We want to find the slope of
the tangent line to the polar curve,
R equals two minus four cosine theta
at theta equals pi over three radians.
Let's begin by determining the point of tangency,
and because we have a polar curve,
we know the points are going to be in the form of
R coma theta, or in our case,
theta is equal to pi over three.
We'll substitute pi over three for theta into our equation,
to determine the first coordinate, R.
We would have R equals two minus four
times cosine pi over three.
Looking at our reference trying to below,
cosine pi over three is equal to the ratio
of the adjacent side of the hypotenuse,
and therefore, cosine pi over three equals one half.
So, we have R equals two minus four times one half,
which is equal to two minus two, which equals zero.
The point of tendency has coordinates
zero coma pi over three,
which means that point of tangency is at the pole here,
and the curve is traced out in this direction here,
where at theta equals zero radians,
we would be at this point,
so because the orientation is in this direction,
at this point, we're looking for the slope
of the tangent line in this direction here,
or the tangent line with a positive slope.
Notice how there's also a tangent line
with a negative slope at this point,
but we're looking for the slope of this tangent line here.
To find the slope of this tangent line,
we will find DY DX,
and evaluate this at theta equals pi over three radians.
Where R equals F of theta,
which is equal to two minus four cosine theta,
and therefore, F prime of theta is equal to,
well, we would have zero,
and then negative four times the derivative of cosine theta,
which equals negative sine theta,
so F prime of theta is equal to positive four sine theta.
Therefore, DY DX is going to be equal to this fraction here,
where the numerator is F prime of theta
times sine theta plus F of theta times cosine theta.
We would have four sine theta times sine theta,
and then we would have plus F of theta,
which is the quantity two minus four
cosine theta times cosine theta.
Then, your denominator is going to be F prime of theta times
cosine theta minus F of theta times sine theta.
We would have four sine theta
times cosine theta,
minus the quantity two minus four cosine theta
times sine theta.
Before we evaluate this at theta equals pi over three,
let's simplify this by clearing the parentheses
in the numerator and denominator.
Let's do this on the next slide.
We will distribute here, as well as here.
We have DY DX is equal to,
well, here we have four sine squared theta,
and then plus two times cosine theta,
so plus two cosine theta,
and then we have plus,
we would have negative for cosine square theta,
or minus four cosine squared theta.
In our denominator, we have four sine theta cosine theta,
and then we have minus two sine theta,
and then we have minus negative four
cosine theta sine theta,
which becomes plus four,
let's write this as sine theta cosine theta.
Notice how our denominator does have two like terms,
here and here.
Let's go ahead and combine those.
The numerator is going to stay the same.
We could perform a double angle substitution here,
but we will go ahead and leave it in this form.
Our denominator is now going to be eight sine theta
cosine theta minus two sine theta.
Now, we will evaluate our derivative function
at theta equals pi over three.
Let's go ahead and do this on the next slide.
Looking at a reference triangle here,
notice that sine pi over three radians
is going to be equal to the ratio
of the opposite side of the hypotenuse,
or square root three divided by two.
As we saw before,
cosine pi over three is equal to one half.
Therefore, DY DX,
evaluated at theta equals pi over three
is going to be equal to,
the numerator is going to be
four times sine pi over three squared,
that would be the square root of three
divided by two squared,
plus two times cosine pi over three is equal to one half,
minus four times cosine pi over three squared
would be one half squared.
In our denominator,
we have eight times sine pi over three,
that's square root three divided by two,
times cosine pi over three, that's one half,
and then we have minus two sine pi over three,
which is square root three divided by two.
Notice here, when we square
square root three divided by two,
we will have a denominator of four,
so this simplifies to square root three squared, or three.
Here, we have plus one.
One half squared is one fourth,
four times one fourth is one, so we have minus one.
In our denominator, these two simplify two one,
this simplifies to a two,
so we have two square root three,
and then here we have minus square root three.
Simplifying one more time,
we have three divided by square root three.
Let's go ahead and get a decimal approximation.
Three divided by square root three
is approximately one point seven three two one.
Again, this is the slope of our tangent line
at theta equals pi over three radians.
Going back to our graph,
notice how this does verify we found
the slope of the tangent line with a positive slope,
in this direction here.
Because the directions don't say round,
we should enter the exact slope as
three divided by square root three.
I hope you found this helpful.