### Video Transcript

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.