POLAR COORDINATES
301
35.91
Find the equation of the tangent line to the cardioid r = 1 — cos 0 at 6 = ir/2.
Hence, by Problem 35.89, the slope m = -r'/r = -\ = -1. When 0 = -rr/2, x = 0 and
y = l. So, an equation of the line is y — 1 = — x, x + y = 1; in polar coordinates, the line is r(cos 0 +
sin 0) = 1 or r = 1 /(cos 0 + sin 0).
35.92
Show that, if 0 1 is such that r=f(01)=0
pole (0, 0J is 0,.
then the direction of the tangent line to the curve r = f(0) at the
At(0,ftj, r = 0 and r'=/'(0i)- If r'^0, then, by the formula of Problem 35.89,
and so,
and again $ = 8 l .
35.93
Find the slope of the three-leaved rose r = cos30 at the pole (see Fig. 35-19).
When r = 0, cos 30 = 0. Then 36 = ir!2,3-rr/2, or 5irl2, and e = ir/6, •nil, or 5ir/6. By Problem
35.92, tan = 1/V3, «>, or -1/V3, respectively.
Fig. 35-19
35.94
Find the slope of r = 1/0 when 6 = ir/3.
For r = 3/7r and r' = -I/O
2 = -9/ir
2 , Problem 35.89 gives
35.95
Investigate r = 1 + sin 6 for horizontal and vertical tangents.
For horizontal tangents, set tan $ = 0 and solve: cos 0 = 0 or 1 + 2 sin 0 = 0. Hence, 0 is 7r/2, 3-ir/2,
For B = it 12, there is a horizontal tangent at (2,7T/2). For 0 = 7ir/6 and llTr/6, there
and
For 0 = 37T/2, the denominator of tan rf> also is 0, and.
are horizontal tangents at
by Problem 35.92, there is a vertical tangent at the pole. For the other vertical tangents, we set the denominator
and obtain the additional cases 6 = ir/6 and 5ir/6. These yield vertical tangents at
and i
See Fig. 35-20.
r' = sin ft
<£ = 0,. K r' = 0,
7ir/6, or lliT/6.
(sin0 + l)(2sin0-l) = 0