Calculus: Analytic Geometry and Calculus, with Vectors

8.1 Trigonometric functions and their derivatives 447

Do this. Then adapt the method of the preceding problem to show that ify

is a function for which (2) holds, then there must be constants A and B for which

(1) holds.

15 Prove that if y and u are functions of t having second derivatives and if

y = e-hour

then

d2y

dt+ 2h da + k2y = e hr [dt +

(k2

16 A right circular cylinder (like a tomato can from which the top and bottom
have been removed) has height h and base radius a. We examine the idea that
good approximations to the area of this surface are obtained by triangulating it
and calculating the sum of the areas of the triangles. The surface is partitioned
into m strips each having height h/m like the
one shown in Figure 8.194. On the top and
bottom circular boundaries of each strip, we
place the vertices of 2n isosceles triangles con-
gruent to the triangle ABC of the figure. The
side r1B of the triangle subtends the angle 27r/n
at the center 0 of the top circle. Of the 2n
triangles in the top strip, n have two vertices
on the top circle and the other n have two
vertices on the lower circle. The total number
of triangles congruent to the triangle ABC is
2mn. The first factor in the product

(1) (a sinn J

/ )2

+ (a - a cos n

)2

is half the length of the base AB of the triangle.

The last factor is the altitude, being the length

Figure 8.194

of the hypotenuse of a triangle one leg of which runs from C to the right angle

on the upper circle and the other leg of which runs (in the direction of 0) from the

right angle to the base AB. The sum Smn of the areas of all of the triangles is

the product of (1) and 2mn. Therefore,

When m and n are both large, the quotients having sin (lr/n) in their numerators

are near 1 and we have the approximation

(3) Smn ^' 2irah +

(2-j-) / 2

h n2