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 𝑦=𝑓(π‘₯), we could express π‘₯ and 𝑦 in terms of a parameter, 𝑑: π‘₯=𝑔(𝑑),𝑦=β„Ž(𝑑).

These parametric equations will describe the exact same curve as 𝑦=𝑓(π‘₯), 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 π‘Ÿ and πœƒ 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 𝑓 and 𝑔 be differentiable functions such that we can form a pair of parametric equations using π‘₯ and 𝑦: 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.

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 β„Ž that is differentiable at π‘₯ and a function 𝑔 that is differentiable at β„Ž(π‘₯), their composition 𝑓=π‘”βˆ˜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 can start with our parametric equations π‘₯=𝑓(𝑑),𝑦=𝑔(𝑑).

And let us suppose that we are able to write these parametric equations in Cartesian form such that 𝑦=𝐢(π‘₯).

We can substitute our parametric equations into this equation to get 𝑔(𝑑)=𝐢(𝑓(𝑑)).

Next, we want to differentiate this equation with respect to 𝑑. On the right-hand side we can see that we have a composite function so we will need to apply the chain rule: 𝑔′(𝑑)=𝑓′(𝑑)𝐢′(𝑓(𝑑)).

We must be careful with the differentiation here, since the derivatives of the lowercase functions denote a differentiation with respect to 𝑑, and the derivative of the uppercase function denotes a derivative with respect to π‘₯. Using this along with our parametric equations, π‘₯=𝑓(𝑑) and 𝑦=𝑔(𝑑), we get 𝑓′(𝑑)=π‘₯𝑑𝑔′(𝑑)=𝑦𝑑.ddanddd

We can now substitute π‘₯ and 𝑦 into our differential equation which will give us dddd𝑦𝑑=π‘₯𝑑𝐢′(π‘₯).

Finally, since 𝐢′(π‘₯) is the derivative of 𝐢 with respect to π‘₯, and 𝑦=𝐢(π‘₯), 𝐢′(π‘₯)=𝑦π‘₯dd, then, dddddd𝑦𝑑=π‘₯𝑑𝑦π‘₯, which can be rearranged to our result dd𝑦π‘₯=.ddddο˜οο—ο

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 𝑦=βˆ’7𝑑+8 and 𝑧=βˆ’7𝑑+3, find the rate of change of 𝑦 with respect to 𝑧.

Answer

We have been asked to find the rate of change of 𝑦 with respect to 𝑧, which can also be written as dd𝑦𝑧. We are given 𝑦 and 𝑧 in terms of a parameter, 𝑑, so we will need to use the formula for finding the derivative of parametric equations, which is as follows: dd𝑦𝑧=.ddddο˜οο™ο

We can start by finding dd𝑦𝑑. 𝑦 is a polynomial in 𝑑, so we can use polynomial differentiation to find this derivative. We multiply each term by its power of 𝑑 and then decrease the power of 𝑑 by one. This gives us dd𝑦𝑑=βˆ’21𝑑.

Similarly, we can find the derivative of 𝑧 with respect to 𝑑 as follows: dd𝑧𝑑=βˆ’14𝑑.

Now that we have dd𝑦𝑑 and dd𝑧𝑑, we can substitute them into our formula to find our solution: dd𝑦𝑧=βˆ’21π‘‘βˆ’14𝑑=32𝑑.

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, 𝑦=𝑓(π‘₯) and 𝑧=𝑔(π‘₯), then we can define the derivative of 𝑓(π‘₯) with respect to 𝑔(π‘₯) as dd𝑦𝑧=.ddddο˜ο—ο™ο—

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 7π‘₯+4π‘₯sin with respect to cosπ‘₯+1 at π‘₯=πœ‹6.

Answer

We can start by defining our functions as 𝑦=7π‘₯+4π‘₯,𝑧=π‘₯+1.sincos

When we write our equations in this form, we can see that the question is asking us to find the derivative of 𝑦 with respect to 𝑧. The formula we can use to find this derivative is dd𝑦𝑧=.ddddο˜ο—ο™ο—

We need to differentiate 𝑦 and 𝑧 with respect to π‘₯. This will involve both trigonometric and polynomial differentiation. For polynomial differentiation, we multiply the term by the power of π‘₯ and then decrease the power of π‘₯ by one. For trigonometric differentiation, we have ddsincosandddcossinπ‘₯(π‘₯)=π‘₯π‘₯(π‘₯)=βˆ’π‘₯.

Using this, we find that ddcos𝑦π‘₯=7+4π‘₯ and ddsin𝑧π‘₯=βˆ’π‘₯.

Substituting these back into our formula, we reach ddcossin𝑦𝑧=7+4π‘₯βˆ’π‘₯.

The question asked us to find the derivative at π‘₯=πœ‹6, so we need to substitute this into dd𝑦𝑧. This give us ddcossin𝑦𝑧|||=7+4βˆ’=7+4.ο—οŠ²οŽ„οŠ¬οŽ„οŠ¬βˆšοŠ©οŠ¨οŠ±οŠ§οŠ¨ο‘½οŽ₯

Simplifying this, we can see that our solution is dd𝑦𝑧|||=14βˆ’4√3.ο—οŠ²ο‘½οŽ₯

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 π‘₯=5πœƒsec and 𝑦=5πœƒtan at πœƒ=πœ‹6.

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 πœƒ: dd𝑦π‘₯=.ddddο˜οΌο—οΌ

Now we need to differentiate these trigonometric equations, which will give us ddsecπ‘¦πœƒ=5πœƒοŠ¨ and ddsectanπ‘₯πœƒ=5πœƒπœƒ.

Putting these back into our formula, we obtain ddsecsectansectancoscossin𝑦π‘₯=5πœƒ5πœƒπœƒ=πœƒπœƒ=πœƒπœƒπœƒ.

This can be simplified further which gives us that our derivative is ddsin𝑦π‘₯=1πœƒ.

In order to find the slope of the tangent at πœƒ=πœ‹6, we need to substitute this into our differential. When we do this, we obtain ddsin𝑦π‘₯|||=1=2.οΌοŠ²οŽ„οŠ¬ο‘½οŽ₯

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 π‘₯ and 𝑦 when πœƒ=πœ‹6. Substituting πœƒ=πœ‹6 into our equations for π‘₯ and 𝑦 gives us π‘₯=5πœ‹6=10√33,𝑦=5πœ‹6=5√33.sectan

Now that we have the slope of our tangent, 2, and a point that the tangent passes through, ο€Ώ10√33,5√33, we are able to form our equation. We can use the formula π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯), where π‘š is the slope of the tangent and (π‘₯,𝑦) is a point that the tangent passes through. This gives us π‘¦βˆ’5√33=2ο€Ώπ‘₯βˆ’10√33=2π‘₯βˆ’20√33.

Next, we can move everything to the left-hand side of the equation: π‘¦βˆ’2π‘₯+15√33=0.

And finally, we simplify to reach our solution: π‘¦βˆ’2π‘₯+5√3=0.

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, π‘š, of the curve at the given point, we must use the fact that slopeofthenormalslopeofthetangent=βˆ’1.

So, in the case that the slope of the tangent is π‘š, the slope of the normal would be βˆ’1π‘š.

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 π‘š at which the curve π‘₯=8π‘š+5π‘š+π‘šβˆ’1, 𝑦=5π‘šβˆ’π‘š+2 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 dd𝑦π‘₯=.ddddο˜ο‰ο—ο‰

This derivative will get larger and larger as ddπ‘₯π‘š gets smaller and smaller. Therefore, our derivative, and the slope of the tangent, will tend to infinity as ddπ‘₯π‘š tends to zero.

Let us find the values of π‘š such that ddπ‘₯π‘š=0. We can start by differentiating π‘₯ with respect to π‘š. We obtain ddπ‘₯π‘š=24π‘š+10π‘š+1.

Now, we need to set this derivative equal to 0 and solve to find the values of π‘š. We have that 24π‘š+10π‘š+1=0.

We can use the quadratic formula which tells us that the solutions to a quadratic equation of the form π‘Žπ‘š+π‘π‘š+𝑐=0 are π‘š=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

In our case, π‘Ž=24, 𝑏=10, and 𝑐=1; therefore, π‘š=βˆ’10±√10βˆ’4Γ—24Γ—12Γ—24=βˆ’10±√100βˆ’9648=βˆ’10Β±248.

We can now split up our two solutions that come from the plus or minus sign. Firstly, taking the plus, π‘š=βˆ’848=βˆ’16.

And, secondly, taking the minus, π‘š=βˆ’1248=βˆ’14.

Therefore, the values of π‘š at which our curve has vertical tangents are when π‘š=βˆ’16 and when π‘š=βˆ’14.

Key Points

  • For a pair of differentiable parametric equations, π‘₯=𝑓(𝑑) and 𝑦=𝑔(𝑑), we define the derivative of the parametric equations as dd𝑦π‘₯=ddddο˜οο—ο when ddπ‘₯𝑑≠0.
  • If we have two functions, 𝑦=𝑓(π‘₯) and 𝑧=𝑔(π‘₯), then we can define the derivative of 𝑓(π‘₯) with respect to 𝑔(π‘₯) as dd𝑦𝑧=.ddddο˜ο—ο™ο—
  • 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 π‘¦βˆ’π‘¦=π‘š(π‘₯βˆ’π‘₯), where (π‘₯,𝑦) is a point on the tangent or normal.

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