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CHAPTER 35
Fig. 35-20
35.96
Find the slope of r = 2 + sin 6 when 6 = Tr/6.
By Problem 35.89, the slope
35.97
Find the slope of r = sin
3 (0/3) when 0 = ir/2.
When 0 = ir/2, r' =sin
2 (0/3)= \ and r=k- Hence, by Problem 35.89, tan = -r'/r = -2.
35.98
Show that the angle t/> from the radius vector OP to the tangent line at a point P(r, 6) (Fig. 35-21) is given by
By Problem 35.89
provided cos 0 ^Q. A limiting process yields the same result as cos 0—>0.
Fig. 35-21
35.99
Find tan ty (see Problem 35.98) for r = 2 + cos 0 at 0 = ir/3.
At 0 = ir/3, /- = 2+| = i, and r' = -sin 6 = -V3/2, so tan i/» = r/r' = -5/V3.
35.100 Find tan At 0 = 7T/4, r = 2(l/V2) = V2, and r' = 6 cos 30 = 6(-1 /V2) = - 3 V2. Hence, tan i/> = rlr' = - j.
35.101 Show that, at each point of r = ae , the radius vector makes a fixed angle with the tangent line. (That is why
the curve r = ae
is called an equiangular spiral.)
tan i/f = rlr', where r' = dr/dO.
r' = ace
ce . Hence, by Problem 35.98, tan i/f = rlr' = 1 /c. So, i/» = tan '(1/c).