# Differentiation of parametric equations with Examples

The derivatives of parametric equations are found by deriving each equation with respect to t. Then, the chain rule is used to obtain a derivative of y with respect to x.

Here, we will learn how to find the derivatives of parametric equations. We will use several examples and practice problems.

##### CALCULUS

Relevant for

Learning about derivatives of parametric equations.

See examples

##### CALCULUS

Relevant for

Learning about derivatives of parametric equations.

See examples

## Differentiation of parametric equations

Let’s learn how to find the derivatives of parametric equations using an example.

Consider the following parametric equations:

$latex x=t+1~~$ $latex ~~y=t^2$

When we derive $latex x$ with respect to $latex t$ and when we derive $latex y$ with respect to $latex t$, we have:

$latex \dfrac{dx}{dt}=1~~$ $latex ~~\dfrac{dy}{dt}=2t$

Using the chain rule, we can write as follows:

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

Therefore, we have:

$$\frac{dy}{dx}=2t(1)=2t$$

Since we have $latex x=t+1$, we can write $latex t=x-1$ and the derivative is:

$$\frac{dy}{dx}=2t$$

$$\frac{dy}{dx}=2x-2$$

## Parametric derivatives – Examples with answers

### EXAMPLE 1

Find $latex \dfrac{dy}{dx}$ in terms of the parameter $latex t$ for:

$latex y=3t^2+2t$

$latex x=1-2t$

We start by finding the derivatives $latex \frac{dy}{dt}$ and $latex \frac{dx}{dt}$:

• When $latex y=3t^2+2t$, we have: $latex \dfrac{dy}{dt}=6t+2$
• When $latex x=1-2t$, we have: $latex \dfrac{dx}{dt}=-2$

Using the chain rule, we can write as follows:

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

$$=(6t+2)\left(-\frac{1}{2}\right)$$

$$\frac{dy}{dx}=-3t-1$$

### EXAMPLE 2

What is the derivative $latex \dfrac{dy}{dx}$ of the following equations? Write it in terms of the parameter $latex t$.

$latex y=(1+2t)^3$

$latex x=t^3$

Let’s find the derivatives $latex \frac{dy}{dt}$ and $latex \frac{dx}{dt}$:

• When $latex y=(1+2t)^3$, we have: $latex \dfrac{dy}{dt}=3(1+2t)^2(2)$. Then: $latex \dfrac{dy}{dt}=6(1+2t)^2$
• When $latex x=t^3$, we have : $latex \dfrac{dx}{dt}=3t^2$

With the chain rule, we have:

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

$$=6(1+2t)^2\times \frac{1}{3t^2}$$

$$\frac{dy}{dx}=\frac{2(1+2t)^2}{t^2}$$

### EXAMPLE 3

Find $latex \dfrac{dy}{dx}$ in terms of $latex t$ for the following equations:

$latex x=3t^4$

$latex y=2t^2-3$

We have to find the derivatives $latex \frac{dy}{dt}$ and $latex \frac{dx}{dt}$:

• When $latex y=2t^2-3$, we have: $latex \dfrac{dy}{dt}=4t$
• When $latex x=3t^4$, we have: $latex \dfrac{dx}{dt}=12t^3$

Now, we can use the chain rule to find $latex \dfrac{dy}{dx}$:

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

$$=4t\times \frac{1}{12t^3}$$

$$\frac{dy}{dx}=\frac{1}{3t^2}$$

### EXAMPLE 4

What is the derivative $latex \dfrac{dy}{dx}$ of the following equations? Write it in terms of $latex t$.

$latex x=2\sqrt{t}$

$latex y=5t-4$

We start by finding the derivatives $latex \frac{dy}{dt}$ and $latex \frac{dx}{dt}$:

• When $latex y=5t-4$, we have: $latex \dfrac{dy}{dt}=5$
• When $latex x=2\sqrt{t}=2t^{\frac{1}{2}}$, we have: $latex \dfrac{dx}{dt}=t^{-\frac{1}{2}}$

Now, we find $latex \dfrac{dy}{dx}$ using the chain rule:

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

$$=5\times \frac{1}{t^{-\frac{1}{2}}}$$

$$\frac{dy}{dx}=5\sqrt{t}$$

### EXAMPLE 5

Find the equation of the tangent line to the curve given parametrically by $latex x=\frac{2}{t}$ and $latex y=3t^2-1$ at the point (2, 2).

We can find the value of the parameter $latex t$ by substituting $latex x=2$, $latex y=2$ into the equations

$latex x=\dfrac{2}{t}~~$ $latex ~~y=3t^2-1$

This gives us $latex t=1$.

When differentiating parametrically, we have:

$latex \dfrac{dy}{dt}=6t~~$ $latex ~~\dfrac{dx}{dt}=-\dfrac{2}{t^2}$

When $latex t=1$, we have:

$$\frac{dy}{dx}\Big|_{t=1}=-3$$

This means that the slope of the tangent line is -3. Then, the equation of the tangent is of the form

$latex y=-3x+c$

Since the tangent line passes through the point (2, 2), we have:

$latex 2=-3(2)+c$

$latex c=8$

The equation of the tangent line is $latex y=-3x+8$.

### EXAMPLE 6

Find the second derivative $latex \dfrac{d^2y}{dx^2}$ in terms of $latex t$ for the following parametric equations:

$latex x=t+1$

$latex y=t^3$

Differentiating each equation with respect to $latex t$, we have:

$latex \dfrac{dy}{dt}=1~~$ $latex ~~\dfrac{dy}{dt}=3t^2$

Using the chain rule, we have

$$\frac{dy}{dx}=\frac{dy}{dt} \frac{dt}{dx}=3t^2$$

Now, we derive each term of this equation with respect to $latex x$ to find the second derivative $latex \dfrac{d^2y}{dx^2}$:

$$\frac{d}{dx}\left( \frac{dy}{dx}\right)=\frac{d}{dx}(3t^2)$$

Since we cannot derive $latex 3t^2$ with respect to $latex x$, we use the chain rule:

$$\frac{d}{dx}(3t^2)=\frac{d}{dt}(3t^2) \frac{dt}{dx}$$

$$=6t \frac{dt}{dx}$$

Substituting this, we have:

$$\frac{d^2y}{dx^2}=6t \frac{dt}{dx}$$

Since $latex \frac{dx}{dt}=1$, we can write:

$$\frac{dx}{dt}=\frac{1}{\frac{dx}{dt}}=1$$

Then, the second derivative is:

$$\frac{d^2y}{dx^2}=6t (1)=6t$$

## Parametric derivatives – Practice problems

Parametric derivatives quiz
You have completed the quiz!

#### The derivative $latex \frac{dy}{dx}$ in terms of t of the equations $latex x=4t(t-2)$ and $latex y=(t-1)^3$ is a fraction. What is the numerator?

Write the numerator in the input box.

$latex ~~=$