Higher Engineering Mathematics, Sixth Edition

118 Higher Engineering Mathematics

From Pythagoras’ theorem,r=

√

32 + 42 =5 (note that

−5 has no meaning in this context). By trigonometric

ratios,θ=tan−^143 = 53. 13 ◦or 0.927rad.

[note that 53. 13 ◦= 53. 13 ×(π/ 180 )rad= 0 .927rad]

Hence (3, 4) in Cartesian co-ordinates corres-

ponds to (5, 53.13◦) or (5, 0.927rad) 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. 12.3.

P y

3

4

0 x

r

 

Figure 12.3

From Pythagoras’ theorem,r=

√

42 + 32 =5.

By trigonometric ratios, α=tan−^134 = 36. 87 ◦ or

0 .644rad.

Henceθ= 180 ◦− 36. 87 ◦= 143. 13 ◦or

θ=π− 0. 644 = 2 .498rad.

Hence the position of pointPin polar co-ordinate

form is (5, 143.13◦) or (5, 2.498rad).

Problem 3. Express (−5,−12) in polar

co-ordinates.

A sketch showing the position (−5,−12) is shown in

Fig. 12.4.

r=

√

52 + 122 = 13

and α=tan−^1

12

5

= 67. 38 ◦or1.176rad

Hence θ= 180 ◦+ 67. 38 ◦= 247. 38 ◦or

θ=π+ 1. 176 = 4 .318rad

y

P

12

5

0 x

r





Figure 12.4

Thus (−5,−12) in Cartesian co-ordinates corres-

ponds to (13, 247.38◦) or (13, 4.318rad) in polar

co-ordinates.

Problem 4. Express (2,−5) in polar

co-ordinates.

A sketch showing the position (2,−5) is shown in

Fig. 12.5.

r=

√

22 + 52 =

√

29 = 5 .385correct to

3decimalplaces

α=tan−^1

5

2

= 68. 20 ◦or 1.190rad

Hence θ= 360 ◦− 68. 20 ◦= 291. 80 ◦or

θ= 2 π− 1. 190 = 5 .093rad

y

(^0) x
5
2
r
P


Figure 12.5
Thus (2,−5) in Cartesian co-ordinates corresponds
to (5.385, 291.80◦) or (5.385, 5.093rad) in polar
co-ordinates.