We want to find the slope of the tangent line
to the parametric curve,
given by x equals four cosine t
and y equals two sin t
at the point t equals pi over six radians.
So to find the slope of the tangent line
at t equals pi over six radians,
we need to find d y d x and evaluate d y d x
at t equals pi over six.
And because we have parametric equations,
d y d x is equal to d y d t divided by d x d t.
Before we find the slope of the tangent line,
let's find the point of tangency
by evaluating the parametric equations
at t equals pi over six radians.
Where the x coordinate at the point of tangency,
would be x of pi over six radians.
And the y coordinate at the point of tangency,
would be y of pi over six radians.
So x of pi over six would be equal
to four times cosine pi over six radians.
And y of pi over six is equal
to two times sin pi over six radians.
Now to find these trig function values,
we can use a reference triangle.
Since pi over six radians is equal to 30 degrees.
Or we could use a unit circle.
And we'll show both.
If this is a 30, 60, 90 reference triangle,
then the 30 degree angle here would be equal
to pi over six radians.
So we can label the opposite side a length of one.
The hypotenuse length two,
and the other leg square root of three.
And therefore, cosine pi over six is equal
to the ratio of the adjacent side of the hypotenuse,
which would be square root three divided by two.
So x equals four times square root of three
divided by two.
Notice how this simplifies
to two square root of three.
Which again is the x coordinate
of the point of tangency.
And then for the y coordinate, we would have
two times sin pi over six
which is equal to the ratio
of the opposite side hypotenuse
which would be 1/2.
And two times 1/2 is equal to one.
Now again we could also find the trig function values,
using the unit circle.
Where notice at pi over six radians,
the x coordinate is square root three divided by two.
Which is the cosine function value.
And the y coordinate is 1/2,
which is the sin function value.
So our point of tangency,
has coordinates 2 square root three, one.
The two square root three makes it hard to plot,
so let's use a decimal approximation.
This would be approximately 3.464, 1.
Looking at our curve,
notice how that would be this point here.
One more thing to mention here,
as t increases, the curve is traced out in this direction
which means our curve has this orientation.
Now our goal here is to find the slope
of the tangent line at this point,
which would be the slope of this line here.
Now we'll find d y d x,
then evaluate d y d x at t equals pi over six radians.
So d y d x,
again is equal to d y d t
divided by d x d t.
Well d y d t would be the derivative of
two sin t with respect to t.
That would be two cosine t.
And d x d t would be four times the derivative
of cosine t with respect to t,
that would be negative four sin t.
Now simplifying, we have negative 2/4
which simplifies to negative 1/2.
And then cosine t divided by sin t
simplifies to cotangent t.
And now we need to evaluate d y d x
at t equals pi over six radians
to find the slope of our tangent line.
Which would be negative 1/2 times
cotangent pi over six.
Looking at our triangle here,
cotangent pi over six is equal to the ratio
of the adjacent side to the opposite side.
Which would be square root three divided by one,
or just square root of three.
So we have negative 1/2 times square root of three.
So the exact slope of the tangent line,
when t is pi over six radians
is negative square root of three divided by two.
Which is approximately negative 0.866.
I also graphed the tangent line using some software
to be more accurate.
And here it is.
And we found the slope of this red tangent line
at this point is negative square root of three
divided by two.
Which is approximately negative 0.866.
I hope you found this helpful.