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 π‘₯ and 𝑦, then we could write parametric equations for these variables in terms of a parameter, 𝑑, as follows: π‘₯=𝑓(𝑑),𝑦=𝑔(𝑑).

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 π‘Ÿ and πœƒ in terms of a parameter.

We can find the derivative of 𝑦 with respect to π‘₯ in terms of the parametric equations using the following definition.

Definition: Derivative of a Parametric Equation

Let 𝑓 and 𝑔 be differentiable functions, such that π‘₯ and 𝑦 are a pair of parametric equations: π‘₯=𝑓(𝑑),𝑦=𝑔(𝑑).

Then we can define the derivative of 𝑦 with respect to π‘₯ as dd𝑦π‘₯=ddddο˜οο—ο when ddπ‘₯𝑑≠0.

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 ddοŠ¨οŠ¨π‘¦π‘₯, 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 𝑑 and dividing ddοŠ¨οŠ¨π‘¦π‘‘ by ddπ‘₯𝑑, 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 𝑦 with respect to π‘₯ by differentiating the first derivative with respect to π‘₯: ddddddοŠ¨οŠ¨π‘¦π‘₯=π‘₯𝑦π‘₯.

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, 𝑑. In order to perform this differentiation with respect to π‘₯, we will need to use the chain rule. Recall the definition of the chain rule.

Definition: The Chain Rule

Given a function β„Ž that is differentiable at π‘₯ and a function 𝑔 that is differentiable at β„Ž(π‘₯), their composition 𝑓=π‘”β‹…β„Ž which is defined by 𝑓(π‘₯)=𝑔(β„Ž(π‘₯)) is differentiable at π‘₯ and its derivative 𝑓′ is given by 𝑓′(π‘₯)=β„Žβ€²(π‘₯)𝑔′(β„Ž(π‘₯)).

We know how to find dd𝑦π‘₯ in terms of 𝑑, since we can find the first derivative of a parametric equation. However, we need to differentiate this with respect to π‘₯ 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: ddddddπ‘₯(𝑦)=𝑒(𝑦)×𝑒π‘₯.

Applying this to our equation for the second derivative, we get ddddddddddddοŠ¨οŠ¨π‘¦π‘₯=π‘₯𝑦π‘₯=𝑑𝑦π‘₯×𝑑π‘₯.

Now, we have dddd𝑑𝑦π‘₯, which is in terms of 𝑑; however, dd𝑑π‘₯ is in terms of π‘₯. Here, we can use the inverse function theorem, which tells us that, for derivatives which are nonzero, dd𝑑π‘₯=1.dd

Substituting this into our equation gives us the following formula for finding the second derivative of parametric equations: ddοŠ¨οŠ¨οο˜ο—ο—οπ‘¦π‘₯=.dddddd

Definition: Second Derivative of a Parametric Equation

Let 𝑓 and 𝑔 be differentiable functions such that π‘₯ and 𝑦 are a pair of parametric equations: π‘₯=𝑓(𝑑),𝑦=𝑔(𝑑). Then, we can define the second derivative of 𝑦 with respect to π‘₯ as ddοŠ¨οŠ¨οο˜ο—ο—οπ‘¦π‘₯=dddddd when ddπ‘₯𝑑≠0.

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 π‘₯=3𝑑+1 and 𝑦=3𝑑+5π‘‘οŠ¨, find ddοŠ¨οŠ¨π‘¦π‘₯.

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 dd𝑦π‘₯=.ddddο˜οο—ο

First, we can differentiate 𝑦 with respect to 𝑑. Since 𝑦 is a polynomial in terms of 𝑑, we use polynomial differentiation. First, multiply each term by its power of 𝑑, and then reduce the power of 𝑑 by one. This gives us dd𝑦𝑑=6𝑑+5.

Similarly, we also need to differentiate π‘₯ with respect to 𝑑. This gives us ddπ‘₯𝑑=6𝑑.

Substituting these back into our formula for the first derivative, we obtain dd𝑦π‘₯=6𝑑+56𝑑=1+56𝑑.

We are now ready to use the formula for the second derivative, which is as follows: ddοŠ¨οŠ¨οο˜ο—ο—οπ‘¦π‘₯=.dddddd

We have already found ddπ‘₯𝑑, so all we need to do is find the derivative of dd𝑦π‘₯ with respect to 𝑑. By differentiating, we obtain dddd𝑑𝑦π‘₯=βˆ’56𝑑.

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 ddοŠ¨οŠ¨οŠ±οŠ«οŠ¬οοŠ©π‘¦π‘₯=6𝑑=βˆ’536𝑑.

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 𝑦=βˆ’5π‘₯βˆ’7 and 𝑧=3π‘₯+16, find ddοŠ¨οŠ¨π‘§π‘¦ at π‘₯=1.

Answer

Since we are trying to find the second derivative of 𝑧 with respect to 𝑦, where 𝑦 and 𝑧 are in parametric form, we can use the following equation: ddοŠ¨οŠ¨ο—ο™ο˜ο˜ο—π‘§π‘¦=ο€Όοˆ.dddddd

We can start by differentiating 𝑦 and 𝑧 with respect to π‘₯. Using polynomial differentiation, we get dd𝑦π‘₯=βˆ’15π‘₯ and dd𝑧π‘₯=6π‘₯.

Using this, we can find dd𝑧𝑦=6π‘₯βˆ’15π‘₯=βˆ’25π‘₯.

Next, we need to differentiate dd𝑧𝑦 with respect to π‘₯. In doing this, we obtain ddddπ‘₯𝑧𝑦=25π‘₯.

We are now ready to find ddοŠ¨οŠ¨π‘§π‘¦. Substituting what we have found into the formula, we obtain ddοŠ¨οŠ¨οŠ¨οŠ«ο—οŠ¨οŠͺ𝑧𝑦=βˆ’15π‘₯=βˆ’275π‘₯.

Now all that we need to do is evaluate our second derivative at the given point, π‘₯=1. When we do this, we reach our solution of ddοŠ¨οŠ¨ο—οŠ²οŠ§π‘§π‘¦|||=βˆ’275.

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

Example 3: Finding the Second Derivative of a Function Defined by Parametric Equations Using Implicit Differentiation

If π‘₯=25𝑧sec and √3𝑦=5𝑧tan, find ddοŠ¨οŠ¨π‘¦π‘₯.

Answer

In this question, we can see that the second parametric equation we have been given is √3𝑦=5𝑧tan. 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: ddοŠ¨οŠ¨ο™ο˜ο—ο—ο™π‘¦π‘₯=.dddddd

This requires the first derivative, which can be found using this formula: dd𝑦π‘₯=.ddddο˜ο™ο—ο™

We can start by differentiating π‘₯ with respect to 𝑧: ddtansecπ‘₯𝑧=105𝑧5𝑧.

The next step will be to differentiate 𝑦 with respect to 𝑧. 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 √3𝑦=5𝑧tan, we may notice that it will be easier to differentiate if we write it as (3𝑦)=5π‘§οŽ οŽ‘tan. Now we can differentiate implicitly with respect to 𝑧, which will give us 32𝑦𝑧(3𝑦)=55𝑧.ddsecοŽͺ

We can rearrange this to make dd𝑦𝑧 the subject: ddsec𝑦𝑧=103(3𝑦)5𝑧.

Now we do still have a 𝑦-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 (3𝑦)=5π‘§οŽ οŽ‘tan in to get ddtansec𝑦𝑧=1035𝑧5𝑧.

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 𝑦: 𝑦=135𝑧.tan

Now we differentiate with respect to 𝑧 to obtain ddtansec𝑦𝑧=1035𝑧5𝑧.

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 dd𝑦𝑧 by ddπ‘₯𝑧 to obtain ddtansectansecsec𝑦π‘₯=5𝑧5𝑧105𝑧5𝑧=135𝑧.

Next, we need to differentiate dd𝑦π‘₯ with respect to 𝑧. In doing this, we will get ddddtansec𝑧𝑦π‘₯=535𝑧5𝑧.

We now have all the components of our formula for the second derivative. We can substitute them in and simplify to reach our solution: ddtansectansecοŠ¨οŠ¨οŠ«οŠ©π‘¦π‘₯=5𝑧5𝑧105𝑧5𝑧=16.

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

Example 4: Finding the Second Derivative of a Function Defined by Parametric Equations

If 𝑦=(π‘₯+4)ο€Ήβˆ’4π‘₯βˆ’1ο…οŠ¨ and 𝑧=(π‘₯βˆ’5)(π‘₯+4), find (2π‘₯βˆ’1)π‘¦π‘§οŠ©οŠ¨οŠ¨dd.

Answer

In order to solve this question, we first need to find ddοŠ¨οŠ¨π‘¦π‘§. We can use the formula ddοŠ¨οŠ¨ο—ο˜ο™ο™ο—π‘¦π‘§=.dddddd

In order to find dd𝑦𝑧, we need to find dd𝑦π‘₯ and dd𝑧π‘₯. We can expand the binomial terms in 𝑦 and 𝑧 to obtain 𝑦=βˆ’4π‘₯βˆ’16π‘₯βˆ’π‘₯βˆ’4,𝑧=π‘₯βˆ’π‘₯βˆ’20.

We differentiate these with respect to π‘₯ to get dddd𝑦π‘₯=βˆ’12π‘₯βˆ’32π‘₯βˆ’1,𝑧π‘₯=2π‘₯βˆ’1.

Using what we have just found, we can say that dd𝑦𝑧=βˆ’12π‘₯βˆ’32π‘₯βˆ’12π‘₯βˆ’1.

In order to find the second derivative, we need to differentiate dd𝑦𝑧 with respect to π‘₯. 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 𝑒𝑣, then we find its derivative using 𝑒𝑣=π‘£π‘’β€²βˆ’π‘’π‘£β€²π‘£.

In our case, 𝑒=βˆ’12π‘₯βˆ’32π‘₯βˆ’1 and 𝑣=2π‘₯βˆ’1. Therefore, 𝑒′=βˆ’24π‘₯βˆ’32 and 𝑣′=2. Substituting these values into our formula, we obtain ddddπ‘₯𝑦𝑧=(2π‘₯βˆ’1)(βˆ’24π‘₯βˆ’32)βˆ’ο€Ήβˆ’12π‘₯βˆ’32π‘₯βˆ’1×2(2π‘₯βˆ’1).

We can simplify the numerator to give us ddddπ‘₯𝑦𝑧=βˆ’24π‘₯+24π‘₯+34(2π‘₯βˆ’1).

Now, we are ready to find ddοŠ¨οŠ¨π‘¦π‘§. Substituting into the formula, we have ddοŠͺο—οŠ°οŠ¨οŠͺο—οŠ°οŠ©οŠͺ(οŠ¨ο—οŠ±οŠ§)οŠ¨οŠ©π‘¦π‘§=(2π‘₯βˆ’1)=βˆ’24π‘₯+24π‘₯+34(2π‘₯βˆ’1).

In order to find the solution to the problem, all we need to do is multiply ddοŠ¨οŠ¨π‘¦π‘§ by (2π‘₯βˆ’1). In doing this, we reach our solution of (2π‘₯βˆ’1)𝑦𝑧=βˆ’24π‘₯+24π‘₯+34.dd

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 ddοŠ¨οŠ¨οο˜ο—ο—οπ‘¦π‘₯=,dddddd where dd𝑦π‘₯=ddddο˜οο—ο when ddπ‘₯𝑑≠0.

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