# Lesson Explainer: Second Derivatives of Parametric Equations Mathematics • Higher Education

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

Parametric equations are a way in which we can express the variables in an equation in terms of another parameter. For example, if we have an equation in terms of the variables and , then we could write parametric equations for these variables in terms of a parameter, , as follows:

### Note

Parametric equations can be used in conjunction with any coordinate system, not only Cartesian. For example, if we wanted to parameterize some polar coordinates, we would express and in terms of a parameter.

We can find the derivative of with respect to in terms of the parametric equations using the following definition.

### Definition: Derivative of a Parametric Equation

Let and be differentiable functions, such that and are a pair of parametric equations:

Then we can define the derivative of with respect to as when .

The first derivative of an equation can be a very useful tool for finding equations of tangents and normals to the curve or calculating gradients along the curve. The second derivative, or , can also tell us useful information about the concavity of the curve.

You may think that we can find the second derivative by finding the second derivatives of the parametric equations with respect to and dividing by , similar to how we did for the first derivative. However, this does not work as you will see below.

We find the second derivative of with respect to by differentiating the first derivative with respect to :

Finding this second derivative in terms of the parametric equations is not simple, since the equation we have for the first derivative is in terms of our parameter, . In order to perform this differentiation with respect to , we will need to use the chain rule. Recall the definition of the chain rule.

### Definition: The Chain Rule

Given a function that is differentiable at and a function that is differentiable at , their composition which is defined by is differentiable at and its derivative is given by

We know how to find in terms of , since we can find the first derivative of a parametric equation. However, we need to differentiate this with respect to to find the second derivative. In order to do this, we will need to use a different form of the chain rule, which is as follows:

Applying this to our equation for the second derivative, we get

Now, we have , which is in terms of ; however, is in terms of . Here, we can use the inverse function theorem, which tells us that, for derivatives which are nonzero,

Substituting this into our equation gives us the following formula for finding the second derivative of parametric equations:

### Definition: Second Derivative of a Parametric Equation

Let and be differentiable functions such that and are a pair of parametric equations: Then, we can define the second derivative of with respect to as when .

Let us now look at an example of how we can find the second derivative of a parametric equation.

### Example 1: Finding the Second Derivative of Parametric Equations

Given that and , find .

### Answer

The first step in finding the second derivative of these parametric equations is to find the first derivative. We can do this by using the formula

First, we can differentiate with respect to . Since is a polynomial in terms of , we use polynomial differentiation. First, multiply each term by its power of , and then reduce the power of by one. This gives us

Similarly, we also need to differentiate with respect to . This gives us

Substituting these back into our formula for the first derivative, we obtain

We are now ready to use the formula for the second derivative, which is as follows:

We have already found , so all we need to do is find the derivative of with respect to . By differentiating, we obtain

We have now found all the parts for our equation for the second derivative; we can substitute them in, which will give us our solution of

We can evaluate the second derivative of a parametric equation at a given point as we can see in the next example.

### Example 2: Evaluating the Second Derivative of Parametric Equations at a Given Point

If and , find at .

### Answer

Since we are trying to find the second derivative of with respect to , where and are in parametric form, we can use the following equation:

We can start by differentiating and with respect to . Using polynomial differentiation, we get and

Using this, we can find

Next, we need to differentiate with respect to . In doing this, we obtain

We are now ready to find . Substituting what we have found into the formula, we obtain

Now all that we need to do is evaluate our second derivative at the given point, . When we do this, we reach our solution of

In the next example, we will see how we can find the second derivative of parametric equations using implicit differentiation.

If and , find .

### Answer

In this question, we can see that the second parametric equation we have been given is . In order to differentiate this, we could either square both sides of the equation and then differentiate or just use implicit differentiation. We will be covering both methods of solving this problem.

In order to find the second derivative of parametric equations, we will need to use the following formula:

This requires the first derivative, which can be found using this formula:

We can start by differentiating with respect to :

The next step will be to differentiate with respect to . As previously mentioned, there are two methods of doing this. We will start with implicit differentiation and then show the other method.

Method 1

When we look at , we may notice that it will be easier to differentiate if we write it as . Now we can differentiate implicitly with respect to , which will give us

We can rearrange this to make the subject:

Now we do still have a -term in here, but this is not a problem, since it is equivalent to the left-hand side of our original equation. Hence, we simply need to substitute in to get

Now we will see the other method of how we could differentiate this parametric equation.

Method 2

We will start by squaring both sides of the parametric equation and then dividing by 3, so the equation is in terms of :

Now we differentiate with respect to to obtain

As we can see, both methods reached the same solution. We are now ready to carry on with our solution. Our next step is to divide by to obtain

Next, we need to differentiate with respect to . In doing this, we will get

We now have all the components of our formula for the second derivative. We can substitute them in and simplify to reach our solution:

In our final example, let us see how we can find a function that involves the second derivative of parametric equations.

If and , find .

### Answer

In order to solve this question, we first need to find . We can use the formula

In order to find , we need to find and . We can expand the binomial terms in and to obtain

We differentiate these with respect to to get

Using what we have just found, we can say that

In order to find the second derivative, we need to differentiate with respect to . In order to do this, we will need to use the quotient rule for differentiation. The quotient rule tells us that if we have a function of the form , then we find its derivative using

In our case, and . Therefore, and . Substituting these values into our formula, we obtain

We can simplify the numerator to give us

Now, we are ready to find . Substituting into the formula, we have

In order to find the solution to the problem, all we need to do is multiply by . In doing this, we reach our solution of

We have now seen how to find the second derivative of parametric equations and how we can use it to find the concavity of a parametric curve. Let us recap some key points.

### Key Points

• We can find the second derivative of parametric equations using the formula where when .

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