Section 5.7 #64

Area = q r² / 2, s = r q , q in radians. (From formulas on the back cover.)

We are asked to write the area of the shaded region as a function of q and to determine the domain of this function. From the drawings you can see that as q increases so does the area of the shaded region. This region is simply the area of the triangle, A_tri, minus the area of the sector, A_sec. The area of the triangle is 1 base times height, and the area of the sector is q r² / 2. We need only determine the height, h, of the triangle to have all the required unkowns. As h/10 = tan q we have that h = 10 tan q .

Now the A_tri = 1 10 (10 tan q ) = 50 tan q .

And the A_sec = q 10² / 2 = 50 q .

So that Area = A_tri - A_sec = 50 tan q - 50 q = 50 (tan q - q ), 0 £ q < p / 2.

Entering this function, 50 (tan q - q ), into y₁ and setting up a table starting q from 0 with increments of 0.3 gives us the following table:

q	0	0.3	0.6	0.9	1.2	1.5
A	0	0.467	4.21	18.0	68.6	630

The area approaches ¥ as q approaches p /2 as one might suspect from looking at the drawings.

See http://www.uncwil.edu/people/tompkinsj/fall98/trig/ArcLengthAreaInverse.htm for a related problem.

Essential Trigonometry, Math Links

by Jack Tompkins