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

162 Chapter 4 Coordinate systems
and so tan −1 (0.8391) could be 40 ◦ or 220 ◦ . Similarly tan105 ◦ = −3.7321 and
tan285 ◦ = −3.7321 and so tan −1 (−3.7321)
( )
could be 105 ◦ or 285 ◦ . The value given
y
on your calculator when calculating tan −1 may not be the actual value of θ we
x
require. In order to clarify the situation it is always useful to sketch the Cartesian coordinates
and the angle θ before embarking on the calculation.
Example4.6 The Cartesian coordinates of P are (4, 7); those of Q are (−5,6). Calculate the polar
coordinates ofPand Q.
Solution Figure 4.13 illustratesthe situation forP.
Then
r = √ 4 2 +7 2 = √ 65 = 8.0623
Note from Figure 4.13 that P is in the first quadrant, that is θ lies between 0 ◦ and 90 ◦ .
Now
( ) ( ) y 7
tan −1 = tan −1
x 4
( ) 7
From a calculator, tan −1 = 60.26 ◦ . Since we know that θ lies between 0 ◦ and 90 ◦
4
then clearly 60.26 ◦ isthe required value.
The polar coordinates of Parer = 8.0623, θ = 60.26 ◦ .
Figure 4.14 illustratesthe situation forQ.
We have
r = √ (−5) 2 +6 2 = √ 61 = 7.8102
From Figure 4.14 wesee that θ lies between 90 ◦ and 180 ◦ . Now
( ) ( ) y 6
tan −1 = tan −1 = tan −1 (−1.2)
x −5
Acalculatorreturnsthevalueof −50.19 ◦ whichisclearlynottherequiredvalue.Recall
that tanθ isperiodic with period 180 ◦ . Hence the required angle is
180 ◦ + (−50.19 ◦ ) = 129.81 ◦
The polar coordinates of Qarer = 7.8102,θ = 129.81 ◦ .
y
P
y
r
7
Q 6
O
u
4
x
–5
O
u
x
Figure4.13
Pis in the firstquadrant.
Figure4.14
Qlies in the second quadrant.