Use of Derivative Functions in Parametric Form

Continuity and Differentiability 2025 ( Maths Continuity and Differentiability )

Jaya Sharma
Updated on Aug 14, 2025 12:08 IST

By Jaya Sharma, Assistant Manager - Content

Parametric equations describe curves in the plane by expressing coordinates x and y as functions of a third variable, which is called the parameter, denoted by t. These equations can represent more complex and varied curves that include those which are not functions in a traditional way. Derivative of a function given in parametric form enables computation of slopes, tangent lines, concavity and other geometric properties directly from parametric representation without converting to Cartesian form.

parametric form

Through this topic from the continuity and differentiability chapter, you will learn about the parametric form in detail. Once you have gone through all the topics of the chapter, continue to practice NCERT exercise of class 12 Math chapter 5.

Table of content
  • What are Parametric Equations?
  • Derivatives in Parametric Form
  • Applications of Derivative in Parametric Form
  • Mistakes to Avoid While Writing in Parametric Equations
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What are Parametric Equations?

These equations define the coordinates of points on a curve as functions of parameter t: x = f(t), y = g(t). Here, t is a real number in an interval. The CBSE board exam asks questions related to this equation on a basic level. Representation in the form of a parametric equation is useful for those curves, like curves and ellipses, that are not easily represented as y = f(x) or x = g(y). A unit circle is expressed in a parametric equation as shown below: x=cost, y=sint. Here, t ranges over [0,2π]. A parabola y = x 2 can be parametrically represented as: x=t, y = t 2

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Derivatives in Parametric Form

 Let us now take a look at the first derivative and second derivative:

1. First Derivative ( d y d x )

The first derivative of y with respect to x for a curve that is defined parametrically is given by d y d x = d y/dt d x/dt

The formula derived from chain rule states that d y d t = d y d x d x d t

After rearranging the above equation, we will get the derivative in a parametric form. Questions based on this will be asked in IIT JAM entrance exam and JEE Main exam.

The derivative ( d y d x ) shows the slope of the tangent line to the curve at any point.

2. Second Derivative ( d 2 y d x 2 )

Second derivative is the derivative of first derivative w.r.t. x. This derivative determines both concavity and curvature of a curve. d 2 y d x 2 = d d x ( d y d x ) = d d t ( d y d x ) d x/dt

By using the quotient rule, it will be:

d 2 y d x 2 = d d t ( d y d t ) d x d t d y d t d 2 x d t 2 ( d x d t ) 3

Complete Class 12 Study Material

NCERT 12 Maths Solutions

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CBSE Sample Papers for 12th Maths

Class 12 Physics Sample Papers

Class 12 CBSE Sample Paper for Chemistry

CBSE Class 12 Maths Previous Year Question Papers

CBSE Class 12 Physics Previous Year Question Papers

CBSE Chemistry Class 12 Question Papers

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Applications of Derivative in Parametric Form

Let us understand the applications of derivatives in parametric form:

  • Finding Slope of Tangent Lines: The first derivative gives slope of tangent line to a parametric curve at any point. This is important for understanding the direction and steepness of curve at any point.
  • Determining concavity: The second derivative d 2 y d x 2 helps in determining the concavity of the curve. If                   d 2        y                  d        x 2         >0 , the curve is concave upwards. On the other hand, if d 2        y                  d        x 2         <0, the curve will concave downwards.
  • Identifying Tangents: Vertical tangents occur when dx d t = 0 and horizontal tangents occur when  dy d t = 0.
  • Analyzing Motion: In Physics, parametric equations describe the position of an object as a function of time t. Derivatives dx d t and dy d t represent the velocity component of objects. 

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Try these practice questions

Q1:

Let a > 0, b > 0. Let a and  respectively be the eccentricity and length of the latus rectum of the hyperbola x 2 a 2 y 2 b 2 = 1 .  Let e’ and l ' respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. I

View Full Question

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Mistakes to Avoid While Writing in Parametric Equations

Let us take a look at the mistakes that must be avoided when using the parametric form. NEET exam aspirants must keep these pointers in mind.

  • Forgetting to divide by d x d t : The derivative d y d x is the ratio d y/dt d x/dt. , not just d y d t .
  • Incorrect differentiation: Ensure proper application of differentiation rules (chain rule, product rule, quotient rule) when computing d x d t and d y d t .
  • Ignoring special cases: Always check for d x d t = 0 or d y d t = 0 to identify vertical or horizontal tangents.
Q&A Icon
Commonly asked questions
Q:  

Why do we use parametric equations?

A: 

We use parametric equations for the following reasons:

  • Parametric equations represent all those curves that are otherwise impossible to be represented as a single function. Circles, cycloids, ellipses and spirals are all described using parametric equations.
  • These equations can easily describe motion of objects over time.
  • Parametric equations help in breaking down complex relationships into simpler components. Rather than dealing with single complex equation, one can describe x and y seperately in terms of t parameter.
  • These equations extend to three dimensions easily and naturally, so that one can describe curves and surfaces in 3D space.
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Maths Continuity and Differentiability Exam

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