2. Morfología
2.15. Los casos gramaticales del estonio y sus equivalentes en español
Robots are now widely used in factories in order to reduce labour costs. They vary in complexity depending on their function. Almost all printed circuit boards found in electronic devices such as computers and mobiles are assembled by robots. Figure 4.9 shows a simple robot that can pick up a surface-mount electronic component in one position and place it in another position.
It consists of a rotating arm the length of which can be extended and contracted. On the end of the arm is a hand which can be closed to pick up a component and opened to release it.
r
zero datum
u
Figure 4.9
A pick and place robot.
It is necessary for a computer to carry out calculations in order to find where a com- ponent is located and where to place a component. Decide upon a suitable coordinate system to use when carrying out these calculations.
Solution
If we examine the geometry of the robot then we see that a polar coordinate system would be the most suitable. The centre of the coordinate system should be on the axis of rotation. The length of the arm is then given by r and the orientation of the arm is given by θ relative to an agreed zero datum mark.
Example 4.4
Plot the points P, Q and R whose polar coordinates are (a) 2, 70◦(b) 4, 160◦
(c) 3, 300◦
Solution
Figure 4.10 shows the three points plotted.300° 3 R O x y 70° 2 O x y 160° 4 Q P O x y (a) (b) (c) Figure 4.10
A point can be located by the values of its polar coordinates.
Consider the arm from the origin to the point. The value of r gives the length of this arm. The value of θ gives the angle between the positive x axis and the arm, measuring anticlockwise from the positive x axis.
By studying △OPA, shown in Figure 4.11, we can see that
cos θ = x
r and so x = r cos θ (4.1)
sin θ = y
r and so y = r sin θ (4.2)
Hence if we know the values of r and θ , that is the polar coordinates of a point, we can use Equations (4.1) and (4.2) to calculate x and y, the Cartesian coordinates of the point.
4.4 Polar coordinates 161 P O A y x r y x u Figure 4.11
The polar coordinates are r, θ ; the Cartesian coordinates are x, y.
210° 4 –2 –3.4641 x y Figure 4.12
The Cartesian coordinates can be calculated from the polar coordinates.
Example 4.5
A point has polar coordinates r = 4, θ = 210◦. Calculate the Cartesian coordinates ofthe point. Plot the point.
Solution
The Cartesian coordinates are given by x = r cos θ = 4 cos 210◦= −3.4641y = r sin θ = 4 sin 210◦= −2
Figure 4.12 illustrates the point.
We now look at the problem of calculating the polar coordinates given the Cartesian coordinates. Equations (4.1) and (4.2) can be arranged so that r and θ can be found from the values of x and y. Consider a typical point P as shown in Figure 4.11.
The Cartesian coordinates are (x, y). Suppose that the values of x and y are known. The polar coordinates are r, θ ; these values are unknown. By applying Pythagoras’s theorem to △OPA we see that
r2=x2+y2
and so
r =px2+y2
Note that since r is the distance from O to P it is always positive and so the positive square root is taken.
We now express θ in terms of the Cartesian coordinates x and y. From Figure 4.11 we see that tan θ =y x and hence θ=tan−1 y x In summary we have r =px2+y2 θ=tan−1 y x
However, we need to exercise a little extra care before calculating tan−1
y x
and reading the result from a calculator. As an illustration note that
and so tan−1(0.8391) could be 40◦ or 220◦. Similarly tan 105◦ = −3.7321 and
tan 285◦ = −3.7321 and so tan−1(−3.7321) could be 105◦or 285◦. The value given
on your calculator when calculating tan−1
y x
may not be the actual value of θ we require. In order to clarify the situation it is always useful to sketch the Cartesian coor- dinates and the angle θ before embarking on the calculation.
Example 4.6
The Cartesian coordinates of P are (4, 7); those of Q are (−5, 6). Calculate the polar coordinates of P and Q.Solution
Figure 4.13 illustrates the situation for P. Thenr =p42+72=
√
65 = 8.0623
Note from Figure 4.13 that P is in the first quadrant, that is θ lies between 0◦and 90◦.
Now tan−1 y x =tan−1 7 4
From a calculator, tan−1
7 4
=60.26◦. Since we know that θ lies between 0◦and 90◦
then clearly 60.26◦is the required value.
The polar coordinates of P are r = 8.0623, θ = 60.26◦.
Figure 4.14 illustrates the situation for Q. We have
r =p(−5)2+62=
√
61 = 7.8102
From Figure 4.14 we see that θ lies between 90◦and 180◦. Now
tan−1 y x =tan−1 6 −5 =tan−1(−1.2)
A calculator returns the value of −50.19◦which is clearly not the required value. Recall
that tan θ is periodic with period 180◦. Hence the required angle is
180◦+ (−50.19◦)=129.81◦
The polar coordinates of Q are r = 7.8102, θ = 129.81◦.
P O 7 4 r y x u Figure 4.13
P is in the first quadrant.
O Q 6 –5 y x u Figure 4.14
4.5 Some simple polar curves 163
EXERCISES 4.4
1 Given the polar coordinates, calculate the Cartesian coordinates of each point.
(a) r = 7, θ = 36◦
(b) r = 10, θ = 101◦
(c) r = 15.7, θ = 3.7 radians (d) r = 1, θ = π
2 radians
2 Given the Cartesian coordinates, calculate the polar coordinates of each point.
(a) (7, 11) (b) (−6, −12) (c) (0, 15) (d) (−4, 6) (e) (4, 0) (f) (−4, 0)
Solutions
1 (a) 5.6631, 4.1145 (b) −1.9081, 9.8163 (c) −13.3152, −8.3184 (d) 0, 1 2 (a) r = 13.0384, θ = 57.53◦ (b) 13.4164, 243.43◦ (c) 15, 90◦ (d) 7.2111, 123.69◦ (e) 4, 0◦ (f) 4, 180◦4.5
SOME SIMPLE POLAR CURVES
Using Cartesian coordinates the equation y = mx describes the equation of a line passing through the origin. The equation of a line through the origin can also be stated using polar coordinates. In addition, it is easy to state the equation of a circle using polar co- ordinates.
Equation of a line
Consider all points whose polar coordinates are of the form r6 45◦. Note that the angle
θis fixed at 45◦but that r, the distance from the origin, can vary. As r increases, a line
at 45◦to the positive x axis is traced out. Figure 4.15 illustrates this.
Thus, θ = 45◦is the equation of a line starting at the origin, at 45◦to the positive
x axis.
In general, θ = θc, where θcis a fixed value, is the equation of a line inclined at θcto
the positive x axis, starting at the origin.
Equation of a circle, centre on the origin
Consider all points with polar coordinates 36 θ. Here r, the distance from the origin, is fixed at 3 and θ can vary. As θ varies from 0◦to 360◦a circle, radius 3, centre on the
origin, is swept out. Figure 4.16 illustrates this.
In general r = rcwhere rcis a fixed value, 0◦6 θ 6360◦describes a circle of radius
rc, centred on the origin.
Example 4.7
Draw the curve traced out by r = 3, 0◦6 θ 6180◦.Solution
Here r is fixed at 3 and θ varies from 0◦to 180◦. As θ varies a semicircle is traced out.Figure 4.17 illustrates this.
45° u = 45° O y x Figure 4.15
When θ is fixed and r varies, a straight line from the origin is traced out.
y
x
3
3
Figure 4.16
When r is fixed and θ varies, a circle is swept out. y x O 3 P Q Figure 4.17 As θ varies from 0◦to 180◦a
semicircle is traced out.
y x O P S R Q Figure 4.18
Surface for Example 4.8.
Example 4.8
Describe the surface defined by 1 6 r 6 2, 0◦6 θ 690◦.Solution
Here r varies from 1 to 2 and θ varies from 0◦to 90◦. Figure 4.18 illustrates the surfaceso formed.
At P, r = 1, θ = 0◦; at Q, r = 2, θ = 0◦; at R, r = 1, θ = 90◦; at S, r = 2, θ = 90◦.
We have seen some simple polar curves in Figures 4.16 and 4.17. In general a polar curve is given by the equation r = f (θ ), where the radius r varies with the angle θ .