Basic Engineering Mathematics, Fifth Edition

216 Basic Engineering Mathematics

24.3 Changing from polar to Cartesian

co-ordinates

y

y

O x Q x

P

r



Figure 24.6

From the right-angled triangleOPQin Figure 24.6,

cosθ=

x

r

and sinθ=

y

r

from trigonometric ratios

Hence,x=rcosθandy=rsinθ.

If lengthsrand angleθare known thenx=rcosθand

y=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◦)isshownin

Figure 24.7.

y

y

O x

x

r 4

 32 

Figure 24.7

Now x=rcosθ=4cos32◦= 3. 39

and y=rsinθ=4sin32◦= 2. 12

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◦)isshownin

Figure 24.8.

B

A O

y

x

r 6

 137 

Figure 24.8

x=rcosθ=6cos137◦=− 4. 388

which corresponds to lengthOAin Figure 24.8.

y=rsinθ=6sin137◦= 4. 092

which corresponds to lengthABin Figure 24.8.

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 quiteso essential to draw a sketch. Use

ofx=rcosθandy=rsinθautomatically produces

the correct values and 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 Figure 24.9.

y

A

B

O x

 5 5.16 rad

r 5 4.5

Figure 24.9

x=rcosθ= 4 .5cos5. 16 = 1. 948

which corresponds to lengthOAin Figure 24.9.

y=rsinθ= 4 .5sin5. 16 =− 4. 057

which corresponds to lengthABin Figure 24.9.

Thus,(1.948,−4.057) in Cartesian co-ordinates cor-

responds to (4.5, 5.16rad) in polar co-ordinates.