### Video Transcript

In this video, weβll learn how we
can use our understanding of polar and Cartesian coordinates to change between
equations in polar and Cartesian form. Weβll consider how these techniques
might help us to recognise graphs of equations written in polar form by converting
to Cartesian or rectangular form and interpreting from there.

Remember, the polar coordinate
system is a way of describing points in a plane by their distance from the origin or
the pole and the angle the line joining this point of the origin makes with the
positive horizontal axis measured in a counterclockwise direction. These are of the form ππ, where
π is the distance from the origin to that point and π is that angle.

We convert from polar to Cartesian
form using the formulae π₯ equals π cos π and π¦ equals π sin π. And these equations are suitable
for all values of π and π. The converse formulae are π
squared equals π₯ squared plus π¦ squared and tan π is equal to π¦ divided by
π₯.

Now in this case, we need to be a
little bit careful establishing the value of π because it works beautifully for
coordinates plotted in the first quadrant. But for other quadrants, the
calculator can give us an incorrect value. And we do have a set of rules that
we can follow for calculating the exact value of π. However, we donβt really need this
formula in this video. Weβre looking to establish how we
can convert between polar equations, one where π is some function of π, and
Cartesian or rectangular equations, where π¦ is some function of π₯. We do, however, use the other three
formulae to perform these conversions. Letβs see what that might look
like.

Convert the equation π₯ squared
plus π¦ squared equals 25 into polar form.

Remember, we convert polar
coordinates to Cartesian or rectangular coordinates using the formulae π₯ equals π
cos π and π¦ equals π sin π. And these are suitable for all
values of π and π. In our original equation, weβve got
π₯ squared and π¦ squared. So letβs use our formulae for π₯
and π¦ to generate expressions for π₯ squared and π¦ squared in terms of π and
π.

Since π₯ is equal to π cos π, it
follows that π₯ squared must be π cos π all squared, which we can distribute and
say that π₯ squared is equal to π squared times cos squared π. Similarly, we see that π¦ squared
must be equal to π sin π all squared, which is equal to π squared sin squared
π.

Now our original equation says that
the sum of these is equal to 25. So we can say that π squared cos
squared π plus π squared sin squared π equals 25. Our next step is to factor π
squared on the left-hand side of this equation. So π squared times cos squared π
plus sin squared π equals 25. But why did we do this?

Well, here is where itβs useful to
know some of our trigonometric identities by heart. We know that cos squared π plus
sin squared π is equal to one for all values of π. So we can replace cos squared π
plus sin squared π in our equation with one. So π squared times one equals
25. Well, we donβt need this one. π squared is simply equal to
25. We solve this equation by taking
the square root of both sides. And we find that π is equal to
five.

Remember, we would usually take
both the positive and negative square root of 25. But since π represents a length,
we donβt need to. π₯ squared plus π¦ squared equals
25 is the same as π equals five in polar form.

Now if we think about what we know
about the equation π₯ squared plus π¦ squared equals 25 and polar coordinates, this
makes a lot of sense. The equation π₯ squared plus π¦
squared equals 25 represents a circle whose centre lies at the origin and whose
radius is the square root of 25, or five. We can also think about what the
equation π equals five means in polar form. Well, it must be all points whose
distance from the origin is five units.

Now of course, if we think right
back to what we know about locus or loci, it follows that this form is a circle
whose centre is the origin and whose radius is five. Letβs now look at converting a
polar equation into rectangular form.

Convert the polar equation π
equals four cos π minus six sin π to the rectangular form.

Remember, we convert from polar
coordinates to Cartesian coordinates or rectangular coordinates using the following
formulae. π₯ is equal to π cos π and π¦ is
equal to π sin π. And these are suitable for all
values of π and π. Our aim is going to be to
manipulate each of our equations so that we have an equation for cos π and sin
π.

Well, if we divide both sides of
our first equation by π, we see that cos π is equal to π₯ over π. Similarly, dividing through by π
in our second equation, and we find sin π equals π¦ over π. We can then replace cos π with π₯
over π and sin π with π¦ over π in our original polar equation. And we see that π is equal to four
times π₯ over π minus six times π¦ over π. This simplifies to four π₯ over π
minus six π¦ over π.

Weβre next going to multiply
everything through by π. And we find that π squared equals
four π₯ minus six π¦. Now weβre obviously not quite
done. We want to convert to rectangular
form. This is usually of the form π¦ is
equal to some function of π₯, although weβre essentially looking for an equation
with π₯ and π¦ as its only variables. So we recall the other conversion
formulae that we use to convert Cartesian to polar coordinates. Thatβs π squared equals π₯ squared
plus π¦ squared. We should be able to see now that
we can replace π squared with π₯ squared plus π¦ squared. So π₯ squared plus π¦ squared
equals four π₯ minus six π¦.

Weβre almost there. You might recognise this
equation. Weβre going to manipulate it by
completing the square. We subtract four π₯ from both sides
and add six π¦. Then weβre going to complete the
square for π₯ and π¦. We halve the coefficient of π₯ β
that gives us negative two β and then subtract negative two squared. So we subtract four. Similarly, we halve the coefficient
of π¦ to get three and then subtract three squared, which is nine. And of course, this is all equal to
zero. Negative four minus nine is
negative 13. So we add 13 to both sides of our
equation. And in rectangular form, our
equation is π₯ minus two all squared plus π¦ plus three all squared equals 13.

Now this is actually a really
useful technique as it tells us something about the shape of the graph. We canβt easily spot what the shape
of the graph given by π equals four cos π minus six sin π looks like. But we do know that a circle whose
centre lies at π, π and whose radius is π has the equation π₯ minus π all
squared plus π¦ minus π all squared equals π squared. So our polar equation, which also
has the rectangular form π₯ minus two all squared plus π¦ plus three all squared
equals 13, must be a circle whose centre is at two, negative three and whose radius
is the square root of 13. Letβs have a look at a similar
example.

Consider the rectangular equation
π₯ squared minus π¦ squared equals 25. Convert the given equation to polar
form. The second part of this question
asks us, which of the following is the sketch of the equation?

So we begin by recalling that we
can convert from polar to Cartesian coordinates using the formulae π₯ equals π cos
π and π¦ equals π sin π. Our equation contains π₯ squared
and π¦ squared. So letβs square each of these
formulae. And when we do, we find that π₯
squared equals π squared cos squared π and π¦ squared equals π squared sin
squared π. We know the difference of these is
equal to 25. Thatβs our rectangular
equation. So π squared cos squared π minus
π squared sin squared π equals 25.

We can then factor π squared. So π squared times cos squared π
minus sin squared π equals 25. But we know that cos two π is
equal to cos squared π minus sin squared π. So letβs replace cos squared π
minus sin squared π with cos two π. That tells us that π squared times
cos two π equals 25. And we can then divide both sides
of this equation by cos two π. Now of course, one over cos π is
sec π. So we find that π squared is equal
to 25 sec two π.

For part two, we need to identify
which of the following is a sketch of the equation. Now we wouldnβt be particularly
easy to sketch the graph of π squared equals 25 sec two π. But we do know the general form of
the graph whose equation is π₯ over π all squared minus π¦ over π all squared
equals one. Itβs a standard hyperbola, centred
at the origin with vertices at plus or minus π, zero and covertices at zero, plus
or minus π.

Letβs rearrange our equation to
equate it to one. To do that, we divide everything
through by 25. And since 25 is five squared, we
can write this as π₯ over five all squared minus π¦ over five all squared equals
one. We know we have a standard
hyperbola with vertices at plus or minus five, zero. And in fact, thereβs only one graph
that satisfies this criteria. Itβs graph A. Of course, it is useful to know
that if we were struggling, we could even try substituting some values of π₯ or π¦
in and sketching the order pairs.

Letβs now have a look at another
example which involves sketching the graph.

Sketch the graph of π equals two
csc π.

Here we have a polar equation. And itβs not particularly easy to
spot what the graph of this function might look like. So instead, weβre going to convert
into rectangular form first. We recall that csc π is one over
sin π. We also know that one of the
formulae we use to convert from polar to Cartesian form is the formula π¦ equals π
sin π. Dividing through by π, and we
obtain the second formula is equivalent to saying sin π equals π¦ divided by
π. So csc π must be equivalent to one
over π¦ over π.

Well, when we divide by a fraction,
we multiply by the reciprocal of that fraction. So we can actually say that csc π
must be equal to π over π¦. By replacing csc π with π over π¦
in our original equation, we find that π is equal to two times π over π¦. Letβs divide through by π. We get one equals two over π¦. Next, weβll multiply by π¦. And we find in rectangular form the
equation is π¦ equals two. And of course, we can now easily
sketch this. Itβs simply the horizontal line
that passes through the π¦-axis at two. This is a great demonstration of
when converting to rectangular form can make sketching a graph given in polar form
much more simple.

In this video, weβve seen that, by
using the formulae for converting between polar and Cartesian coordinates, we can
convert quite easily between polar and rectangular equations. We also saw that this process can
help us sketch more complicated graphs given in polar form.