### Video Transcript

Second derivatives of parametric
equations. In this video, we will learn how to
find the second derivatives and higher order derivatives of parametric equations by
applying the chain rule. And we would also be covering a
variety of examples. Now, letβs say weβve been given a
pair of parametric equations. π₯, which is equal to some function
π of π‘, and π¦, which is equal to some function π of π‘. We know that we can find dπ¦ by dπ₯
with the following equation. We have that dπ¦ by dπ₯ is equal to
dπ¦ by dπ‘ over dπ₯ by dπ‘. Letβs now consider how weβd find
the second derivative of π¦ with respect to π₯. So thatβs d two π¦ by dπ₯
squared. And we can in fact find this by
taking the first derivative of dπ¦ by dπ₯ with respect to π₯.

Now, there is a slight issue
here. And we can see what this issue is
by looking at the equation for dπ¦ by dπ₯. dπ¦ by dπ₯ is equal to dπ¦ by dπ‘ over
dπ₯ by dπ‘. Now, both dπ¦ by dπ‘ and dπ₯ by dπ‘
are equations in terms of π‘. Therefore, dπ¦ by dπ₯ will also be
an equation in terms of π‘. Therefore, we will not be able to
differentiate it directly with respect to π₯. We can, however, get around this
problem by using the chain rule.

The chain rule tells us that if we
have some differential dπ by dπ₯, then this is equal to dπ by dπ‘ multiplied by
dπ‘ by dπ₯, where π‘ is another variable. And we can apply this to our
derivative. We obtain that d two π¦ by dπ₯
squared is equal to d by dπ‘ of dπ¦ by dπ₯ multiplied by dπ‘ by dπ₯. Now, we could find dπ‘ by dπ₯ by
rearranging π₯ is equal to π of π‘ to π‘ is equal to π inverse of π₯ and then
differentiating with respect to π₯. However, this would be in terms of
π₯. And the other term in our equation
d by dπ‘ of dπ¦ by dπ₯ is in terms of π‘. So ideally, weβd want this second
term to also be in terms of π‘.

We can find this in terms of π‘ by
using a differential rule for inverse functions. This differential rule tells us
that dπ‘ by dπ₯ is equal to one over dπ₯ by dπ‘. And we can substitute this in to
our equation. This will give us our formula for
finding the second derivative of parametric equations. Which tells us that d two π¦ by dπ₯
squared is equal to d by dπ‘ of dπ¦ by dπ₯ over dπ₯ by dπ‘. And this, of course, only works
when dπ₯ by dπ‘ is nonzero. Weβre now ready to use this formula
to help us find second derivatives of parametric equations. Letβs quickly note that we mustnβt
forget the equation for the first derivative of parametric equations since this is
used within the equation 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 d two π¦ by
dπ₯ squared.

In this question, weβve been given
a pair of parametric equations. And weβve been asked to find d two
π¦ by dπ₯ squared, which is the second derivative of π¦ with respect to π₯. Now, we in fact have an equation to
help us find the second derivative of parametric equations. The equation tells us that d two π¦
by dπ₯ squared is equal to d by dπ‘ of dπ¦ by dπ₯ over dπ₯ by dπ‘. In order to use this equation, we
also need to find dπ¦ by dπ₯. And we also have an equation for
finding dπ¦ by dπ₯ when given parametric equations. This equation tells us that dπ¦ by
dπ₯ is equal to dπ¦ by dπ‘ over dπ₯ by dπ‘.

We can start our solution by
finding dπ¦ by dπ‘ and dπ₯ by dπ‘. We have that π₯ is equal to three
π‘ squared plus one. Using the power rule for
differentiation, we multiply by the power and decrease the power by one. And so, differentiating the term
three π‘ squared, we obtain six π‘. Since one is a constant and we
differentiate it, it will go to zero. Therefore, we have that dπ₯ by dπ‘
is equal to six π‘. Weβve also been given that π¦ is
equal to three π‘ squared plus five π‘. This can again be differentiated
using the power rule. Differentiating the first term, we
again get six π‘. And when we differentiate the five
π‘, we simply get five. Therefore, we have that dπ¦ by dπ‘
is equal to six π‘ plus five.

Weβve now found all the components
in order to find dπ¦ by dπ₯. We obtain that dπ¦ by dπ₯ is equal
to six π‘ plus five over six π‘. When we look at our formula for the
second derivative of π¦ with respect to π₯, we spot that we have to find d by dπ‘ of
dπ¦ by dπ₯. So thatβs d by dπ‘ of six π‘ plus
five over six π‘, which is a quotient. Therefore, we can use the quotient
rule to help us differentiate here. We find that the derivative of a
quotient of some functions, π’ over π£, is equal to π£ multiplied by dπ’ by dπ₯
minus π’ multiplied by dπ£ by dπ₯ all over π£ squared.

Now, in our case, our numerator is
equal to six π‘ plus five. Therefore, itβs equal to π’. And our denominator is six π‘. And so itβs equal to π£. We can differentiate six π‘ plus
five with respect to π‘ to find that dπ’ by dπ‘ is equal to six. And differentiating six π‘ with
respect to π‘, we find that dπ£ by dπ‘ is also equal to six. Now, weβre ready to substitute π’,
π£, dπ’ by dπ‘, and dπ£ by dπ‘ into the formula given to us by the quotient
rule. What we get is six π‘ times six
minus six π‘ plus five times six all over six π‘ squared. Multiplying through, we obtain 36π‘
minus 36π‘ minus 30 over 36π‘ squared. Therefore, we can cancel the 36π‘
with the minus 36π‘. And we can cancel through our
factor of five to obtain that d by dπ‘ of dπ¦ by dπ₯ is equal to negative five over
six π‘ squared.

So if we look back at our formula
for d two π¦ by dπ₯ squared, we have now found d by dπ‘ of dπ¦ by dπ₯. And itβs equal to negative five
over six π‘ squared. We also found dπ₯ by dπ‘
earlier. And itβs equal to six π‘. Using our formula, we have that d
two π¦ by dπ₯ squared is equal to negative five over six π‘ squared over six π‘,
which simplifies to give us a solution that d two π¦ by dπ₯ squared is equal to
negative five over 36π‘ cubed.

This method can be useful in
helping us to find the concavity of parametric equations, as we will see in the next
example.

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

Here, we have been asked about the
concavity of a curve. We know that, in order to determine
the concavity of a curve, we need to consider the second derivative of π¦ with
respect to π₯. We know that when d two π¦ by dπ₯
squared is greater than zero, our curve is concave down. And when itβs less than zero, our
curve is concave up. From this, we can see that, in
order to consider the concavity of our curve, we first need to find d two π¦ by dπ₯
squared. Now, weβve been given our curve in
terms of parametric equations. Therefore, we can use the following
formula to find d two π¦ by dπ₯ squared. This formula tells us that d two π¦
by dπ₯ squared is equal to d by dπ of dπ¦ by dπ₯ over dπ₯ by dπ. In order to use this, we first need
to find dπ¦ by dπ₯, which we also know a formula for. We have that dπ¦ by dπ₯ is equal to
dπ¦ by dπ over dπ₯ by dπ. Therefore, we can start by finding
dπ¦ by dπ and dπ₯ by dπ.

Weβve been given that π₯ is equal
to cos π. And π¦ is equal to sin π. Differentiating cos π with respect
to π, we obtain that dπ₯ by dπ is equal to negative sin π. And differentiating sin π with
respect to π, we obtain that dπ¦ by dπ is equal to cos π. And substituting these into our
equation for dπ¦ by dπ₯, we obtain that dπ¦ by dπ₯ is equal to negative cos π over
sin π or negative cot π. Next, we need to differentiate dπ¦
by dπ₯ with respect to π, which is equivalent to d by dπ of negative cot π. Now, we know that cot π
differentiates to give negative csc squared π. And so we can see that negative cot
π will differentiate to csc squared π. Now, we have found d by dπ of dπ¦
by dπ₯ and dπ₯ by dπ. And so we obtain that d two π¦ by
dπ₯ squared is equal to csc squared π over negative sin π. Since csc π is equal to one over
sin π, we have that d two π¦ by dπ₯ squared is equal to negative csc cubed π.

At this stage, we found d two π¦ by
dπ₯ squared. We just need to evaluate it at π
is equal to π by six. At π is equal to π by six, we
have that it is equal to negative csc cubed of π by six. And we can use the fact that csc is
equal to one over sin, giving us that this is equal to negative one over sin cubed
of π by six. We know that sin of π by six is
equal to one-half. So the denominator of our fraction
is equal to one-half cubed. Now, one-half cubed is
one-eighth. And one over one-eighth is simply
eight. So weβve evaluated d two π¦ by dπ₯
squared at π is equal to π by six to be equal to negative eight. Since negative eight is less than
zero, this tells us that, at π is equal to π by six, our curve is concave up.

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 that this is equal to d
by dπ₯ of d two π¦ by dπ₯ squared. And we already know how to find d
two π¦ by dπ₯ squared since itβs given by this equation. However, this gives us it in terms
of π‘. And so when weβre trying to
differentiate it with respect to π₯, it will be easier to simply apply the chain
rule. This gives us that d three π¦ by
dπ₯ cubed is equal to d by dπ‘ of d two π¦ by dπ₯ squared multiplied by dπ‘ by
dπ₯. Similarly to before, we can apply
the fact that dπ‘ by dπ₯ is equal to one over dπ₯ by dπ‘. And so we obtain that d three π¦ by
dπ₯ cubed is equal to d by dπ‘ of d two π¦ by dπ₯ squared over dπ₯ by dπ‘.

Looking at the formulas for the
third derivative of π¦ with respect to π₯ and the second derivative of π¦ with
respect to π₯, we may notice a pattern start to emerge. And if we in fact continue to
differentiate these derivatives to find higher order derivatives, we would notice
this pattern continue. Therefore, we can write an equation
for the πth derivative of π¦ with respect to π₯. We can say that d ππ¦ by dπ₯ to
the π is equal to d by dπ‘ of d π minus one π¦ by dπ₯ to the π minus one over dπ₯
by dπ‘. And this formula can be used to
find any higher order derivative of parametric equations.

Letβs now move on to our final
example.

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

So weβve been given some parametric
equations. And weβve been asked to find the
third derivative of π¦ with respect to π₯. We have a formula for finding the
πth derivative of π¦ with respect to π₯. And it tells us that itβs equal to
d by dπ‘ of d π minus one π¦ by dπ₯ to the π minus one over dπ₯ by dπ‘. In our case, π is equal to
three. So it can be substituted in. We obtain that d three π¦ by dπ₯
cubed is equal to d by dπ‘ of d two π¦ by dπ₯ squared over dπ₯ by dπ‘. Now, we also have a formula for d
two π¦ by dπ₯ squared, since this is required in the formula for the third
derivative. And thereβs one final formula weβll
need, which is for the first derivative of π¦ with respect to π₯. Weβll need all three of these
formulas in order to find the third derivative. Weβll start by finding dπ¦ by
dπ₯. Therefore, weβll need dπ¦ by dπ‘
and dπ₯ by dπ‘.

Weβve been given π₯ and π¦ in terms
of π‘ in the question. We can differentiate these. And we obtain that dπ₯ by dπ‘ is
equal to four. And dπ¦ by dπ‘ is equal to two π
to the π‘ minus three π‘ squared. We can combine these two together
to say that dπ¦ by dπ₯ is equal to two π to the π‘ minus three π‘ squared over
four. We can then differentiate this with
respect to π‘ to obtain that d by dπ‘ of dπ¦ by dπ₯ is equal to two π to the π‘
minus six π‘ all over four. Here, we notice we can cancel a
factor of two, simplifying our result to π to the π‘ minus three π‘ over two. In order to find the second
derivative of π¦ with respect to π₯, we need to divide this result by dπ₯ by dπ‘,
which is equal to four. We obtain that d two π¦ by dπ₯
squared is equal to π to the π‘ minus three π‘ over eight.

For the numerator of our formula
for the third derivative, we need to differentiate this with respect to π‘. By differentiating d two π¦ by dπ₯
squared with respect to π‘, we obtain that itβs equal to π to the π‘ minus three
all over eight. And for our final step in finding
the third derivative of π¦ with respect to π₯, we simply need to divide this by dπ₯
by dπ‘, which is again equal to four. This is where we reach our
solution, which is that d three π¦ by dπ₯ cubed is equal to π to the π‘ minus three
over 32.

We have now seen how to find second
and higher order derivatives of parametric equations. Letβs recap some key points of this
video. Key points, we can find the second
derivative of parametric equations with the formula d two π¦ by dπ₯ squared is equal
to d by dπ‘ of dπ¦ by dπ₯ over dπ₯ by dπ‘, where dπ¦ by dπ₯ is equal to dπ¦ by dπ‘
over dπ₯ by dπ‘. And dπ₯ by dπ‘ is nonzero. This formula can be useful for
finding the concavity of a function defined by parametric equations. We can find higher order
derivatives of parametric equations using the following formula. The πth derivative of π¦ with
respect to π₯ is equal to d by dπ‘ of d π minus one π¦ by dπ₯ to the π minus one
over dπ₯ by dπ‘, where dπ₯ by dπ‘ is nonzero.