Derivative of Functions in Parametric Forms

Derivatives of the functions express the rate of change in the functions. We know how to calculate the derivatives for standard functions. Chain rule, product rule, and division rule are used to calculate the derivatives of the complex functions which are made up of composition from two or more functions. These functions have two variables that are related to each other in an implicit or an explicit manner. Sometimes we encounter functions in which variables are not related to each other implicitly or explicitly, instead, they are related to each other through a third variable. Let’s see how to calculate the derivatives for such functions in detail.

How to Find Derivatives of Functions in Parametric Forms?

Let’s say we have two variables x and y, usually, such variables are related to each other in an implicit or an explicit manner. But in some cases, these variables are related to each other through a third variable. This form is called the parametric form of the equation and the variable is called a parameter.

x = f(t) and y = g(t) here, t is a parameter.

The derivative of such functions is given by chain rule,

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

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ wherever $\frac{dx}{dt} \ne 0$

Thus, $\frac{dy}{dx}$ = $\frac{g'(t)}{f'(t)}$ (as $\frac{dy}{dx} = g'(t)$ and $\frac{dx}{dt} = f'(t)$ )

Sample Problems

Question 1: Find $\frac{dy}{dx}$ , if x = acos( $\theta$ ) , y = asin( $\theta$ ).

Solution:

x = acos(θ) and y = asin(θ)
$\frac{dx}{d\theta} = -sin(\theta)$
$\frac{dy}{d\theta} = acos(\theta)$
Now, let’s find out
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
⇒ $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$
⇒ $\frac{dy}{dx} = \frac{acos(\theta)}{-asin(\theta)}$
⇒ $\frac{dy}{dx} = -cot(\theta)$

Question 2: Find $\frac{dy}{dx}$ , if x = acos²( $\theta$ ) , y = asin²( $\theta$ ).

Solution:

x = acos²( $\theta$ ) , y = asin²( $\theta$ )
$\frac{dx}{d\theta} = -2acos(\theta)sin(\theta)$
$\frac{dx}{d\theta} = 2acos(\theta)sin(\theta)$
Now, let’s find out
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
⇒ $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$
⇒ $\frac{dy}{dx} = \frac{-2acos(\theta)sin(\theta)}{2acos(\theta)sin(\theta)}$
⇒ $\frac{dy}{dx} = -1$

Question 3: Find $\frac{dy}{dx}$ , if x = at² + 2t, y = t at t = 0.

Solution:

x = at² + 2t, y = t
$\frac{dx}{dt} = 2at + 2$
$\frac{dy}{dt} = 2$
Now, let’s find out
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
⇒ $\frac{dy}{dx} = \frac{2}{2at + 2}$
⇒ $\frac{dy}{dx} = \frac{2}{ 2}$
⇒ $\frac{dy}{dx} = 1$

Question 4: Find $\frac{dy}{dx}$ , if x = at³ + 2t², y = t² at t = 1.

Solution:

x = at³ + 2t², y = t²
$\frac{dx}{dt} = 3at^2 + 4t$
$\frac{dy}{dt} = 2t$
Now, let’s find out
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
⇒ $\frac{dy}{dx} = \frac{2t}{3at^2 + 4}$
⇒ $\frac{dy}{dx} = \frac{2}{3a + 4}$

Question 5: Find $\frac{dy}{dx}$ , if x = 4t, y = $\frac{1}{t}$ at t = 1.

Solution:

x = 4t, y = $\frac{1}{t}$
$\frac{dx}{dt} = 4$
$\frac{dy}{dt} = -\frac{1}{t^2}$
Now, let’s find out
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
⇒ $\frac{dy}{dx} = \frac{-\frac{1}{t^2}}{4}$
⇒ $\frac{dy}{dx} = -\frac{1}{4t^2}$
At t = 1
$\frac{dy}{dx} = -\frac{1}{4}$

Question 6: Find $\frac{dy}{dx}$ , if x = 4e^t, y = cos(t)at t = 1.

Solution:

x = 4e^t, y =cos(t)
$\frac{dx}{dt} = 4e^t$
$\frac{dy}{dt} = -sin(t)$
Now, let’s find out
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
⇒ $\frac{dy}{dx} = \frac{-sin(t)}{4e^t}$
At t = 1
⇒ $\frac{dy}{dx} = -\frac{sin(1)}{4e}$

Question 7: Find $\frac{dy}{dx}$ , if x = e^t + sin(t), y = t², at = 0.

Solution:

x = e^t + sin(t), y =t²
$\frac{dx}{dt} = e^t + cos(t)$
$\frac{dy}{dt} = 2t$
Now, let’s find out
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
⇒ $\frac{dy}{dx} = \frac{2t}{e^t + cos(t)}$
At t = 0
⇒ $\frac{dy}{dx} = 0$

Question 8: Find $\frac{dy}{dx}$ , if x = tsin(t), y = cos(t), at t = $\frac{\pi}{2}$ .

Solution:

x = tsin(t), y = cos(t)
$\frac{dx}{dt} = tcos(t) + sin(t)$
$\frac{dy}{dt} = -sin(t)$
Now, let’s find out
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
⇒ $\frac{dy}{dx} = \frac{-sin(t)}{tcos(t) + sin(t)}$
At t = $\frac{\pi}{2}$
⇒ $\frac{dy}{dx} = \frac{-sin(\frac{\pi}{2})}{\frac{\pi}{2}cos(\frac{\pi}{2}) + sin(\frac{\pi}{2})}$
⇒ $\frac{dy}{dx} = -1$

Question 9: Find $\frac{dy}{dx}$ , if x = 4t² + 10, y =t²at t = 1.

Solution:

x = 4t² + 10, y =t²
$\frac{dx}{dt} = 8t$
$\frac{dy}{dt} = 2t$
Now, let’s find out
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
⇒ $\frac{dy}{dx} = \frac{2t}{8t}$
⇒ $\frac{dy}{dx} = \frac{1}{4}$
At t = 1
$\frac{dy}{dx} = \frac{1}{4}$

Question 10: Find $\frac{dy}{dx}$ , if x = at², y =2at at t = 1.

Solution:

x = at², y =2at
$\frac{dx}{dt} = 2at$
$\frac{dy}{dt} = 2a$
Now, let’s find out
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
⇒ $\frac{dy}{dx} = \frac{2a}{2at}$
⇒ $\frac{dy}{dx} = \frac{1}{t}$
At t = 1
$\frac{dy}{dx} = 1$

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Last Updated : 16 May, 2021

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