# Video: Second Derivatives of Parametric Equations

In this video, we will learn how to find second derivatives and higher-order derivatives of parametric equations by applying the chain rule.

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### 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, letβs say weβve 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‘. Letβs now consider how weβd find the second derivative of π¦ with respect to π₯. So thatβs 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, weβd 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. Weβre now ready to use this formula to help us find second derivatives of parametric equations. Letβs quickly note that we mustnβt 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, weβve been given a pair of parametric equations. And weβve 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‘. Weβve 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.

Weβve 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 thatβs 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, itβs equal to π’. And our denominator is six π‘. And so itβs 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, weβre 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 itβs equal to negative five over six π‘ squared. We also found dπ₯ by dπ‘ earlier. And itβs 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 itβs 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, weβve 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.

Weβve 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 weβve 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, letβs consider what happens when we try to find higher order derivatives of parametric equations. Letβs 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 itβs given by this equation. However, this gives us it in terms of π‘. And so when weβre 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 πth derivative of π¦ with respect to π₯. We can say that d ππ¦ 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.

Letβs 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 weβve been given some parametric equations. And weβve been asked to find the third derivative of π¦ with respect to π₯. We have a formula for finding the πth derivative of π¦ with respect to π₯. And it tells us that itβs 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 thereβs one final formula weβll need, which is for the first derivative of π¦ with respect to π₯. Weβll need all three of these formulas in order to find the third derivative. Weβll start by finding dπ¦ by dπ₯. Therefore, weβll need dπ¦ by dπ‘ and dπ₯ by dπ‘.

Weβve 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 itβs 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. Letβs 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 πth 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.