POLAR COORDINATES
303
35.102 Show that the angle is one-half the angle 0 that the radius vector makes with the polar axis.
35.103 For the spiral of Archimedes, r = ad (6 a 0), show that the angle line is 7T/4 when 0 = 1 and i^-»7r/2 as 0-*+<*>.
35.104 Prove the converse of Problem 35.102.
35.105 Find the intersection points of the curves r = sin 6 and r = cos 0.
I Setting sin 6 = cos 0, we obtain the intersection point (V2/2, 77/4). Substituting (-r, 0 + IT) for(r, 0)
in either equation yields no additional points. However, both curves pass through the pole, which is, therefore, a
second intersection point. [In fact, the curves are two circles of radius |, with centers (0, \) and (|,0),
respectively.]
35.106 Find the angle at which the curves r = sin 0 and r = cos 0 intersect at the point
Clearly
where
and
are the angles between the common radius vector and the two
tangent lines. Hence,
For r = sin 6, r' =
and, therefore, bv Problem 35.98,
For r = cos 0, r' = -sin 0 and
Hence
Therefore.
35.107 Find the angles of intersection of the curves r = 3cos0 and r = l+cos0
Solving the two equations simultaneously yields cos 0 =
and, therefore,
with r =
No other points lie on both curves except the pole. For r = 3 cos 0, r' = -3 sin 0 and tan (li, = —cot 0.
r' = a. By Problem 35.98, tan = rlr' = 0. When 0 = 1, tan +ae,
tan+°° and ifi—nr/2.
r' = asin0.
By Problem 35.98,
tan $ = rlr' = (1 - cos 0)/sin 0 = 2sin
2 (0/2)/[2sin (0/2) cos (0/2)] =
tan (0/2). Hence ^ = 61/2.
If tA=|0, so that r/r' = tan j0 = (1 - cos 0)/sin 0, then
[a = const.], r = a(l-cos0).
dO, Inr = ln[a(l-cos0)]
COS0
0 = 7r/3,57r/3,
For r = 1 + cos 0, r' = —sin 0 and tan 2 = -(1 + cos 0)/sin 6. Here, i/fj and i/> 2 are, as usual, the angles
between the radius vector and the tangent lines. The angle £ between the curves is ij/l- ifi2, and, therefore,
tan £ = (tan i/», - tan i/r 2 ) /(I + tan tfi 1 tan t/»,). Now, at
0 = ir/3, tan i/f, = -1 /V3, tan i/», = V5, and
tan£ = [(-l/V3) + -v
/ 3]/[l + (-l/V3)(-V3)] = l/V3. Hence, £ = ir/6. By symmetry, the angle at
(§,5u73) also is 77/6. The circle r = 3cos0 passes through the pole when 0= ir/2, and the cardioid
passes through the pole when 0 = IT. Hence, by Problem 35.92, those are the directions of the tangent lines,
and, therefore, the curves are orthogonal at the pole.
35.108 For a curve r = /(0), show that the curvature K = [r
2 + 2(r')
2 - rr"]/[r
2 + (r')
2 ]
3 '
2 , where r' =
drldO and r" = d2r/d02.
By definition, K = d/ds. But, = 0 + (Fig. 35-21) and) and
We know by Problem 35.98 that tan i/f = rlr'. Hence, by differentiation,
But, we know that ds/dO =
Hence,
303
35.102 Show that the angle is one-half the angle 0 that the radius vector makes with the polar axis.
35.103 For the spiral of Archimedes, r = ad (6 a 0), show that the angle line is 7T/4 when 0 = 1 and i^-»7r/2 as 0-*+<*>.
35.104 Prove the converse of Problem 35.102.
35.105 Find the intersection points of the curves r = sin 6 and r = cos 0.
I Setting sin 6 = cos 0, we obtain the intersection point (V2/2, 77/4). Substituting (-r, 0 + IT) for(r, 0)
in either equation yields no additional points. However, both curves pass through the pole, which is, therefore, a
second intersection point. [In fact, the curves are two circles of radius |, with centers (0, \) and (|,0),
respectively.]
35.106 Find the angle at which the curves r = sin 0 and r = cos 0 intersect at the point
Clearly
where
and
are the angles between the common radius vector and the two
tangent lines. Hence,
For r = sin 6, r' =
and, therefore, bv Problem 35.98,
For r = cos 0, r' = -sin 0 and
Hence
Therefore.
35.107 Find the angles of intersection of the curves r = 3cos0 and r = l+cos0
Solving the two equations simultaneously yields cos 0 =
and, therefore,
with r =
No other points lie on both curves except the pole. For r = 3 cos 0, r' = -3 sin 0 and tan (li, = —cot 0.
r' = a. By Problem 35.98, tan
tan+°° and ifi—nr/2.
r' = asin0.
By Problem 35.98,
tan $ = rlr' = (1 - cos 0)/sin 0 = 2sin
2 (0/2)/[2sin (0/2) cos (0/2)] =
tan (0/2). Hence ^ = 61/2.
If tA=|0, so that r/r' = tan j0 = (1 - cos 0)/sin 0, then
[a = const.], r = a(l-cos0).
dO, Inr = ln[a(l-cos0)]
COS0
0 = 7r/3,57r/3,
For r = 1 + cos 0, r' = —sin 0 and tan
between the radius vector and the tangent lines. The angle £ between the curves is ij/l- ifi2, and, therefore,
tan £ = (tan i/», - tan i/r 2 ) /(I + tan tfi 1 tan t/»,). Now, at
0 = ir/3, tan i/f, = -1 /V3, tan i/», = V5, and
tan£ = [(-l/V3) + -v
/ 3]/[l + (-l/V3)(-V3)] = l/V3. Hence, £ = ir/6. By symmetry, the angle at
(§,5u73) also is 77/6. The circle r = 3cos0 passes through the pole when 0= ir/2, and the cardioid
passes through the pole when 0 = IT. Hence, by Problem 35.92, those are the directions of the tangent lines,
and, therefore, the curves are orthogonal at the pole.
35.108 For a curve r = /(0), show that the curvature K = [r
2 + 2(r')
2 - rr"]/[r
2 + (r')
2 ]
3 '
2 , where r' =
drldO and r" = d2r/d02.
By definition, K = d
We know by Problem 35.98 that tan i/f = rlr'. Hence, by differentiation,
But, we know that ds/dO =
Hence,