### Video Transcript

Thereβll be times where we might
wish to find the slope of the tangent line to a polar curve given in the form π
equals π of π. In this video, weβre going to learn
how we can extend our understanding of finding the derivative of parametric
equations to help us calculate the derivatives of polar curves and, hence, the slope
of a polar curve.

We begin by recalling that we can
convert polar coordinates into Cartesian coordinates using the following equations,
π₯ equals π cos π and π¦ equals π sin π. Weβre going to be considering a
tangent line to a polar curve with the equation π is equal to π of π. So, letβs replace π in these
equations with π of π so that π₯ is equal to π of π cos π and π¦ is equal to π
of π times sin of π. Notice, we now have what looks a
lot like a pair of parametric equations. We have equations for π₯ and π¦ in
terms of a third parameter π.

We recall that for two functions π₯
and π¦ in terms of π‘, their derivative dπ¦ by dπ₯ is dπ¦ by dπ‘ divided by dπ₯ by
dπ‘. So, we replace π‘ with π, and we
see that we can find the slope by calculating dπ¦ by dπ₯. Itβs dπ¦ by dπ divided by dπ₯ by
dπ. But letβs have a look at our
expressions for π₯ and π¦. Theyβre both actually the product
of two functions in π, so we can use the product rule to find their
derivatives.

Remember, the product rule says
that the derivative of the product of two differentiable functions π’ and π£ is π’
times dπ£ by dπ₯ plus three times dπ£ by dπ₯. This means dπ₯ by dπ is the
derivative of π times cos π minus π times the derivative of cos π, which is sin
π. So, thatβs π prime of π cos π
minus π of π sin π. Now, letβs replace π of π with π
and π prime of π with dπ dπ, and we see that dπ₯ by dπ is equal to dπ dπ cos
π minus π sin π.

By performing a similar process for
dπ¦ by dπ, we see that itβs equal to dπ by dπ times sin π plus π cos π. And since we know that dπ¦ by dπ₯
is equal to dπ¦ by dπ divided by dπ₯ by dπ, we can say that dπ¦ by dπ₯ must be
equal to dπ dπ sin π plus π cos π divided by dπ dπ cos π minus π sin
π. Now, what this also means is that
since we find horizontal tangents by finding the points where dπ¦ by dπ₯ is equal to
zero, we can find horizontal tangents by letting dπ¦ by dπ equal to zero, provided
that dπ₯ by dπ does not also equal zero. Similarly, we locate vertical
tangents at the point where dπ₯ by dπ equals zero, assuming that π¦ by dπ₯ is not
equal to zero at these points. Letβs now have a look at the
application of these formulae.

Find the slope of the tangent line
to the curve π equals one over π at π equals π.

Remember, when finding the slope,
we need to evaluate the derivative of that curve at a specific point. The formula weβre interested in for
a polar curve of the form π equals π of π is dπ¦ by dπ₯ equals dπ¦ by dπ divided
by dπ₯ by dπ. And then, we have specific
equations for dπ¦ by dπ and dπ₯ by dπ in terms of π and π. We can see that our equation is
given as π equals one over π. So, weβll begin by working out the
value of dπ dπ.

Letβs start by writing one over π
as π to the power of negative one. And of course, to differentiate a
simple polynomial terms like this one, we multiply by the exponent and then reduce
that exponent by one. So, thatβs negative one times π to
the power of negative two, which is the same as negative one over π squared. Weβre also given that π is equal
to π. So, all we need to do is substitute
everything we know into our formula. dπ¦ by dπ is equal to negative one over π
squared times sin π plus one over π times cos π.

And then, we obtain dπ₯ by dπ to
be equal to negative one over π squared cos π minus one over π sin π. dπ¦ by dπ₯
is the quotient of these too. And of course, we could simplify
this somewhat by factoring one over π, for example. However, weβre going to evaluate
this when π is equal to π, so we may as well go ahead and substitute that in. When we do, we find that the slope
is equal to negative one over π squared times sin π plus one over π times cos π
over negative one over π squared cos π minus one over π sin π.

And in fact, sin of π is zero and
cos of π is negative one, so this simplifies really nicely to negative one over π
over one over π squared. Due to the nature of dividing
fractions, this simplifies further to negative π squared over π, which is simply
negative π. And we found the value of the
derivative at π equals π. And therefore, the slope of the
tangent line to the curve at this point is negative π.

In our next example, weβll look at
how we can use this formulae to locate horizontal and vertical tangent lines.

Find the points at which π equals
four cos π has a horizontal or vertical tangent line.

Remember, the formula for the slope
of the polar curve π is equal to π of π is dπ¦ by dπ₯ equals dπ¦ by dπ over dπ₯
by dπ, where dπ¦ by dπ is equal to dπ by dπ sin π plus π cos π and dπ₯ by dπ
is equal to dπ dπ cos π minus π sin π. Horizontal tangents will occur when
the derivative is equal to zero, in other words, where the numerator is equal to
zero. So, we locate horizontal tangents
by finding the points where dπ¦ by dπ equals zero, assuming that dπ₯ by dπ does
not also equal zero.

Vertical tangents occur where our
denominator dπ₯ by dπ is equal to zero and the numerator dπ¦ by dπ is not equal to
zero. In some sense, we can think of our
gradient as positive or negative infinity here. But strictly speaking, we say that
dπ¦ by dπ₯ is undefined, in other words, where the denominator dπ₯ by dπ is equal
to zero, provided that d π¦ by dπ does not also equal to zero. So, weβre going to need to work out
dπ¦ by dπ and dπ₯ by dπ.

Weβre told that π is equal to four
cos π. We quote the general result for the
derivative of cos π as being negative sin π. And we see that dπ by dπ must be
negative four sin π. This means dπ¦ by dπ is equal to
negative four sin π times sin π plus π times cos π. And π is cos four cos π, so
thatβs plus four cos π cos π. That simplifies to negative four
sin squared π plus four cos squared π. And weβll set this equal to zero to
find the location of any horizontal tangents.

We begin to solve for π by
dividing through by four and then adding sin squared π to both sides of the
equation. Then, we divide through by cos
squared π. And this is really useful because
we know that sin squared π divided by cos squared π is equal to tan squared
π. By finding the square root of both
sides of this equation, remembering to take by the positive and negative square root
of one, we obtain tan π to be equal to plus or minus one.

Taking the inverse tan will give us
the value of π. So, π is equal to the inverse tan
of negative one or the inverse tan of one. That gives us values of π of
negative π by four and π by four. Now, remember, weβre trying to find
the points at which these occur, so we do need to substitute our values of π back
into our original function for π. That gives us π is equal to four
cos of negative π by four or four cos of π by four, which is two root two in both
cases.

And so, we see that we have
horizontal tangent lines at the points two root two π by four and two root two
negative π by four. Weβre now going to work out the
vertical tangent lines. So, weβre going to repeat this
process for dπ₯ by dπ, substituting π and dπ by dπ into our formula for dπ₯ by
dπ. And we see that itβs equal to
negative four sin π cos π minus four cos π sin π. We set this equal to zero, and we
see that we can divide through by negative four.

So, we obtain zero to be equal to
sin π cos π plus cos π sin π, or two sin π cos π. Now, actually, we know that sin of
two π is equal to two sin π cos π, so we can solve this by setting zero equal to
sin of two π. So, two π is equal to the inverse
tan of zero. Now, sin is periodic, and we need
to consider that this is also the case for two π must be equal to π. And since π is greater than
negative π and less than or equal to π, these are actually the only options we
consider.

Dividing through by two, we obtain
π to be equal to zero and π by two. Remember, we substitute this back
into our original equation for π. And we get that π equals four cos
π or four cos π by two, which gives us values of π as four and zero. And we found the points at which π
equals four cos π has horizontal and vertical tangent lines. The horizontal tangent lines are at
the point two root two, π by four and two root two, negative π by four and the
vertical tangent lines are at the points four, zero and zero, π by two.

Weβll now consider how we can find
the slopes of the tangent lines to a polar curve at the tips of the leaves.

Find the slopes of the tangent
lines to π equals two sin three π at the tips of the leaves.

Remember the leaves in our polar
curves are the repeating loops. And here, we recall that the graph
of π equals sin of ππ β and in fact, multiples of sin ππ β has π loops when π
is odd, and two π loops when π is an even integer. In our case, π is three; itβs
odd. So, the graph of π equals two sin
of three π is going to have three loops, or petals. So, how do we find the loops? Well, we begin by finding the two
smallest non-negative values of π for which two sin of three π equals zero. These will describe the beginning
and end point of each of our loops, of course, when π is equal to zero.

By dividing through by two, we see
that sin of three π equals zero. And then, we take the inverse sin
of zero, and we see that three π is equal to zero or, alternatively, π. This means that π must be equal to
zero or π by three. One leaf is, therefore, described
by letting π go from zero to π by three. That might look a little something
like this. This means the tip of its petal
will occur when π is exactly halfway between π equals zero and π equals π by
three. Thatβs π equals π by six.

Since we have three loops, we
continue by adding a third of a full turn. Thatβs two π by three. So, we find that π is equal to π
by six, π by six plus two π over three, which is five π by six, and then we also
subtract two π by three from π by six. And that gives us negative π by
two. Itβs important to substitute these
values into our original equation for π, and we find that π is equal to two
throughout. We can say that the tips of the
leaves are at two, π by six; two, five π by six; and two, negative π by two.

Now, we can alternatively redefine
our final point as negative two, π by two. And this is because a negative
value of π takes us through the origin and out the other side at exactly the same
distance. So, these are actually the exact
same points. Remember, we need to find the
slopes of the tangent lines to our curve at these points. And we recall that we can find the
derivative and, therefore, the slope by using the given formula. The derivative of sin of three π
is three cos of three π. So, dπ by dπ is six cos of three
π.

We can substitute this into our
formula for dπ¦ by dπ₯. And we see that itβs equal to six
cos of three π sin π plus two sin of three π cos π over six cos three π cos π
minus two sin of three π sin π. Weβll evaluate this at the tip of
our first leaf by substituting π equals π by six. And that gives us a slope of
negative root three. We then substitute in π is equal
to five π by six, and we find the slope here is root three. And finally, we substitute in π
equals π by two, which gives us a slope of zero.

And itβs important to realise that
have we substituted the π equals negative π by two, we would have got the same
answer. So, we find the slope at the first
tip is a negative root three, and thatβs at the point two, π by six. The second leaf tip is at two, five
π by six. And the slope there is root
three. And our third is at negative two,
π over two or two, negative π over two. And the slope there is zero.

We can also use the techniques
outlined in this video to find an equation for the tangent or normal to a polar
curve at a point. Letβs see what that might look
like.

Find the equation of the tangent
line to the polar curve with equation π equals cos two π at π is equal to π by
four.

Remember, the equation for a
straight line is given by π¦ minus π¦ one equals π times π₯ minus π₯ one. Here, π is the slope, whereas π₯
one, π¦ one are the point where the tangent line hits the curve. So, weβre going to need to begin by
working out the slope of our tangent line. The formula we can use is dπ¦ by
dπ₯ equals dπ by dπ sin π plus π cos π over dπ dπ cos π minus π sin π.

Well, our π is equal to cos two
π, so dπ by dπ must be equal to negative two sin of two π. We can substitute this into our
equation for dπ¦ by dπ₯, and we get negative two sin two π sin π plus cos two π
cos π all over negative two sin two π cos π minus cos two π sin π. Remember, weβre looking to find the
slope when π is equal to π by four. So, letβs evaluate this when π is
equal to π by four. When we replace π with π by four,
we find dπ¦ by dπ₯ to be equal to one. So, we know the slope of the curve,
but what about π₯ one, π¦ one?

Well, we use the fact that π₯ is
equal to π cos π and π¦ is equal to π sin π. And since π is equal to cos two
π, we find that π₯ is equal to cos two π cos π and π¦ is equal to cos two π sin
π. Then, when π is equal to π by
four, we find that π₯ is equal to cos of two times π by four times cos of π by
four, which is simply equal to zero. Similarly, π¦ is equal to cos of
two times π by four times sin of π by four, which is also zero. So, the equation of the tangent
line to our curve at π is equal to π by four is π¦ minus zero equals one times π₯
minus zero, which is simply π¦ equals π₯.

In this video, weβve seen that we
can use the formula dπ¦ by dπ₯ equals dπ¦ by dπ over dπ₯ by dπ to find the
derivative of a polar curve given in the form π equals π of π, where dπ¦ by dπ
and dπ₯ by dπ are as shown. By setting dπ¦ by dπ equals zero,
we can establish the existence of horizontal tangents. And by setting the dπ₯ by dπ equal
to zero, we find the vertical tangents. We can also combine this formula
with the equation of a straight line to help us find the equation of the tangent and
normal to a curve.