In this problem we're given
the parametric equations of a plane curve,
asked to find the first
derivative of Y with respect to X,
and the second derivative
of Y with respect to X.
Then find the slope of the tangent line
and the concavity at the given value of T.
So, we can find the
slope of the tangent line
by evaluating the fist
derivative at T equals zero,
and then we can find the concavity
by determining the sine
of the second derivative
at T equals zero.
So to begin let's determine
what point we're referring to
on the curve when T equals zero.
To do this, we'll substitute T equals zero
into the parametric equations.
Which means we'd have X
equals two times sine,
three times zero, which is zero,
and Y would be equal to four
cosine two times zero, which is zero.
Well, the sine of zero is zero,
so X is equal to two times zero, or zero,
and Y is equal to four times cosine zero,
cosine zero is equal to
one, four times one is four.
So the point we're referring to
when we're determining the slope
of the tangent line and the concavity,
would be the point with
coordinates zero four.
Now let's find the first
derivative of Y with respect to X.
Because we have parametric equations,
remember DY DX is equal to
DY DT
divided by DX DT.
So, DY DX,
is equal to, again DY DT
divided by DX DT,
well here's Y, the derivative
of four cosine two T,
with respect to T is going
to require the chain rule.
So we'll have negative four
sine two T times the derivative of two T,
which is two, so times two,
and DX DT is going to be equal to
two cosine three T times three.
Notice how this does simplify,
we have a common factor of two here.
So this gives us negative four sine two T,
divided by three cosine three T.
Now before we find the second derivative,
let's go and evaluate the first
derivative at T equals zero
to determine the slope of the tangent line
at this point here.
At the end we'll verify our results.
So, DY DX,
evaluated at T equals zero,
would be negative four sine zero,
divided by three cosine zero.
Well again, sine zero is zero,
so the numerator would be zero,
the denominator would be three times one
since cosine zero is one.
Zero divided by three is equal to zero,
which means the slope of the tangent line
at T equals zero, or at this point,
would be zero meaning we have
a horizontal tangent line.
Now let's work on finding
the second derivative of Y
with respect to X.
To find the second derivative
of Y with respect to X,
we want to find the derivative
of the first derivative
with respect to X.
But because DY DX is in terms of T,
we'll have to find the derivative
of the first derivative
with respect to T and
then divide by DX DT.
So this is going to take
quite a bit of work.
The second derivative
of Y with respect to X
is going to be equal to the derivative
of the first derivative with respect to T,
which is negative four sine two T
divided by three cosine three T.
Then we'll divide this by DX DT.
So notice how this is going to require
the quotient rule as
well as the chain rule.
So let's begin to set this up,
let's start with our
denominator here, or DX DT.
We found DX DT on the previous slide,
it's going to be, two
cosine three T times three,
or six cosine three T.
Now let's find the
derivative of this quotient
with respect to T.
So we'll have another
fraction in the numerator.
Let's begin with our denominator here,
which remember, when
applying to quotient rule,
is just the denominator squared,
so we'll have nine cosine squared three T.
Now for the numerator we'll have
the denominator times the
derivative of the numerator,
minus the numerator times the
derivative of the denominator.
So again, we'll start
with three cosine three T
times the derivative of
negative four sine two T,
which will be negative four
cosine two T times two.
Applying the chain rule
that'll be negative eight
cosine two T.
Then we'll have minus the numerator
which is negative four sine two T,
times the derivative of the denominator,
the derivative of three cosine three T
would be negative three
sine three T times three,
or negative nine sine three T.
Now let's begin to simplify this,
remember this fraction
bar here means division,
so we're dividing by six cosine three T,
which is the same as multiplying
by one over six cosine three T.
So this factor here is
going to join the factor
of nine cosine squared
three T in the denominator.
So our denominator is going to be 54
cosine cubed three T.
Now looking at the numerator,
notice how the angles of
these cosines are different.
So we'll have negative 24
cosine three T,
cosine two T,
and here we have minus
and then a negative here
and a negative here,
so it's still going to be minus 36
and then sine two T,
sine three T.
Now looking at the
coefficients, 24, 36, and 54,
share a common factor of six,
so for the next step I'm going to go ahead
and factor out a negative
six from the numerator.
So I factor out negative six,
that'll leave us with four cosine three T
cosine two T,
and then because we're
factoring a negative out
it will be plus six
sine two T
sine three T
divided by, 54 is equal to six times nine,
so I'll this as six times
nine cosine cubed three T.
In this form we can see
the sixes simplify out,
leaving us with our second derivative.
Where our denominator is
nine cosine cubed three T,
and the numerator would be,
negative and then the
quantity four cosine three T,
cosine two T
plus six
sine two T
sine three T.
Remember our goal here was to
evaluate this at T equals zero
to determine the concavity
at the point, zero four.
So let's go ahead and
evaluate the second derivative
at T equals zero on the next slide.
Notice when T equals
zero all of the angles
of the trig functions
would be equal to zero.
So we'll have negative then the quantity,
four times cosine zero times cosine zero,
and cosine zero is equal to one.
Then here we'll have six times
sine zero times sine zero,
which is zero.
We'll divide this by nine
times cosine cubed zero,
again cosine zero is equal to one.
So this is nine times one cubed.
If we simply this notice
how we have negative four
over nine.
So notice how the second
derivative is negative,
or less than zero, at T equals zero,
which means the curve is concave down
at the point when T equals zero.
Remember T equals zero corresponds
to the point zero four.
So we can say the curve
is concave down
at the point
zero four,
or when T equals zero.
Now let's go ahead and
verify this graphically.
Here's the graph of our parametric curve
on the coordinate plane.
We already found that when T equals zero,
the point on the curve is this point here,
zero comma four,
and notice how at that
point the tangent line
would be a horizontal line,
meaning the slope is zero,
or the first derivative
would be equal to zero
at this point, which is what we found.
Notice also at this point
the function is concave down,
verifying that at T equals zero,
the second derivative would be negative.
I hope you found this explanation helpful.
