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

In this explainer, we will learn how to find the first derivative of a curve defined by parametric equations and find the equations of tangents and normals to the curves.

Parametric equations are a way of expressing the variables in our equation in terms of a parameter. For example, if we have a Cartesian equation of the form , we could express and in terms of a parameter, :

These parametric equations will describe the exact same curve as , just in a different form.

### 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.

Writing equations in parametric form has many different uses. It can make writing one-to-many relations a lot easier, for example, equations of ellipses, cardioids, and limacons, which can typically be more difficult to write in Cartesian form.

There is a method we can use to find the derivative of an equation in parametric form without having to convert the parametric equations back to Cartesian form. The formula we can use is as follows.

### Definition: Derivative of a Parametric Equation

Let and be differentiable functions such that we can form a pair of parametric equations using and :

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

Let us discuss how we can come about reaching this equation. In doing this, we are going to need to use the chain rule, so we should remind ourselves of its definition.

### 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 can start with our parametric equations

And let us suppose that we are able to write these parametric equations in Cartesian form such that

We can substitute our parametric equations into this equation to get

Next, we want to differentiate this equation with respect to . On the right-hand side we can see that we have a composite function so we will need to apply the chain rule:

We must be careful with the differentiation here, since the derivatives of the lowercase functions denote a differentiation with respect to , and the derivative of the uppercase function denotes a derivative with respect to . Using this along with our parametric equations, and , we get

We can now substitute and into our differential equation which will give us

Finally, since is the derivative of with respect to , and , , then, which can be rearranged to our result

Now that we have seen the derivation of the formula for the derivative of parametric equations, let us look at an example of how we can use it.

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

Given that and , find the rate of change of with respect to .

### Answer

We have been asked to find the rate of change of with respect to , which can also be written as . We are given and in terms of a parameter, , so we will need to use the formula for finding the derivative of parametric equations, which is as follows:

We can start by finding . is a polynomial in , so we can use polynomial differentiation to find this derivative. We multiply each term by its power of and then decrease the power of by one. This gives us

Similarly, we can find the derivative of with respect to as follows:

Now that we have and , we can substitute them into our formula to find our solution:

We are in fact able to find the derivative of a function with respect to another function using derivatives of parametric equations.

### Definition: The Derivative of a Function with respect to Another Function

If we have two functions, and , then we can define the derivative of with respect to as

Let us look at an example of how this can be used.

### Example 2: Finding the Derivative of a Function with respect to Another Function

Find the derivative of with respect to at .

### Answer

We can start by defining our functions as

When we write our equations in this form, we can see that the question is asking us to find the derivative of with respect to . The formula we can use to find this derivative is

We need to differentiate and with respect to . This will involve both trigonometric and polynomial differentiation. For polynomial differentiation, we multiply the term by the power of and then decrease the power of by one. For trigonometric differentiation, we have

Using this, we find that and

Substituting these back into our formula, we reach

The question asked us to find the derivative at , so we need to substitute this into . This give us

Simplifying this, we can see that our solution is

We know that we can use derivatives to find the slope of a line at a given point, and by using this slope we can find the tangent or normal to the line at that point. We can do a similar thing using the derivatives of parametric equations.

In the next example, we will see how the derivatives of some parametric equations can be used to find the tangent to the curve at a given point.

### Example 3: Finding the Tangent to a Curve Using Derivatives of Parametric Equations

Find the equation of the tangent to the curve and at .

### Answer

We have been given a pair of parametric equations and asked to find the tangent to the curve at a given point. In order to find the equation of the tangent, we first need to find the slope of the tangent. We can do this by evaluating the derivative of the parametric equations at the given point. We will need to use the formula for finding the derivative of a parametric equation, remembering that our parameter is :

Now we need to differentiate these trigonometric equations, which will give us and

Putting these back into our formula, we obtain

This can be simplified further which gives us that our derivative is

In order to find the slope of the tangent at , we need to substitute this into our differential. When we do this, we obtain

Now that we have found the slope of the tangent, we need to find a point that the tangent passes through. We can do this by finding the values of and when . Substituting into our equations for and gives us

Now that we have the slope of our tangent, 2, and a point that the tangent passes through, , we are able to form our equation. We can use the formula where is the slope of the tangent and is a point that the tangent passes through. This gives us

Next, we can move everything to the left-hand side of the equation:

And finally, we simplify to reach our solution:

We could use a similar method to find the normal to a parametric curve at a given point. The only thing we need to be careful of is that, once we have found the slope, , of the curve at the given point, we must use the fact that

So, in the case that the slope of the tangent is , the slope of the normal would be .

In our final example, we will see how we can identify the points at which a parametric curve has vertical tangents.

### Example 4: Finding the Value of a Variable That Makes a Curve with Parametric Equations Have a Vertical Tangent

Find the value of at which the curve , has a vertical tangent.

### Answer

In order for a curve to have a vertical tangent, its gradient must be infinite. While this may sound slightly abstract, our formula for the derivative of a parametric equation can help us. We know that

This derivative will get larger and larger as gets smaller and smaller. Therefore, our derivative, and the slope of the tangent, will tend to infinity as tends to zero.

Let us find the values of such that . We can start by differentiating with respect to . We obtain

Now, we need to set this derivative equal to 0 and solve to find the values of . We have that

We can use the quadratic formula which tells us that the solutions to a quadratic equation of the form are

In our case, , , and ; therefore,

We can now split up our two solutions that come from the plus or minus sign. Firstly, taking the plus,

And, secondly, taking the minus,

Therefore, the values of at which our curve has vertical tangents are when and when .

### Key Points

• For a pair of differentiable parametric equations, and , we define the derivative of the parametric equations as when .
• If we have two functions, and , then we can define the derivative of with respect to as
• We can find the equations of tangents and normals to a parametric curve by finding the slope using the derivative and then applying the formula , where is a point on the tangent or normal.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.