### Video Transcript

Second Derivatives of Parametric
Equations

In this video, we will learn how to
find the second derivative of curves defined parametrically by applying the chain
rule. To do this, letβs start with a pair
of parametric equations: π₯ is equal to the function π of π‘ and π¦ is equal to the
function π of π‘. We want to find an expression for
the second derivative of π¦ with respect to π₯. We canβt do this directly because
π¦ is not given as a function in π₯. If it was, we could just
differentiate π¦ with respect to π₯ twice. So instead, because π¦ is given as
a function in π‘ and π₯ is given us a function in π‘, we can apply the chain rule to
find an expression for dπ¦ by dπ₯.

We recall if π is a differentiable
function in π‘ and π is a differentiable function in π‘, then by applying the chain
rule and the inverse function theorem, we can show the derivative of π¦ with respect
to π₯ is equal to dπ¦ by dπ‘ divided by dπ₯ by dπ‘, provided dπ₯ by dπ‘ is
nonzero. We want to use this to find an
expression for the second derivative of π¦ with respect to π₯. Remember, we can find this by
differentiating dπ¦ by dπ₯ with respect to π₯. And once again we have the same
problem. We can note that dπ¦ by dπ₯ is a
function in π‘, since dπ¦ by dπ‘ is a function in π‘ β particularly itβs π prime of
π‘ β and dπ₯ by dπ‘ is also a function in π‘ β itβs π prime of π‘. So we canβt just directly
differentiate this with respect to π₯. Weβre once again going to need to
use the chain rule.

To help us apply the chain rule to
this example, we can start by recalling that the chain rule tells us if capital πΉ
is a function in π‘ and π‘ in turn is a function in π₯, then the derivative of
capital πΉ with respect to π₯ is equal to the derivative of capital πΉ with respect
to π‘ multiplied by the derivative of π‘ with respect to π₯. In this case, weβre going to set
our function capital πΉ to be dπ¦ by dπ₯. And this is a function in π‘. And weβre differentiating this with
respect to π₯. Applying the chain rule to this
then gives us the derivative of dπ¦ by dπ₯ with respect to π‘ multiplied by dπ‘ by
dπ₯.

And now we can see weβve simplified
this expression slightly. dπ¦ by dπ₯ is a function in π‘, and now weβre
differentiating this with respect to π‘. So we can attempt to do this by
using our derivative results. However, we can notice weβre trying
to differentiate π‘ with respect to π₯, but weβre not given π‘ as a function in
π₯. Instead, weβre given π₯ as a
function in π‘. But weβve seen how to get around
this problem before when we tried to evaluate dπ¦ by dπ₯ by using the chain rule and
the inverse function theorem. Applying the inverse function
theorem to dπ‘ by dπ₯, we can show itβs equal to one divided by dπ₯ by dπ‘, provided
the denominator is not equal to zero.

We can now rewrite dπ‘ by dπ₯ by
using the inverse function theorem to find an expression for d two π¦ by dπ₯
squared. This gives us the second derivative
of π¦ with respect to π₯ is equal to the derivative of dπ¦ by dπ₯ with respect to π‘
divided by dπ₯ by dπ‘, provided dπ₯ by dπ‘ is nonzero. And this is a very useful
result. If π¦ and π₯ are given
parametrically, then we can find an expression for d two π¦ by dπ₯ squared by
differentiating dπ¦ by dπ₯ with respect to π‘ and dividing this by the derivative of
π₯ with respect to π‘. And then itβs worth noting to use
this formula, we need to find an expression for dπ¦ by dπ₯. And to do this, we need to use our
other formula. Letβs now see some examples of
applying this formula to determine the second derivatives of parametric
equations.

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 asked to
find the second derivative of π¦ with respect to π₯. And we can notice something
interesting. π¦ is not given as a function in
π₯. Instead, weβre given a pair of
parametric equations in terms of the variable π‘. This means weβre going to need to
differentiate this by using parametric differentiation.

And before we recall our formula
for finding the second derivative of parametric equations, we can note that we know
π₯ is a differentiable function in π‘ and π¦ is a differentiable function in π‘
because theyβre both polynomials. And this helps justify that we can
use this formula to determine the second derivative of π¦ with respect to π₯, which
we recall is equal to the derivative of dπ¦ by dπ₯ with respect to π‘ divided by dπ₯
by dπ‘. And this is, of course, provided
that dπ₯ by dπ‘ is nonzero.

To use this to determine the second
derivative of π¦ with respect to π₯, we first need to find an expression for dπ¦ by
dπ₯. And we can recall if π₯ and π¦ are
given as a pair of parametric equations in terms of a variable π‘, then the
derivative of π¦ with respect to π₯ is equal to the derivative of π¦ with respect to
π‘ divided by the derivative of π₯ with respect to π‘. And once again, this is provided
that π¦ and π₯ are differentiable functions in π‘ and dπ₯ by dπ‘ is not equal to
zero.

Weβre now ready to start finding
our expression for d two π¦ by dπ₯ squared. We see we first need to find
expressions for dπ¦ by dπ‘ and dπ₯ by dπ‘. Letβs start by finding dπ¦ by
dπ‘. This is equal to the derivative of
three π‘ squared plus five π‘ with respect to π‘. Since this is a polynomial, we can
do this term by term by using the power rule for differentiation. We multiply by the exponent of π‘
and then reduce this exponent by one. We get six π‘ plus five. We can follow the same process to
determine dπ₯ by dπ‘. Thatβs the derivative of three π‘
squared plus one with respect to π‘. Once again, we apply the power rule
for differentiation term by term. This time, we get six π‘.

We can now substitute these into
our formula for dπ¦ by dπ₯. This then gives us that dπ¦ by dπ₯
is equal to six π‘ plus five divided by six π‘. And we could leave this expression
like this. However, we know in our formula for
d two π¦ by dπ₯ squared, weβre going to need to differentiate this with respect to
π‘. And although we could evaluate this
derivative by using the quotient rule, itβs easier to divide both terms in our
numerator by six π‘. And since six π‘ divided by six π‘
is equal to one and five divided by six π‘ can be rewritten as five-sixths π‘ to the
power of negative one, weβve shown dπ¦ by dπ₯ is equal to one plus five-sixths π‘ to
the power of negative one.

And then we can just differentiate
this by using the power rule for differentiation. And this is much easier to
differentiate because we can just do this by using the power rule for
differentiation. We could now start substituting our
expressions into this formula. However, itβs easier to evaluate
the numerator of the fraction on the right-hand side of the equation separately. We want to find the derivative of
dπ¦ by dπ₯ with respect to π‘. Thatβs the derivative of
five-sixths π‘ to the power of negative one with respect to π‘. And we can evaluate this term by
term by using the power rule for differentiation. We get negative five-sixths π‘ to
the power of negative two.

And we can simplify this slightly
by using our laws of exponents. Multiplying by π‘ to the power of
negative two is the same as dividing by π‘ squared. So we can rewrite this as negative
five over six π‘ squared. Weβre now ready to substitute this
and our expression for dπ₯ by dπ‘ into our formula for d two π¦ by dπ₯ squared. This then gives us the second
derivative of π¦ with respect to π₯ is equal to negative five over six π‘ squared
all divided by six π‘. And we can then simplify this. Dividing by six π‘ is the same as
multiplying by one over six π‘. And we can then evaluate this. Six π‘ squared times six π‘ is 36π‘
cubed. So this gives us negative five over
36π‘ cubed, which is our final answer.

Therefore, we were able to show if
π₯ is equal to three π‘ squared plus one and π¦ is equal to three π‘ squared plus
five π‘, then the second derivative of π¦ with respect to π₯ is equal to negative
five divided by 36π‘ cubed.

Letβs now see an example where we
need to evaluate the second derivative of a pair of parametric equations at a given
point.

If π¦ is equal to negative five π₯
cubed minus seven and π§ is equal to three π₯ squared plus 16, find the second
derivative of π§ with respect to π¦ at π₯ is equal to one.

In this question, weβre asked to
find the second derivative of π§ with respect to π¦. And weβre asked to do this when π₯
is equal to one because weβre not given π§ as a function in π¦. Instead, weβre given π§ and π¦ as a
pair of parametric equations. So weβre going to want to do this
by using parametric differentiation. And to do this, we should first
note that π¦ and π§ are polynomials, so theyβre both differentiable functions in
π§. We can then recall if π§ is a
differentiable function in π₯ and π¦ is a differentiable function in π₯, then the
second derivative of π§ with respect to π¦ is equal to the derivative of dπ§ by dπ¦
with respect to π₯ divided by dπ¦ by dπ₯.

However, we canβt just directly use
this formula to answer this question, since we canβt differentiate π§ with respect
to π¦ directly. Instead, weβre going to need to use
the chain rule, which, if we use in conjunction with the inverse function theorem,
tells us dπ§ by dπ¦ will be equal to dπ§ by dπ₯ divided by dπ¦ by dπ₯. And this will now allow us to
answer this question. We need to find expressions for dπ¦
by dπ₯ and dπ§ by dπ₯. Letβs start by finding an
expression for dπ¦ by dπ₯. Thatβs the derivative of negative
five π₯ cubed minus seven with respect to π₯. This is a polynomial expression in
π₯, so we can do this term by term by using the power rule for differentiation. We multiply by the exponent of π₯
and then reduce this exponent by one. This gives us dπ¦ by dπ₯ is
negative 15π₯ squared.

We can follow the same process to
find dπ§ by dπ₯. Itβs the derivative of three π₯
squared plus 16 with respect to π₯. And we apply the power rule for
differentiation term by term to get six π₯. Now that we have expressions for
dπ§ by dπ₯ and dπ¦ by dπ₯, we can substitute these into our formula to find dπ§ by
dπ¦. Doing this, we get that dπ§ by dπ¦
is equal to six π₯ divided by negative 15π₯ squared, and we can simplify this. First, we can cancel a shared
factor of π₯ in the numerator and denominator. Next, we can cancel a shared factor
of three in the numerator and denominator. This then leaves us with negative
two divided by five π₯.

Now we could substitute these
expressions into our formula for d two π¦ by dπ₯ squared. However, itβs easier to calculate
the expression in the numerator of the right-hand side of the equation
separately. This then gives us the derivative
of dπ§ by dπ¦ with respect to π₯ is equal to the derivative of negative two over
five π₯ with respect to π₯. And we can evaluate this derivative
by rewriting negative two over five π₯ as negative two-fifths π₯ to the power of
negative one. We can then differentiate this by
using the power rule for differentiation. We multiply by the exponent of π₯
and reduce this exponent by one. This gives us two-fifths π₯ to the
power of negative two, which we can rewrite as two over five π₯ squared.

We can now substitute these
expressions into our formula for d two π§ by dπ¦ squared. This then gives us the second
derivative of π§ with respect to π¦ is equal to two divided by five π₯ squared all
divided by negative 15π₯ squared. And we can simplify this. Dividing by negative 15π₯ squared
is the same as multiplying by the reciprocal of negative 15π₯ squared. And since five π₯ squared
multiplied by 15π₯ squared is 75π₯ to the fourth power, this simplifies to give us
negative two divided by 75π₯ to the fourth power.

But weβre not done yet. Remember, the question wants us to
determine the second derivative of π§ with respect to π¦ evaluated when π₯ is equal
to one. And we can evaluate this by
substituting π₯ is equal to one into our expression for d two π§ by dπ¦ squared. We get negative two divided by 75
times one to the fourth power, which we can calculate is equal to negative two
divided by 75. Therefore, we were able to show if
π¦ is equal to negative five π₯ cubed minus seven and π§ is equal to three π₯
squared plus 16, then the second derivative of π§ with respect to π¦ at π₯ is equal
to negative one evaluates to give negative two over 75.

In our next example, weβll have to
use parametric differentiation to determine the second derivative of π¦ with respect
to π₯. However, one of our parametric
equations will be defined implicitly.

If π₯ is equal to two times the sec
of five π§ and the square root of three π¦ is equal to the tan of five π§, find the
second derivative of π¦ with respect to π₯.

In this question, weβre asked to
find the second derivative of π¦ with respect to π₯. But we can see weβre not given π¦
as a function in π₯. Instead, weβre given π₯ as a
function in π§. And weβre also given π¦ defined
implicitly in terms of π§. This means to find the second
derivative of π¦ with respect to π₯, weβre going to need to apply two results. Weβre going to need to use
parametric differentiation to determine an expression for d two π¦ by dπ₯
squared.

And thereβs two different methods
we could use to find an expression for dπ¦ by dπ§. Either we could square both sides
of the equation and rearrange to find an equation for π¦ in terms of π§ or we could
use implicit differentiation. Either method would work. We can start by recalling the
following formula for the second derivative of π¦ with respect to π₯. Itβs equal to the derivative of dπ¦
by dπ₯ with respect to π§ divided by dπ₯ by dπ§, provided π₯ and π¦ are defined
parametrically in terms of π§. And to use this result, we need to
find an expression for dπ¦ by dπ₯ in terms of π§. And we can find this by using the
chain rule and the inverse function theorem, which tells us that dπ¦ by dπ₯ would be
equal to dπ¦ by dπ§ divided by dπ₯ by dπ§.

Therefore, we can find an
expression for the second derivative of π¦ with respect to π₯ by finding an
expression for dπ¦ by dπ§ and dπ₯ by dπ§. Letβs start by finding an
expression for dπ₯ by dπ§. Thatβs the derivative of two sec of
five π§ with respect to π§. We can do this by recalling the
derivative of the sec of π times π for any real constant π is equal to π times
the sec of ππ multiplied by the tan of ππ. So, dπ₯ by dπ§ is 10 sec of five π§
tan of five π§.

We now need to find an expression
for dπ¦ by dπ§. And as stated before, thereβs two
ways we could do this. And itβs personal preference which
one we want to use. Weβll differentiate both sides of
this equation with respect to π§. To differentiate root three π¦ with
respect to π§, weβre going to need to use implicit differentiation. To make this easier, letβs start by
rewriting root three π¦ as root three multiplied by π¦ to the power of one-half. We can now differentiate this with
respect to π§ by using the chain rule.

First, π¦ is a function in π§, so
we need to differentiate π¦ with respect to π§. Then we multiply this by the
derivative of this expression with respect to π¦. We evaluate this by using the chain
rule. We multiply by the exponent of π¦,
which is one-half, and reduce this exponent by one. This gives us dπ¦ by dπ§ multiplied
by root three over two times π¦ to the power of negative one-half. We then need to differentiate the
right-hand side of this equation with respect to π§. Thatβs the derivative of the tan of
five π§ with respect to π§. And we can evaluate this. Itβs five times the sec squared of
five π§.

Remember, we want to find an
expression for dπ¦ by dπ§. So weβre going to need to rearrange
this equation to make dπ¦ by dπ§ the subject. We can multiply both sides of the
equation through by root π¦. We can multiply both sides of the
equation through by two and divide through by root three. This gives us dπ¦ by dπ§ is 10 over
root three multiplied by root π¦ times the sec squared of five π§. But remember, weβre going to need
to differentiate our expression for dπ¦ by dπ₯ with respect to π§. So we donβt want any π¦-terms in
our expression. And we can get around this by
noting we can find an expression for root π¦ in terms of π§. Weβre told that root three π¦ is
equal to the tan of five π§. If we divide both sides of this
equation through by root three, we get that root π¦ is equal to one over root three
times the tan of five π§.

We can substitute this into our
expression for dπ¦ by dπ§. This gives us 10 over root three
multiplied by one over root three times the tan of five π§ multiplied by the sec
squared of five π§. And we can simplify this
slightly. Root three multiplied by root three
is equal to three. So dπ¦ by dπ§ is ten-thirds the tan
of five π§ multiplied by the sec squared of five π§. We can substitute this expression
and our expression for dπ₯ by dπ§ to determine an expression for dπ¦ by dπ₯. This then gives us that the
derivative of π¦ with respect to π₯ is equal to ten-thirds tan of five π§ times the
sec squared of five π§ divided by 10 sec of five π§ tan of five π§. And we can simplify this. We can cancel a shared factor of
sec of five π§ in the numerator and denominator. We can cancel a shared factor of
tan of five π§ in the numerator and denominator. And we can also cancel a shared
factor of 10 in the numerator and denominator. This just leaves us with one-third
times the sec of five π§.

Weβre now almost ready to find an
expression for second derivative of π¦ with respect to π₯. Letβs start by evaluating the
numerator on the right-hand side of our formula. Letβs start by first clearing some
space. We want to differentiate our
expression for dπ¦ by dπ₯ with respect to π§. Thatβs the derivative of one-third
sec of five π§ with respect to π§. And we can evaluate this. Itβs equal to five-thirds times the
sec of five π§ multiplied by the tan of five π§.

We can now substitute this into our
formula for d two π¦ by dπ₯ squared. This then gives us the second
derivative of π¦ with respect to π₯ is equal to five-thirds times the sec of five π§
tan of five π§ all divided by 10 sec of five π§ multiplied by the tan of five
π§. And we can evaluate this. We can cancel the shared factor of
sec of five π§ times the tan of five π§ in the numerator and denominator. And we can also cancel a shared
factor of five in the numerator and denominator. This gives us one-third divided by
two, which we can just evaluate is equal to one-sixth, which is our final
answer. Therefore, we were able to show if
π₯ is two times the sec of five π§ and the square root of three π¦ is equal to the
tan of five π§, then the second derivative of π¦ with respect to π₯ is equal to
one-sixth.

Letβs now go over the key points of
this video. We were able to show if π¦ is a
differentiable function π in π‘ and π₯ is a differentiable function π of π‘, then
we can find an expression for the second derivative of π¦ with respect to π₯ by
applying the chain rule and inverse function theorem. We showed 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 this was provided that dπ₯ by
dπ‘ was nonzero and we can find dπ¦ by dπ₯ by dividing dπ¦ by dπ‘ by dπ₯ by dπ‘.