Video Transcript
A coordinate system represents a
point in a plane by an ordered pair of numbers called coordinates. Weβre most used to working with
Cartesian coordinates in an π₯π¦-plane. These are directed distances from
two perpendicular axes. But what if this sort of coordinate
system isnβt convenient? Well, as far back as 130 BC, there
are references to a Greek astronomer and mathematician, named Hipparchus, using
angles and chord lengths from a pole to establish the position of stars. This system was formalised by Isaac
Newton, though its name polar coordinates is attributed to Gregorio Fontana in the
18th century.
In this video, weβll learn how to
define and plot points given in polar coordinates and convert between the Cartesian
and polar coordinates of a point. Letβs have a look at the system
itself. We choose a point in the plane
thatβs called the pole. This is a bit like the origin. And we label it π. Then, we draw a half line or a ray
starting at π, called the polar axis. This axis is usually drawn
horizontally to the right and corresponds to the positive π₯-axis in a Cartesian
plane. We add this point π. Then, π is the distance between π
and π. Then, π is the angle, the line
segment ππ makes with the polar axis, measured in a counterclockwise
direction. π is then represented by the
ordered pair π and π. These are the polar coordinates of
π.
What this means is that an angle
measured in a clockwise direction will be negative. We also say that if π is negative,
the points π, π and negative π, π lie on the same line through the origin. And theyβre the same distance π
from π, but on opposite sides. We can see that negative π, π can
also be described as π, π plus π. And in fact, we try to avoid using
negative values for π wherever possible. So it, of course, describes the
length from the origin. Itβs also worth noting that, due to
the nature of the system, we say that two points are coincident if they lie at the
same point. For instance, the points two, 80
degrees and two, 440 degrees are coincident. Their value of π is the same. And 440 is 360 degrees greater than
80. Essentially, we move from 80 to 440
by completing a full turn, thereby ending up at the same point. Letβs use this information to
simply write the polar coordinates of some points.
Consider the points plotted on the
graph. Write down the polar coordinates of
πΆ, giving the angle π in the range π is greater than negative π and less than or
equal to π.
Weβre interested in the point
πΆ. And we want to know its polar
coordinates. Remember, these are of the form π,
π. Letβs add a half line or a ray from
the pole to point πΆ. Our job is going to be to work out
the value of π, thatβs the length of our half line, and π, the angle that this
half line makes with the positive π₯-axis. And since weβre told that π must
be greater than negative π and less than or equal to π, weβre going to travel in a
clockwise direction.
Now, π is quite easy to
calculate. We follow the grid around. And we see that the point is
located exactly one unit from the pole. So π must be equal to one. But what about the angle π? We know that a full turn is two π
radians. And half a turn is π radians. This half a turn is split into 12
subintervals. So each subinterval must represent
π by 12 radians. Our half line travels three of
these subintervals. Thatβs three lots of π by 12,
which is π by four. But weβre travelling in a clockwise
direction. So our value of π for the polar
coordinates of πΆ is negative π by four. And the polar coordinates of πΆ are
therefore one, negative π by four. Notice that had we travelled in a
counterclockwise direction, weβd have, of course, an angle of seven, π by four. But thatβs outside of the range of
π given.
Letβs just see what this might have
looked like for, say, point πΈ. This time, our point lies two units
from the pole. So π must be equal to two. Measuring in a counterclockwise
direction from the positive π₯-axis, we travel eight lots of π by 12, which is two
π by three radians. And therefore, the polar
coordinates are two, two π by three. And thatβs all fine and well. But now, suppose we want to convert
between polar and Cartesian coordinates. What then?
Imagine we have a point π, given
by Cartesian coordinates π₯, π¦ and polar coordinates π, π. We can form a right angle triangle
with this information. We know that the length of the
hypotenuse is π-units. The height of this one is
π¦-units. And its width is π₯-units. Weβll use right angle trigonometry
to form expressions for π₯ and π¦ in terms of π and π. We know that cos of π is equal to
adjacent over hypotenuse. In this case, itβs equal to π₯ over
π. Multiplying both sides of this
equation by π and we find π₯ to be equal to π cos π. Similarly, sin π is equal to
opposite over hypotenuse. This time, when we multiply by π,
we find that π¦ is equal to π sin π. These are the equations we use to
convert from polar coordinates to Cartesian. And although they have been deduced
from our diagram, where π is a positive value and π is greater than zero and less
than π by two, these equations are valid for all values of π and π.
For instance, suppose we wanted to
convert the point with polar coordinates two, negative π by three into Cartesian
coordinates. We see that π is equal to two. And π is equal to negative π by
three. This means π₯ must be equal to two
cos of negative π by three, which is equal to one. And π¦ is equal to two sin of
negative π by three, which is negative root three. In Cartesian form then, our point
is given by the coordinates one, negative root three.
But what about converting in the
other direction? If we want to go the other way, we
use the Pythagorean theorem to find an expression for π in terms of π₯ and π¦. Remember, the sum of the squares of
the smaller two sides in a right angle triangle is equal to the square of the larger
side. So π squared is equal to π₯
squared plus π¦ squared. Or π is equal to the square root
of π₯ squared plus π¦ squared. We know that tan of π is equal to
opposite over adjacent. In this case, thatβs π¦ divided by
π₯. To solve for π, we find the
inverse tan of both sides. So π is equal to the inverse tan
of π¦ over π₯. We do, however, need to be a little
bit careful when converting from polar to Cartesian form. This formula works beautifully for
coordinates plotted in the first quadrant. But for the other quadrants, a
calculator will give us an incorrect value. And so we have a couple of
options. We could plot the coordinates out
and go from there. Or we can quote the following
standard results for values of π between negative π and π or negative 180 and
180.
For points plotted in the first or
fourth quadrant, we use the value of π generated by a calculator. Thatβs π is equal to the inverse
tan of π¦ divided by π₯. For coordinates plotted in the
second quadrant, remember, thatβs the top left quadrant, we add π radians or 180
degrees to the value of π we get from our calculator. And for points plotted in the third
quadrant, we subtract π radians or 180 degrees from the calculator value. Letβs see what this might look
like.
Convert negative two, five to polar
coordinates. Give the angle in radians and round
to three significant figures throughout.
Remember, polar coordinates are
given in the form π, π. There are some general formulae
that we can use to evaluate these. But we do need to be extra careful
of our value for π. This is because we usually measure
in a counterclockwise direction from the positive π₯-axis. And if we draw a sketch of our
Cartesian coordinates, we see we end up with a point plotted in the second
quadrant. But we have a set of rules that can
help us. For points plotted in the first
quadrant and the fourth quadrant, we use the calculator values. For points plotted in the second
quadrant, we add π to the value we get on our calculator. And for points plotted in the third
quadrant, we subtract π from our calculator value.
So letβs begin by working out the
value of π. This is essentially the length of
the line segment that joins our point to the pole or the origin. Itβs given by the square root of π₯
squared plus π¦ squared. And our value for π₯ is negative
two. Our value for π¦ is five. So π is equal to the square root
of negative two squared plus five squared, which is root 29. Correct to three significant
figures, that gives us a value of π as 5.39. We calculate π by evaluating the
inverse tan of five divided by negative two, which is negative 1.190 and so on. Our coordinate lies in the second
quadrant. So we add π radians to this
value. And that gives us 1.9513, which
correct to three significant figures is 1.95 radians. And so in polar form, our
coordinates are 5.39, 1.95.
Weβre now going to consider how to
find the distance between two points given as polar coordinates. We might recall the distance
formula for two Cartesian coordinates π₯ one, π¦ one and π₯ two, π¦ two. Itβs the square root of π₯ one
minus π₯ two squared plus π¦ one minus π¦ two squared. So how can we adapt this formula to
find the distance between two polar coordinates?
Well, for two points π one and π
two given by π one cos π one, π one sin π one and π two cos π two, π two sin
π two. We can see that the distance
between them will be given by the square root of π one cos π one minus π two cos
π two squared plus π one sin π one minus π two sin π two squared. We distribute the parentheses and
recall the fact that cos squared π plus sin squared π is equal to one. So we find that the distance is
equal to the square root of π one squared plus π two squared minus two times π
one times π two times cos π one cos π two plus sin π one sin π two. But then, we know that we can say
that cos of π one minus π two is the same as cos π one cos π two plus sin π one
sin π two. So we can rewrite our formula as
the square root of π one squared plus π two squared minus two times π one times
π two times cos of π one minus π two. Letβs have a look at the
application of this formula.
Find the distance between the polar
coordinates two, π and three, negative three π over four. Give your answer accurate to three
significant figures.
Remember, to find the distance
between two polar coordinates given by π one, π one and π two, π two, we use the
formula the square root of π one squared plus π two squared minus two times π one
π two times cos of π one minus π two. Letβs let π one be two and π one
be π. So π two is three and π two is
negative three π by four. Then, we substitute straight into
this formula. And we see the distance between
them is the square root of two squared plus three squared minus two times two times
three times cos of π minus negative three π by four. That gives us the square root of 13
minus 12 of cos of seven π by four, which is equal to 2.124 and so on. So correct to three significant
figures, we see the distance between our polar coordinates is 2.12 units.
In this video, weβve learned that a
polar coordinate is one of the form π, π, where π is the distance of that point
away from the pole or origin and π is the angle measured in a counterclockwise
direction from the positive π₯-axis. We saw that we can convert from
polar to Cartesian form using the formulae π₯ equals π cos π and π¦ equals π sin
π and the given formulae to convert from Cartesian back into polar form. But we also saw that we need to be
really careful to establish which quadrant our point lies in to make sure we get the
correct value of π. Finally, we saw that the distance
between two points with polar coordinates π one, π one and π two, π two is given
by the square root of π one squared plus π two squared minus two times π one π
two cos of π one minus π two.