functions. For any positive integer n, define the Dirac sequence 1 (f n ) n ≥1 by

f n (x) = α

Then # 1 −1 f n (x)dx = 1. Moreover, these functions concentrate (see Figure 10.1), for increasing n, more and more of their “mass” at the origin, in the following sense:

for any ε > 0 there exist δ ∈ (0,1) and an integer N such that for all n ≥ N,

f n (x)dx < 1 and

f n (x)dx +

f n (x)dx < ε .

be continuous and nondecreasing functions such that for each x ∈ [a,b] ,

10.5.1. Let f , g : [a, b] → [0,∞)

f (t) dt ≤

g (t) dt and

f (t) dt =

g (t) dt .

Prove that # b # b

a 1 + f (t)dt ≥ a 1 + g(t)dt .

# Solution. Let F # x x (x) =

a f (t) dt and G(x) = a g(t)dt. The functions F and

G are convex, F (a) = 0 = G(a), and F(b) = G(b), by the hypothesis. We have to prove that

1 + (F ′ (t)) 2 dt ≥

1 + (G ′ (t)) 2 dt .⊓ ⊔

Fig. 10.1 Mass concentration of the Dirac sequence (n = 1, 2, 3). 1 Introduced by the British scientist Paul Adrien Maurice Dirac (1902–1984), Nobel Prize in

Physics 1933, “for the discovery of new productive forms of atomic theory.”

408 10 Applications of the Integral Calculus This is equivalent to showing that the arc length of the graph of F is greater than

or equal to the arc length of the graph of G. This is clear, since both functions are convex, their graphs have common endpoints, and the graph of F is below the graph of G.

10.5.2. Let f be a real-valued function with continuous nonnegative derivative. Assume that f (0) = 0 , f (1) = 1 , and let ℓ

be the length of f on the interval [0, 1] . Prove that

Solution. We apply the elementary inequalities

(a + b) ≤ a 2 +b 2 ≤ a + b,

for any a , b ≥ 0. We have

1 +(f ′ (x)) 2 dx .

Therefore, since f ′ (x) ≥ 0, √

(1 + f ′ (x)) dx ≤ ℓ ≤

(1 + f ′ (x)) dx = 2 .

√ Thus, 2 ≤ ℓ ≤ 2. If ℓ = 2, then necessarily,

1 +(f ′ (x)) 2 =1+f ′ (x), for all x ∈ [0,1]. This implies f ′ (x) = 0, for any x ∈ [0,1]; hence f is constant. The latter contradicts

the conditions f (0) = 0, f (1) = 1. ⊓ ⊔

10.5.3. Let f : [0, 1]→[0,1]

be a concave function of class C 1 such that f (0) =

f (1) = 0 . Prove that the length of the graph of f does not exceed 3 .

Solution. By Rolle’s theorem, there exists c ∈ (0,1) such that f ′ (c) = 0. Since f is concave, it follows that f ′ is decreasing, so f increases in (0, c) and is decreasing in (c, 1). The length of the graph on f in [0, c] is given by

c n −1 $

L [0,c] =

1 +f ′ (x) 2 dx = lim

∑ +f ( ξ k ) 2 0 , → n

k =0

where ξ k ∈ (kc/n,(k + 1)c/n). By the Lagrange mean value theorem, we can assume that

((k + 1)c/n) − f (kc/n)

k )=

c /n

10.6 Independent Study Problems 409 Hence

n −1 $ L [0,c] = lim

n → ∞ ∑ k =0

(c/n) 2 + ( f ((k + 1)c/n) − f (kc/n)) 2

n −1

≤ lim n

→ ∞ ∑ [(c/n) + ( f ((k + 1)c/n) − f (kc/n))] = c + f (c), k =0

since f is increasing. A similar argument shows that L [c,1] ≤ 1 − c + f (c). Therefore L [0,1] ≤ c + f (c) + 1 − c + f (c) ≤ 3. ⊓ ⊔