Higher Engineering Mathematics

CARTESIAN AND POLAR CO-ORDINATES 135

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 triangleOPQin Fig. 13.6.

cosθ=

x

r

and sinθ=

y

r

, from

trigonometric ratios

Hence x=rcosθ and y=rsinθ

Figure 13.6

If lengthsrand angleθare known thenx=rcosθ
andy=rsinθare 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=rcosθ=4 cos 32◦= 3. 39

and y=rsinθ=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=rcosθ=6 cos 137◦=− 4. 388

which corresponds to lengthOAin Fig. 13.8.

y=rsinθ=6 sin 137◦= 4. 092

which corresponds to lengthABin 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 ofx=rcosθandy=rsinθauto-

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.

x=rcosθ= 4 .5 cos 5. 16 = 1. 948