There are two ways in which the position of a point in a plane can be represented. These are
(a) by Cartesian co-ordinates, i.e. (x, y), and (b) by polar co-ordinates, i.e. (r,θ), where r is a
‘radius’ from a fixed point andθis an angle from a fixed point.
13.2
Changing from Cartesian into
polar co-ordinates
In Fig. 13.1, if lengths x and y are known, then the length of r can be obtained from Pythagoras’ theo- rem (see Chapter 12) since OPQ is a right-angled triangle. Hence r2=(x2+y2)
from which, r=
*
x2+y2
Figure 13.1
From trigonometric ratios (see Chapter 12), tanθ= y
x
from which θ=tan−1 y
x
r=x2+y2 and θ= tan−1 y
x are the two for- mulae we need to change from Cartesian to polar co-ordinates. The angleθ, which may be expressed in degrees or radians, must always be measured from the positive x-axis, i.e. measured from the line OQ in Fig. 13.1. It is suggested that when changing from Cartesian to polar co-ordinates a diagram should always be sketched.
Problem 1. Change the Cartesian co-ordinates (3, 4) into polar co-ordinates.
A diagram representing the point (3, 4) is shown in Fig. 13.2.
Figure 13.2
From Pythagoras’ theorem, r=√32+42=5 (note that−5 has no meaning in this context). By trigono- metric ratios,θ= tan−1 43=53.13◦or 0.927 rad. [note that 53.13◦=53.13×(π/180) rad=0.927 rad] Hence (3, 4) in Cartesian co-ordinates corres- ponds to (5, 53.13◦) or (5, 0.927 rad) in polar co-ordinates.
Problem 2. Express in polar co-ordinates the position (−4, 3).
A diagram representing the point using the Cartesian co-ordinates (−4, 3) is shown in Fig. 13.3.
Figure 13.3
From Pythagoras’ theorem, r=√42+32=5. By trigonometric ratios, α=tan−1 34=36.87◦ or 0.644 rad.
Henceθ=180◦−36.87◦=143.13◦or
θ=π−0.644=2.498 rad.
Hence the position of point P in polar co-ordinate form is (5, 143.13◦) or (5, 2.498 rad).
Problem 3. Express (−5,−12) in polar co-ordinates.
A sketch showing the position (−5,−12) is shown in Fig. 13.4. r =52+122 =13 and α=tan−112 5 =67.38◦or 1.176 rad Hence θ=180◦+67.38◦=247.38◦or θ=π+1.176=4.318 rad Figure 13.4
Thus (−5,−12) in Cartesian co-ordinates corres- ponds to (13, 247.38◦) or (13, 4.318 rad) in polar co-ordinates.
Problem 4. Express (2,−5) in polar co-ordinates.
A sketch showing the position (2,−5) is shown in Fig. 13.5. r =22+52=√29=5.385 correct to 3 decimal places α=tan−15 2 =68.20 ◦or 1.190 rad Henceθ =360◦−68.20◦=291.80◦or θ =2π−1.190=5.093 rad Figure 13.5
Thus (2, −5) in Cartesian co-ordinates corres- ponds to (5.385, 291.80◦) or (5.385, 5.093 rad) in polar co-ordinates.
Now try the following exercise.
Exercise 61 Further problems on changing from Cartesian into polar co-ordinates In Problems 1 to 8, express the given Carte- sian co-ordinates as polar co-ordinates, correct to 2 decimal places, in both degrees and in radians. 1. (3, 5) [(5.83, 59.04◦) or (5.83, 1.03 rad)] 2. (6.18, 2.35) (6.61, 20.82◦) or (6.61, 0.36 rad) 3. (−2, 4) (4.47, 116.57◦) or (4.47, 2.03 rad)
B
4. (−5.4, 3.7) (6.55, 145.58◦) or (6.55, 2.54 rad) 5. (−7,−3) (7.62, 203.20◦) or (7.62, 3.55 rad) 6. (−2.4,−3.6) (4.33, 236.31◦) or (4.33, 4.12 rad) 7. (5,−3) (5.83, 329.04◦) or (5.83, 5.74 rad) 8. (9.6,−12.4) (15.68, 307.75◦) or (15.68, 5.37 rad)13.3
Changing from polar into
Cartesian co-ordinates
From the right-angled triangle OPQ in Fig. 13.6. cosθ= x
rand sinθ= y r, from
trigonometric ratios Hence x = r cosθ and y = r sinθ
Figure 13.6
If lengths r and angleθare known then x=r cosθ
and y=r sinθ are the two formulae we need to change from polar to Cartesian co-ordinates.
Problem 5. Change (4, 32◦) into Cartesian co-ordinates.
A sketch showing the position (4, 32◦) is shown in Fig. 13.7.
Now x=r cosθ=4 cos 32◦=3.39 and y=r sinθ =4 sin 32◦ =2.12
Figure 13.7
Hence (4, 32◦) in polar co-ordinates corresponds to (3.39, 2.12) in Cartesian co-ordinates.
Problem 6. Express (6, 137◦) in Cartesian co-ordinates.
A sketch showing the position (6, 137◦) is shown in Fig. 13.8.
x=r cosθ=6 cos 137◦= −4.388 which corresponds to length OA in Fig. 13.8.
y=r sinθ=6 sin 137◦=4.092 which corresponds to length AB in Fig. 13.8.
Figure 13.8
Thus (6, 137◦) in polar co-ordinates corresponds to (−4.388, 4.092) in Cartesian co-ordinates. (Note that when changing from polar to Cartesian co-ordinates it is not quite so essential to draw a sketch. Use of x=r cosθ and y=r sinθ auto- matically produces the correct signs.)
Problem 7. Express (4.5, 5.16 rad) in Cartesian co-ordinates.
A sketch showing the position (4.5, 5.16 rad) is shown in Fig. 13.9.
Figure 13.9
which corresponds to length OA in Fig. 13.9. y=r sinθ=4.5 sin 5.16= −4.057 which corresponds to length AB in Fig. 13.9. Thus (1.948, −4.057) in Cartesian co-ordinates corresponds to (4.5, 5.16 rad) in polar co-ordinates.
13.4
Use of R
→P and P→R
functions on calculators
Another name for Cartesian co-ordinates is rect- angular co-ordinates. Many scientific notation cal- culators possess R→P and P→R functions. The R is the first letter of the word rectangular and the P is the first letter of the word polar. Check the operation manual for your particular calculator to determine how to use these two functions. They make changing from Cartesian to polar co-ordinates, and vice-versa, so much quicker and easier.
Now try the following exercise.
Exercise 62 Further problems on changing polar into Cartesian co-ordinates
In Problems 1 to 8, express the given polar co- ordinates as Cartesian co-ordinates, correct to 3 decimal places. 1. (5, 75◦) [(1.294, 4.830)] 2. (4.4, 1.12 rad) [(1.917, 3.960)] 3. (7, 140◦) [(−5.362, 4.500)] 4. (3.6, 2.5 rad) [(−2.884, 2.154)] 5. (10.8, 210◦) [(−9.353,−5.400)] 6. (4, 4 rad) [(−2.615,−3.207)] 7. (1.5, 300◦) [(0.750,−1.299)] 8. (6, 5.5 rad) [(4.252,−4.233)] 9. Figure 13.10 shows 5 equally spaced holes on an 80 mm pitch circle diameter. Calculate their co-ordinates relative to axes 0x and 0y in (a) polar form, (b) Cartesian form. Calculate also the shortest distance between the centres of two adjacent holes.
x O y Figure 13.10 [(a) 40∠18◦, 40∠90◦, 40∠162◦, 40∠234◦, 40∠306◦, (b) (38.04+j12.36), (0+j40), (−38.04+j12.36), (−23.51− j32.36), (23.51−j32.36) 47.02 mm]