We want to find the points
on the curve given by the parametric equations
where the tangent lines are horizontal
or vertical, and then check our work graphically.
Well we know the value of dy dx gives us a slope
with a tangent line, and since we have parametric equations,
dy dx is equal to dy dt divided by dx dt.
Horizontal tangent lines occur
where the derivative's equal to zero,
which in parametric form would be
when dy dt is equal to zero.
So when dy dt equals zero, and dx dt does not equal zero,
we'll have horizontal tangent lines.
And then where dy dx is undefined,
we would have vertical tangent lines
and notice the derivative would be undefined
when the denominator or dx dt equals zero.
So we can find the vertical tangent lines
by determining where dx dt equals zero again
as long as dy dt does not also equal zero.
Let's start by determining
where we have horizontal tangent lines
by determining where dy dt equals zero.
Notice that y is equal to two sine two t in there
for dy dt
will require the chain rule.
This will be equal to two cosine two t times two
or four cosine two t.
We want to determine the values of t
that make this derivative equal to zero.
You will notice if you divide both sides by four,
we can just determine where cosine two t equals zero.
Well we know from the unit circle
that the cosine function value is zero
at pi over two radians and three pi over two radians.
But because we have a double angle here,
cosine two t will be equal to zero
when two t equals pi over two
and when the angle two t is equal to three pi over two.
So if we solve this for t,
we would multiply both sides by 1/2.
So t is equal to pi over four,
and here we'd have t equals three pi over four.
But we're not done
because this curve is going to be traced
over the interval from zero to two pi,
where we'll find all the values of t
from zero to two pi where dy dt equal zero.
What we can do now is go back
to where we had two t equals pi over two,
and cosine two t would be equal to zero
when two t is equal to pi over two,
plus the period of cosine
where pi over two plus two pi would be
equal to five pi over two.
Multiplying both sides by 1/2,
you would have t equals five pi over four.
Next, two t can also equal three pi over two
plus two pi radians
which would be seven pi over two radians.
Multiplying by 1/2 we have t equals seven pi over four.
Notice how this is still less than two pi radians.
So all these values of t would make dy dt equal to zero,
and therefore, we should have horizontal tangent lines
at the points with these given values of t.
First, go ahead and find the points
at these values of t using our parametric equations.
Let's start with t equals pi over four
which means x is equal to four cosine pi over four
and y is equal to two sine,
two times pi over four would be pi over two,
so here we'll have four times cosine pi over four is equal
to square root two over two.
If we need to, we can verify it here on the unit circle,
if our cosine data is equal to the x coordinate.
So four times square root two over two
would be two square root two,
and then y equals two sine pi over two
where sine pi over two is equal to one.
So we have two times one equals two.
So one point where we have a horizontal tangent line
would have an x coordinate of two square root two
and a y coordinate of positive two.
Now let's consider t equals three pi over four.
Again, subbing into our parametric equations
we have x equals four cosine three pi over four,
and y equals two sine three pi over four times two
would be three pi over two.
So cosine three pi over four, going back to our unit circle,
would be negative square root two over two.
So now we have four times negative square root two over two
giving us negative two square root two,
and then sine three pi over two is equal to negative one.
So we have two times negative one equals negative two.
So the next point where we have a horizontal tangent line
would have coordinates negative two square root two
comma negative two.
Next, we have t equals five pi over four.
So we have x equals four times cosine five pi over four,
and y equals two times sine two times pi over four
or five pi over two.
Cosine five pi over four, as you see here,
is negative square root two over two.
So we have four times negative square root two over two
which equals negative two square root two,
and then we have sine five pi over two.
Five pi over two is coterminal to pi over two.
Just notice the sine function value would be equal to one.
So we have two times one which is equal to two.
So the next point would be negative two square root two
comma two.
We can probably figure out the next point just
by looking in the pattern here.
Let's go ahead and find it.
If t equals seven pi over four,
then x equals four times cosine seven pi over four,
and y equals two times sine of two times seven pi over four
or seven pi over two.
Cosine seven pi over four, going back to our unit circle,
is equal to positive square root two over two.
So, four times positive square root two over two
is two square root two,
and then for two times sine seven pi over two,
seven pi over two is coterminal to three pi over two,
and therefore, the sine function value
would be negative one.
We have two times negative one equals negative two.
So the last part we have a horizontal tangent line,
would have an x coordinate of positive two square root two
and a y coordinate of negative two.
Okay now we just have to determine
where we have vertical tangent lines
so now we determine where dx dt equals zero.
Let's go ahead and do this on the next slide.
This will actually be much faster.
Dx dt is going to be equal to negative four sine t,
and we want to determine where this is equal to zero.
And if we divided both sides by negative four,
we really just want to determine where sine t equals zero.
So gang, quickly going back to the unit circle,
since sine data is equal to y,
notice the y coordinate is zero at zero radians,
as well as pi radians.
So, our solution's here are t equals zero,
and t equals pi radians.
So to find the points where we have vertical tangent lines,
we'll substitute these values of t
into the parametric equations.
So, when t equals zero,
notice that x would be equal to four cosine zero,
and y would be equal to two sine zero.
Cosine zero is equal to one,
so the x coordinate would be four.
Sine zero is equal to zero
so the y coordinate would be zero.
So the location, so one point
where we have a vertical tangent line
would have coordinates for zero.
And then lastly, when t equals pi radians,
we would have x equals four times cosine pi,
and y equals two sine two times pi or two pi.
Cosine pi is equal to negative one,
four times negative one is negative four,
and sine two pi is zero so the y coordinate here is zero.
So their location would be at the point negative four zero.
So we found there are two points
where we expect a vertical tangent line,
and we have four points
where we expect horizontal tangent lines.
So let's go ahead and verify this graphically.
Here's the graph of the plane curve
using the parametric equations.
Notice how there are four points
where we'd have horizontal tangent lines.
Notice there are four points
where we'd have horizontal tangent lines.
Here, here, here and here.
And there are two points
where we'd have vertical tangent lines here and here.
So this graph does verify our work.
Let's go ahead and label the coordinates.
So this graph does verify our work,
and just in case you were wondering,
two square root two is approximately 2.83
or a little less than three
as we see here on the graph
from the points that we plotted.
I hope you found this explanation helpful.