### Video Transcript

Second Derivatives of Parametric
Equations

In this video, weβll learn how to
find the second derivatives of parametric equations with respect to π₯ and how to
find higher-order derivatives of parametric equations by using the chain rule. And weβll also cover a variety of
examples and uses of the second derivatives of parametric equations. So letβs start with a pair of
parametric equations. Letβs say that π₯ is equal to some
function π of π‘ and π¦ is equal to some function π of π‘. And we already know how to find an
expression for dπ¦ by dπ₯ assuming π of π‘ and π of π‘ are differentiable and dπ₯
by dπ‘ is not equal to zero.

We saw by applying the chain rule
and the inverse function theorem to our parametric equations π₯ is equal to π of π‘
and π¦ is equal to π of π‘, we get dπ¦ by dπ₯ is equal to dπ¦ by dπ‘ divided by dπ₯
by dπ‘ provided dπ₯ by dπ‘ is not equal to zero. But we want to find an expression
for the second derivative of π¦ with respect to π₯. Of course, the second derivative of
π¦ with respect to π₯ is the derivative of dπ¦ by dπ₯ with respect to π₯. Normally. we would just
differentiate this directly. However, in this case, thereβs a
little problem. We see from our formula for dπ¦ by
dπ₯ that the numerator dπ¦ by dπ‘ is a function in π‘ and the denominator dπ₯ by dπ‘
is also a function in π‘. So in this case, dπ¦ by dπ₯ will be
a function in π‘.

So when we try to evaluate d two π¦
by dπ₯ squared by differentiating dπ¦ by dπ₯ with respect to π₯, weβll have a
problem. Weβre trying to differentiate a
function in π‘ with respect to π₯. However, we can get around this
problem by applying the chain rule. We need to recall the chain rule
tells us if capital πΉ is a function in π‘ and π‘ is a function in π₯, then we can
calculate dπΉ by dπ₯ by calculating dπΉ by dπ‘ and multiplying this by dπ‘ by
dπ₯. In our case, capital πΉ will be the
differential dπ¦ by dπ₯. So by applying the chain rule, we
now have that d two π¦ by dπ₯ squared is equal to the derivative of dπ¦ by dπ₯ with
respect to π‘ multiplied by dπ‘ by dπ₯.

And now we know that dπ¦ by dπ₯ is
a function in π‘. So we can differentiate this with
respect to π‘ by using one of our derivative rules. However, we donβt have π‘ as a
function in π₯. Instead, weβre given π₯ as a
function in π‘. But in actual fact, weβve seen how
to get around this problem before when we found dπ¦ by dπ₯ by using the inverse
function theorem and the chain rule. When we were finding this
expression, instead of finding dπ‘ by dπ₯ directly, we used the inverse function
theorem to find dπ₯ by dπ‘. We can do this in this case as
well. We set π‘ to be the inverse
function π of π₯. Then we can find an expression for
dπ‘ by dπ₯ by using the inverse function theorem. We get that dπ‘ by dπ₯ is equal to
one divided by dπ₯ by dπ‘ provided dπ₯ by dπ‘ is not equal to zero.

Now, all we need to do is
substitute this into our equation. And this gives us a formula for
finding d two π¦ by dπ₯ squared for parametric equations in this form. We 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‘
provided that dπ₯ by dπ‘ is not equal to zero. Weβre now ready to start using this
formula to find the second derivative of π¦ with respect to π₯ for parametric
equations. However, itβs worth reiterating we
are still going to need to use our formula for finding the first derivative of π¦
with respect to π₯ in parametric equations, since this is directly used in our
formula for the second derivative.

Given that π₯ is equal to three π‘
squared plus one and π¦ is equal to three π‘ squared plus five π‘, find the second
derivative of π¦ with respect to π₯.

In this question, weβre given a
pair of parametric equations. And we need to use these to find
the second derivative of π¦ with respect to π₯. And we know we have a formula for
finding the second derivative of π¦ with respect to π₯ for parametric equations. But to use this formula, we do need
to check that our function for π₯ in terms of π‘ and our function for π¦ in terms of
π‘ are both differentiable. In this case, theyβre both
polynomial, so we do in fact know theyβre differentiable. So letβs now recall this
formula. We have 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 this is only valid when dπ₯ by
dπ‘ is not equal to zero.

But to use this formula, we see we
first need to find an expression for dπ¦ by dπ₯. But of course, weβre given π₯ in
terms of π‘ and π¦ in terms of π‘. So to find an expression for dπ¦ by
dπ₯, weβre going to need to use another formula. Once again, for a pair of
parametric equations, π₯ given in terms of π‘ and π¦ given in terms of π‘, we know
dπ¦ by dπ₯ will be equal to dπ¦ by dπ‘ divided by dπ₯ by dπ‘ provided that dπ₯ by
dπ‘ is not equal to zero. So to find dπ¦ by dπ₯, weβre going
to need to first find dπ¦ by dπ‘ and dπ₯ by dπ‘.

Letβs start by finding dπ₯ by
dπ‘. First, we know that π₯ is equal to
three π‘ squared plus one. So dπ₯ by dπ‘ will be the
derivative of three π‘ squared plus one with respect to π‘. Of course, since this is a
polynomial, we can do this term by term by using the power rule for
differentiation. We want to multiply by our exponent
of π‘ and then reduce this exponent by one. This gives us dπ₯ by dπ‘ is equal
to six π‘. And of course, we know that the
constant one doesnβt change as π‘ changes. So its rate of change with respect
to π‘ is equal to zero. So we have dπ₯ by dπ‘ is equal to
six π‘.

We now need to find an expression
for dπ¦ by dπ‘. We know π¦ is equal to three π‘
squared plus five π‘. This means dπ¦ by dπ‘ will be the
derivative of three π‘ squared plus five π‘ with respect to π‘. Once again, we can do this term by
term by using the power rule for differentiation. And of course, when we do this, we
get six π‘ plus five. So weβve shown dπ¦ by dπ‘ is equal
to six π‘ plus five. We now have expressions for both
dπ₯ by dπ‘ and dπ¦ by dπ‘. So we can use these in our formula
to find an expression for dπ¦ by dπ₯. Substituting in our expressions for
dπ¦ by dπ‘ and dπ₯ by dπ‘, we get dπ¦ by dπ₯ is equal to six π‘ plus five divided by
six π‘. And of course, we know this is not
valid when dπ₯ by dπ‘ is equal to zero.

So this wonβt be valid when our
denominator six π‘ is equal to zero. And of course, six π‘ being equal
to zero means that π‘ is equal to zero. In other words, we know that our
formula for dπ¦ by dπ₯ is valid for all values of π‘ except when π‘ is equal to
zero. But in this question, we donβt
actually need to find the values of π‘ where our second derivative will be
valid. So we donβt need to include
this. However, itβs worth thinking
about. Remember, we found that expression
for dπ¦ by dπ₯ to help us find an expression for d two π¦ by dπ₯ squared.

We can see in this expression, we
need to find the derivative of dπ¦ by dπ₯ with respect to π‘. So weβre going to need to
differentiate our expression for dπ¦ by dπ₯ with respect to π‘. Using this, we get the derivative
of dπ¦ by dπ₯ with respect to π‘ is equal to the derivative of six π‘ plus five all
divided by six π‘ with respect to π‘.

Since this is the quotient of two
differentiable functions, we might be tempted to use the quotient rule here. However, remember, our value of π‘
is not allowed to be equal to zero. So in fact, we could just divide
both terms in our numerator by six π‘. So by dividing each term in our
numerator separately by the denominator, we now have the derivative of six π‘
divided by six π‘ plus five divided by six π‘ with respect to π‘. Once again, we know that π‘ is not
equal to zero. So we can simplify our first
fraction to just give us one. And since we need to differentiate
this, it might be easier to rewrite π‘ from the denominator in our numerator by
using our laws of exponents. Instead of dividing by π‘, weβll
multiply by π‘ to the power of negative one.

Now we can evaluate this derivative
term by term by using the power rule for differentiation. First, the derivative of the
constant one with respect to π‘ is equal to zero. Next, the derivative of five times
π‘ to the power of negative one divided by six is equal to negative five π‘ to the
power of negative two divided by six. And once again, weβll use our laws
of exponents to simplify this expression. Instead of multiplying by π‘ to the
power of negative two, weβll instead divide by π‘ squared. So weβve now found the derivative
of dπ¦ by dπ₯ with respect to π‘. We got negative five divided by six
π‘ squared.

Weβre now ready to use this to find
an expression for d two π¦ by dπ₯ squared. Weβve already found an expression
for both the numerator and the denominator of this formula. The numerator, the derivative of
dπ¦ by dπ₯ with respect to π‘, is equal to negative five divided by six π‘
squared. And the denominator dπ₯ by dπ‘ is
equal to six π‘. Now, by substituting these two
expressions into our formula, we get d two π¦ by dπ₯ squared is equal to negative
five divided by six π‘ squared all divided by six π‘. And once again, we know this wonβt
be valid when π‘ is equal to zero. And we can simplify this
expression. Instead of dividing by six π‘, we
can instead multiply by the reciprocal of six π‘. So by using this, we get d two π¦
by dπ₯ squared is equal to negative five divided by six π‘ squared all multiplied by
one over six π‘.

And now we can just simplify this
expression. First, in our denominator, six
multiplied by six simplifies to give us 36. And by using our laws of exponents,
π‘ squared multiplied by π‘ simplifies to give us π‘ cubed. And this gives us our final
answer. If π₯ is equal to three π‘ squared
plus one and π¦ is equal to three π‘ squared plus five π‘, we were able to show that
d two π¦ by dπ₯ squared is equal to negative five divided by six π‘ cubed. This formula can be useful in
helping us find the concavity of curves defined by parametric equations as weβll see
in our next example.

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

Here, weβve been asked about the
concavity of a curve. And we know to check the concavity
of a curve, weβre going to want to find an expression for the second derivative of
π¦ with respect to π₯. We know when the second derivative
of π¦ with respect to π₯ is greater than zero, then the slope of our tangent lines
are increasing. This means that these tangent lines
will be below our curve, and so our curve will be concave upward on this
interval. And we also know when the second
derivative of π¦ with respect to π₯ is less than zero, the slope of our tangent
lines will be decreasing. This means that our tangent lines
will all lie above our curve, so our curve will be concave downward.

So one way to determine the
concavity of our curve is to find an expression for d two π¦ by dπ₯ squared. And we can see that weβre given π₯
and π¦ as a pair of parametric equations. So to find d two π¦ by dπ₯ squared,
weβre going to need to recall our formula for finding these parametric
equations. We have 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 we know this formula wonβt be
valid when dπ₯ by dπ is equal to zero.

And we can see to use this formula,
weβre going to first need to find an expression for dπ¦ by dπ₯. And once again, we can do this by
using our formulas involving parametric curves. We get dπ¦ by dπ₯ is equal to dπ¦
by dπ divided by dπ₯ by dπ. And once again, this formula will
only be valid provided dπ₯ by dπ is not equal to zero. And now we can see that weβre given
π¦ in terms of π and π₯ in terms of π. So we can find expressions for dπ¦
by dπ and dπ₯ by dπ.

Letβs start by finding dπ₯ by
dπ. First, weβre told that π₯ is equal
to the cos of π. So dπ₯ by dπ will be the
derivative of the cos of π with respect to π. And this is a standard
trigonometric derivative result. We know this is equal to negative
the sin of π. So weβve shown dπ₯ by dπ is
negative the sin of π. We can do the same to find dπ¦ by
dπ. Weβre told π¦ is the sin of π, so
dπ¦ by dπ will be the derivative of the sin of π with respect to π. And we know that this is equal to
the cos of π. Now, we could substitute both of
these into our expression for dπ¦ by dπ₯. And this gives us that dπ¦ by dπ₯
is equal to the cos of π divided by negative the sin of π. And we can simplify this using our
trigonometric identities to give us negative the cot of π.

And remember, we found this to find
an expression for d two π¦ by dπ₯ squared. In this numerator, we now need to
differentiate dπ¦ by dπ₯ with respect to π. So we need to differentiate
negative the cot of π with respect to π. And once again, this is a standard
trigonometric derivative result we should commit to memory. The derivative of the cot of π
with respect to π is equal to negative the cosec squared of π.

So this tells us that the negative
cot of π differentiates to give us the cosec squared of π. And now weβre ready to find an
expression for d two π¦ by dπ₯ squared. Weβve already found an expression
for the derivative of dπ¦ by dπ₯ with respect to π, and weβve already found an
expression for the derivative of π₯ with respect to π. So by substituting these
expressions in, we get d two π¦ by dπ₯ squared is equal to the cosec squared of π
divided by negative the sin of π. And remember, by using our
trigonometric identities, dividing by the sin of π is the same as multiplying by
the cosec of π. So we can rewrite this as negative
the cosec cubed of π.

Now we found an expression for d
two π¦ by dπ₯ squared. We just need to evaluate this at π
is equal to π by six. Substituting π is equal to π by
six, we get d two π¦ by dπ₯ squared at π by six is equal to negative the cosec
cubed of π by six. And we can use the fact that the
cosec of π is one divided by the sin of π to make this easier. Doing this, we get negative one
divided by the sin cubed of π by six. And in fact, we know the sin of π
by six is just equal to one-half. So the denominator of our fraction
simplifies to give us one-half cubed. And, of course, one-half cubed is
equal to one-eighth, and negative one divided by one-eighth is equal to negative
eight.

So weβve shown d two π¦ by dπ₯
squared at π is equal to π by six is equal to negative eight. Since this is negative, we can
conclude our curve must be concave downward at this value of π. So we were able to show when π is
equal to π by six, our curve is concave downward.

Now letβs consider what happens
when we try to find higher-order derivatives of parametric equations. Letβs start with d three π¦ by dπ₯
cubed. We can say this will be equal to
the derivative of d two π¦ by dπ₯ squared with respect to π₯. And of course, we already found an
expression for d two π¦ by dπ₯ squared for a pair of parametric equations. And now we have the same problem we
had before. Our numerator is a function in π‘
and our denominator is a function in π‘. So d two π¦ by dπ₯ squared is a
function in π‘, which means to evaluate this derivative, weβre going to need to use
the chain rule. This gives us d three π¦ by dπ₯
cubed is equal to the derivative of d two π¦ by dπ₯ squared with respect to π‘
multiplied by dπ‘ by dπ₯. And just as we did before, we can
apply the inverse function theorem to rewrite dπ‘ by dπ₯ as one divided by dπ₯ by
dπ‘.

And itβs worth pointing out since
we used our formula for d two π¦ by dπ₯ squared, we already know dπ₯ by dπ‘ is not
equal to zero. Now, by applying this, we have d
three π¦ by dπ₯ cubed is equal to the derivative of d two π¦ by dπ₯ squared with
respect to π‘ divided by dπ₯ by dπ‘. And now weβre starting to see a
pattern emerging. In our expression for d two π¦ by
dπ₯ squared, in the numerator, we have the derivative of dπ¦ by dπ₯ with respect to
π‘. And in the denominator, we have dπ₯
by dπ‘. And in our expression for d three
π¦ by dπ₯ cubed, we still have dπ₯ by dπ‘ in our denominator. However, in our numerator, instead
of the first derivative, we have the second derivative. And in fact, this pattern continues
for higher-order derivatives. We get d π π¦ by dπ₯ π is equal
to the derivative of d π minus one π¦ by dπ₯ π minus one with respect to π‘
divided by dπ₯ by dπ‘.

And just like before, we need to be
careful when dπ₯ by dπ‘ is equal to zero. We can then use this formula to
find higher-order derivatives of parametric equations. Weβll see how to do this in our
final example.

If π₯ is equal to four π‘ plus
three and π¦ is equal to two π to the power of π‘ minus π‘ cubed, find d three π¦
by dπ₯ cubed.

Weβve been given a pair of
parametric equations and asked to find the third derivative of π¦ with respect to
π₯. We first need to recall our formula
for finding higher-order derivatives of pairs of parametric equations. We recall the following formula,
which is valid so long as dπ₯ by dπ‘ is not equal to zero. We want to find the third
derivative of π¦ with respect to π₯. So we need our value of π equal to
three. So we can just substitute π is
equal to three into this expression.

To use this, we need to find an
expression for d two π¦ by dπ₯ squared. And we also have a formula to find
d two π¦ by dπ₯ squared. However, to use this formula, we
also need an expression for dπ¦ by dπ₯. Luckily, we also have a formula to
find this for parametric equations which are differentiable. So to find our third derivative, we
first need to find expressions for dπ¦ by dπ‘ and dπ₯ by dπ‘. Letβs start with dπ₯ by dπ‘. Thatβs the derivative of four π‘
plus three with respect to π‘. This is a linear function, so its
derivative will be the coefficient of π‘ which is four. We also need to find dπ¦ by
dπ‘. Thatβs the derivative of two π to
the power of π‘ minus π‘ cubed with respect to π‘. And we can do this term by
term.

First, the derivative of the
exponential function is equal to itself. So our first term differentiates to
give us two π to the power of π‘. Next, we can differentiate negative
π‘ cubed by using the power rule for differentiation. We get negative three π‘
squared. We can substitute these expressions
into our formula for dπ¦ by dπ₯. We get two π to the power of π‘
minus three π‘ squared all divided by four. We can then differentiate this with
respect to π‘ to get d by dπ‘ of dπ¦ by dπ₯ is equal to two π to the power of π‘
minus six π‘ all divided by four. And we can cancel a shared factor
of two.

To find d two π¦ by dπ₯ squared, we
need to divide this by dπ₯ by dπ‘, which we found was four. This gives us d two π¦ by dπ₯
squared is equal to π to the power of π‘ minus three π‘ all divided by eight. To find d three π¦ by dπ₯ cubed, we
need to differentiate this with respect to π‘. And if we do this, we get π to the
power of π‘ minus three all divided by eight. Finally, in our formula for d three
π¦ by dπ₯ cubed, we need to divide this by dπ₯ by dπ‘. So we divide through by four,
giving us π to the power of π‘ minus three all divided by 32.

Letβs now go through the key points
of this video. First, we were able to find an
expression for d two π¦ by dπ₯ squared for a curve defined by a pair of parametric
equations. And we saw this was useful for
helping us to determine the concavity of a curve defined by a pair of parametric
equations. Finally, we also saw how we could
extend this definition to higher-order derivatives in our pairs of parametric
equations.