### Video Transcript

Consider the parametric curve π₯ is
equal to one plus the sec of π and π¦ is equal to one plus the tan of π. Determine whether this curve is
concave up, down, or neither at π is equal to π by six.

The question gives us a curve
defined by a pair of parametric equations π₯ is some function of π and π¦ is some
function of π. We need to determine the concavity
of this curve at the point where π is equal to π by six. Letβs start by recalling what we
mean by the concavity of a curve. The concavity of a curve tells us
whether the tangent lines to the curve lie above or below our curve.

In particular, we can find this
information by using the second derivative of π¦ with respect to π₯. We know if d two π¦ by dπ₯ squared
is positive, then our curve is concave upward at this point. We also know if d two π¦ by dπ₯
squared is negative, then our curve is concave downward at this point. So, to find the concavity of our
curve, we just need to find an expression for d two π¦ by dπ₯ squared. However, we canβt do this directly
since weβre given a parametric curve. So, weβre going to need to recall
some of our rules for differentiating parametric curves.

First, we recall we can use the
chain rule to find an expression for dπ¦ by dπ₯. If π¦ is a function of π and π₯ is
a function of π, then dπ¦ by dπ₯ is equal to dπ¦ by dπ divided by dπ₯ by dπ. And this is only valid when the
denominator dπ₯ by dπ is not equal to zero. But we want to find an expression
for d two π¦ by dπ₯ squared. So, we need to differentiate this
with respect to π₯. And we could do this again by using
the chain rule. We would get that d two π¦ by dπ₯
squared is equal to the derivative of dπ¦ by dπ₯ with respect to π divided by dπ₯
by dπ.

And of course, we still require our
denominator dπ₯ by dπ is not equal to zero. So, to find an expression for d two
π¦ by dπ₯ squared on our parametric curve, we need to find dπ¦ by dπ₯. Doing this, we need to find dπ¦ by
dπ and dπ₯ by dπ. And π₯ and π¦ are already given as
functions of π, so we can do this.

Letβs start with finding dπ₯ by
dπ. Thatβs the derivative of one plus
the sec of π with respect to π. To do this, we recall one of our
standard derivative results for trigonometric functions. The derivative of the sec of π
with respect to π is equal to the sec of π times the tan of π. So, because the derivative of the
constant one is equal to zero, we get dπ₯ by dπ is equal to the sec of π times the
tan of π.

We can now do the same to find an
expression for dπ¦ by dπ. Thatβs the derivative of one plus
the tan of π with respect to π. Once again, we need to use one of
our standard derivative results for trigonometric functions. The derivative of the tan of π
with respect to π is equal to the sec squared of π. So, this gives us that dπ¦ by dπ
is equal to the sec squared of π. Now that we found expressions for
dπ¦ by dπ and dπ₯ by dπ, we can use our formula to find an expression for dπ¦ by
dπ₯.

We get dπ¦ by dπ₯ is equal to the
sec squared of π divided by the sec of π times the tan of π. Remember, to find d two π¦ by dπ₯
squared, we need to differentiate this expression with respect to π. So we should simplify this into a
form which is easy to differentiate. To start, weβll cancel the shared
factor of the sec of π in our numerator and our denominator. But this is still the quotient of
two functions. We could differentiate this by
using the quotient rule. However, we could simplify this to
make it even easier.

To help us simplify this, we need
to recall two of our trigonometric identities. The sec of π is equivalent to one
divided by the cos of π, and the tan of π is equivalent to the sin of π divided
by the cos of π. Using our trigonometric identities,
we get dπ¦ by dπ₯ is equal to one over the cos of π all divided by the sin of π
divided by the cos of π. And either by using our rules for
rearranging fractions or by multiplying the numerator and the denominator by the cos
of π, we can just simplify this to get one divided by the sin of π.

And, of course, we know that one
divided by the sin of π is equivalent to the csc of π. And itβs important to note that
this is a much easier function to differentiate than the sec of π divided by the
tan of π. Weβre now almost ready to find our
expression for d two π¦ by dπ₯ squared. Letβs start by finding an
expression for our numerator. Thatβs the derivative of dπ¦ by dπ₯
with respect to π. And we already showed dπ¦ by dπ₯ is
equal to the csc of π. So, we need to differentiate the
csc of π with respect to π.

And this is a standard
trigonometric derivative result. We know the derivative of the csc
of π with respect to π is equal to negative the cot of π times the csc of π. So, we found an expression for the
numerator of d two π¦ by dπ₯ squared. We got negative the cot of π times
the csc of π. Now, all we need to do is divide
this by dπ₯ by dπ. And we already found dπ₯ by
dπ. We got that this was equal to the
sec of π times the tan of π.

So, by using our formula, we get d
two π¦ by dπ₯ squared is equal to negative the cot of π times the csc of π divided
by the sec of π multiplied by the tan of π. Now, thereβs a few different ways
we could proceed with this question. For example, we only need to find
the concavity of our curve at the point where π is equal to π by six. So, at this point, we could just
directly substitute this value into our expression. However, weβre going to simplify
this expression first.

First, weβre going to use the fact
that the cot of π is the same as dividing by the tan of π. So, weβll remove the cot of π in
our numerator and then square the tan of π in our denominator. Now, we want to simplify the csc of
π divided by the sec of π. We know we can rewrite this as one
over the sin of π divided by one over the cos of π. But we can then simplify this by
multiplying the numerator and the denominator by the cos of π.

In our denominator, we can cancel
the cos of π divided by the cos of π to give us one. And in our numerator, we get the
cos of π divided by the sin of π, but this is also equivalent to one divided by
the tan of π. So, by using this, we can rewrite d
two π¦ by dπ₯ squared as negative one divided by the tan cubed of π. And now, itβs easy to substitute
our value of π equal to π by six into this expression.

Doing this, we get d two π¦ by dπ₯
squared of π is equal to π by six is equal to negative one divided by the tan
cubed of π by six. And if we evaluate this expression,
we get negative three times the square root of three. So, weβve shown the second
derivative of π¦ with respect to π₯ is negative when π is equal to π by six. Therefore, our curve must be
concave downwards when π is equal to π by six. Therefore, given the parametric
curve π₯ is equal to one plus the sec of π and π¦ is equal to one plus the tan of
π. We were able to show that this
curve was concave downwards when π was equal to π by six.