082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

4.5 Some simple polar curves 163
EXERCISES4.4
1 Given the polar coordinates, calculate the Cartesian
coordinatesofeach point.
(a) r=7,θ=36 ◦
(b) r = 10, θ = 101 ◦
(c) r = 15.7, θ = 3.7 radians
(d) r=1,θ= π 2 radians
2 Given the Cartesian coordinates, calculatethe polar
coordinatesofeachpoint.
(a) (7,11) (b) (−6,−12) (c) (0,15)
(d) (−4,6) (e) (4,0) (f) (−4,0)
Solutions
1 (a) 5.6631,4.1145 (b) −1.9081,9.8163
(c) 15, 90 ◦
(c) −13.3152, −8.3184 (d) 0, 1
(d) 7.2111,123.69 ◦
(e) 4,0 ◦
2 (a) r = 13.0384, θ = 57.53 ◦
(f) 4,180 ◦
4.5 SOMESIMPLEPOLARCURVES
UsingCartesiancoordinatestheequationy =mxdescribestheequationofalinepassing
through the origin. The equation of a line through the origin can also be stated using
polar coordinates. In addition, it is easy to state the equation of a circle using polar coordinates.
Equationofaline
Consider all points whose polar coordinates are of the formr̸ 45 ◦ . Note that the angle
θ is fixed at 45 ◦ but thatr, the distance from the origin, can vary. Asr increases, a line
at45 ◦ tothe positivexaxis istraced out.Figure 4.15 illustratesthis.
Thus, θ = 45 ◦ is the equation of a line starting at the origin, at 45 ◦ to the positive
x axis.
Ingeneral, θ = θ c
,where θ c
isafixedvalue,istheequationofalineinclinedat θ c
to
the positivexaxis, startingatthe origin.
Equationofacircle,centreontheorigin
Consider all points with polar coordinates 3̸ θ. Herer, the distance from the origin, is
fixed at 3 and θ can vary. As θ varies from 0 ◦ to 360 ◦ a circle, radius 3, centre on the
origin, issweptout.Figure 4.16 illustratesthis.
Ingeneralr =r c
wherer c
isafixedvalue,0 ◦ θ 360 ◦ describesacircleofradius
r c
, centred on the origin.
Example4.7 Draw the curve traced outbyr = 3,0 ◦ θ 180 ◦ .
Solution Hereris fixed at 3 and θ varies from 0 ◦ to 180 ◦ . As θ varies a semicircle is traced out.
Figure 4.17 illustratesthis.
AtP, θ =0 ◦ ,whileatQ, θ =180 ◦ .