## Why do we use parametric differentiation?

Parametric differentiation is used to describe the slope and concavity of parametric curves which are defined by parametric equations. These types of equations often describe curves that inter-lap at several points and are hard to describe with Cartesian equations. The parametric derivatives can then be used to construct the equations of tangents and the normals to curves.

## What are the steps for parametric differentiation?

Let's break this down.

### Apply the standard chain rule

The first step is to apply the standard Chain Rule as we are still looking for the derivative of *dy/dx*, but we need to find the correct expression for the given parametric equations. The Chain Rule is as follows:

$\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$

Let's look at an example containing the parametric equations of a circle:

$x\left(t\right)=\mathrm{cos}\left(t\right)\phantom{\rule{0ex}{0ex}}y\left(t\right)=\mathrm{sin}\left(t\right)$Using the chain rule, we now have the following expression :

$\frac{dy}{dx}=\frac{dy}{dt}\xb7\frac{dt}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

The chain rule can also be expressed as a Ratio of the two derivatives in respect to the common parameter, by rearranging the expression as shown above – provided that the denominator is not zero. This is a trick that simplifies the process, also called the reverse chain rule.

### Differentiation of each parametric equation separately

As a second step, we must carry out the differentiation of each equation. We will now proceed with differentiating the x function in respect to t, and then repeat the process for the y function. This will then be used to create the Ratio shown in step 2, to find the parametric derivative *dy/dx*, by dividing the two derivatives.

Continuing with the first example, the derivatives of x, y Functions with respect to t are as follows:

$\frac{dx}{dt}=-\mathrm{sin}\left(t\right)\phantom{\rule{0ex}{0ex}}\frac{dy}{dt}=\mathrm{cos}\left(t\right)$### Create a ratio of the parametric derivatives

The third and last step involves the substitution of each parametric derivative into the ratio expression obtained by the application of the chain rule. Proceeding with the example, we divide the y derivative over the x derivative.

$\frac{dy}{dx}\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\mathrm{cos}\left(t\right)}{-\mathrm{sin}\left(t\right)}=-cot\left(t\right)$

- Find the ratio dy/dx if dy/dθ is x
^{5}+6 and dx/dθ is x^{3}.

Solution:

We divide dy/dθ over dx/dθ to get dy/dx.

$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{{x}^{5}+6}{{x}^{3}}$

- Find the ratio du/dz is du/da is tan(a) and dz/da is cos
^{2}(a).

Solution: We divide du/da over dz/da so that the da component of the ratio is eliminated.

$\frac{du}{dz}=\frac{\frac{du}{da}}{\frac{dz}{da}}=\frac{\mathrm{tan}\left(a\right)}{{\mathrm{cos}}^{2}\left(a\right)}$

## Finding the equation of a tangent of a curve using parametric differentiation

The tangent of a curve is a straight line that touches the surface of a curve at a specific point$({x}_{1},{y}_{1})$ as seen in the figure** **below. Since a tangent is a straight line, the equation of the tangent has the form of *y = ax +c * where a is the gradient or slope and *c * is a constant.

The slope of the tangent line is equal to the derivative of the curve. If the point of tangency is known as well as the parametric equations, the equation of the tangent can be found using the parametric differentiation with the following formula, where * ${x}_{1},{y}_{1}$* are the coordinates of the point of tangency.

$y-{y}_{1}=a(x-{x}_{1})$

Find the equation of a tangent of a curve at the point (4, -6).$x={t}^{2},y={t}^{2}-5t$

Solutions:

First, perform the parametric differentiation since parametric equations are given. Using the reverse chain rule, we divide *dx/dt *over *dy/dt**.** ** *

$\frac{dx}{dt}=2t\phantom{\rule{0ex}{0ex}}\frac{dy}{dt}=2t-5\phantom{\rule{0ex}{0ex}}\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}=\frac{2t-5}{2t}$

Now, we need to find the values of t at the tangent point by substituting the coordinates given (4,-6) in one of the parametric equations.

We will substitute the y coordinate into the y parametric equation for this particular example in order to show what happens when two possible t values occur. In general, only one substitution is necessary to find a value for t at the point of the tangent.

$-6={t}^{2}-5t\Rightarrow {t}^{2}-5t+6=0\Rightarrow t=3,t=2$

In order to check which of the two values of x is the valid one, we need to check the corresponding x value if it is true for the chosen t value, as we want a value of t that is consistent and true for both x and y values.

When t=3 then:

$x={t}^{2}\Rightarrow x={3}^{2}=9\Rightarrow x=9$

but the real value of x is 4.

Therefore the value of t=3 is not valid for the point of tangency.

When t=2 then:

$x={t}^{2}\Rightarrow x={2}^{2}=4\Rightarrow x=4$

Therefore the value of t=2 is valid.

We have found the value of t that is true, so we can now continue with finding the corresponding slope by substituting the value of t into the derivative found. The last step is to substitute the slope and coordinates into the given formula:

$\frac{dy}{dx}=\frac{2t-5}{2t}=\frac{2\times 2-5}{2\times 2}=\frac{-1}{4}\phantom{\rule{0ex}{0ex}}y-{y}_{1}=a(x-{x}_{1})\Rightarrow y-(-6)=\frac{-1}{4}(x-4)\phantom{\rule{0ex}{0ex}}y+6=\frac{-1}{4}x+\frac{-1}{4}\xb7(-4)\phantom{\rule{0ex}{0ex}}y=-\frac{1}{4}x-5$## How to find the equation of a normal to a curve

A normal to a curve is also a straight line that is perpendicular to the tangent of the curve at a specific point*$({x}_{1},{y}_{1})$* as seen in the figure below. Similarly to the equation of a tangent, the equation of the normal can also be found using the derivative at the specified point. However, the slope of the normal must be the negative reciprocal of the derivative of the curve like shown in the formula below where T and N indicate the slope of the tangent and normal respectively.

$Slop{e}_{T}\times Slop{e}_{N}=-1$

Find the normal to the curve of the example above.

Solutions:

Continuing the example above, the equation of the normal to the curve can be found by applying the formula given. The slope of the tangent was found to be -1/4 at the coordinates given.

$-\frac{1}{4}\xb7slop{e}_{N}=-1\Rightarrow Slop{e}_{N}=4$

Using the formula of the equation of a straight line and substituting the coordinates and slope we find the equation of the normal to the curve.

$y-{y}_{1}=a(x-{x}_{1})\Rightarrow y-(-6)=4(x-4)\phantom{\rule{0ex}{0ex}}y=4x-22$## Parametric Differentiation - Key takeaways

Parametric differentiation is used to differentiate parametric equations.

Parametric equations can be differentiated applying the reverse chain rule

The parametric derivatives are divided to remove the common parameter and find $\frac{dy}{dt}\xf7\frac{dx}{dt}$

The derivative or slope of a curve can be used to find the equation of the tangent of the curve if the point of tangency is known.

Similarly, the equation of a normal to a curve can be found using the derivative or slope as well as the points of touch.

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##### Frequently Asked Questions about Parametric Differentiation

What are parametric equations in differentiation?

Parametric equations in differentiation are equations that contain one dependent variable and one common parameter.

How do you differentiate parametric equations?

You use the reverse chain rule which is the ratio that is found by dividing the two parametric equations.

What is dy/dx in parametric equations?

dy/dx in parametric equations is the rate of change of y with respect to x. This is the derivative that is eliminated from the common parameter t, and only contains y and x.

What is the relationship between dx/ dt and dy/dt?

dx/dt is the relationship that describes the rate of change of x with respect to parameter t, while dy/dt is the relationship that describes the rate of change of y with respect to parameter t.

What is the formula for parametric differentiation?

dy/dx= (dy/dt)/(dx/dt)

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