Derivatives of Function in Parametric Form & Sample Questions

Jasmine Grover logo

Jasmine Grover

Content Strategy Manager

A mathematical function is an expression commonly used to state a relation between two variables, one of these two variables is independent or dependent. A function between two variables x and y is represented as y = f(x). This correlates two variables x and y, but at times this correlation may get complicated and thus often a third variable is introduced to ease up the relation and calculation. This third variable is known as the parameter and the way of correlation is called the parametric form.

Read Also: Linear Equation in Two Variables

Key Takeaways: Function, Variables, Parameters, Parametric Formula, Derivatives of Function in Parametric Form, Relation, Calculation 


What are Functions?

[Click Here for Sample Questions]

A mathematical function is an expression, law, or rule, commonly used to establish a relation between two variables, these two variables are independent and dependent. The function of two variables is represented as: 

y = f(x)

This means the y variable is equal to a function of the x variable. This denotes that for every specific value of the variable x, a specific value would be present for the variable y too. The concept of functions was first explained by the renowned mathematician Peter Dirichlet in 1837.

The video below explains this:

Derivatives of Function in Parametric Form Detailed Video Explanation:

Read more: Continuity and Differentiability


What is Parametric Equation?

[Click Here for Sample Questions]

At times, the use of the function to correlate an independent and a dependent variable becomes complex. In such cases, a third variable is introduced to ease the calculation. This third variable is known as a parameter and the expression using the third variable is known as the parametric equation.

Thus, now there are three variables - x, y, and t. Since t is the parameter, it becomes the independent variable, and x and y become the dependent variables. Hence, x and y are represented as equations of t.

Also Read:


Derivative of the Function in Parametric Form

[Click Here for Sample Questions]

The derivative of a function in parametric form is derived in two parts; the first derivative and the second derivative. To derive the equation, let us suppose there are two dependent variables x and y and one independent variable t.

Hence, x = (x)t, and y = (y)t

First Derivative

We know that,

dy = dy/dt and dx = dx/dt

We need to take a ratio of dy and dx, thus,

\(\frac{dy}{dx}\) = \(\frac{dy/dt}{dx/dt}\)

Thus, using the chain rule of derivatives we obtain,

\(\frac{dy}{dt}\)= \(\frac{dy}{dx}\) X\(\frac{dt}{dx}\)

Divide both the sides of the equation with dx/dt

\(\frac{dy}{dt}\)X \( \frac{dt}{dx}\)= \(\frac{dy}{dx}\) X \(\frac{dx}{dt}\) X \(\frac{dt}{dx}\)

\(\frac{dy}{dx}\) = \(\frac{dy}{dx}\)

Thus, it can be simply written as \(\frac{dy}{dx}\) (t)

Second Derivative

This is derived using the quotient rule of derivatives. Here the ratio of dy/dx is squared. This is the final formula representing the derivative formula of parametric form.

\(\frac{d^2y}{dx^2}\) = \(\frac{d}{dx}(\frac{d}{dx})\)x

\(\frac{d^2y}{dx^2}\) =\(\frac{\frac{d}{dx}\frac{dy}{dx}}{dx-dt} \)

Thus, the derivative of parametric form is:

\(\frac{d^2y}{dx^2}\) =\(\frac{\frac{d}{dx}\frac{dy}{dx}}{dx-dt} \)


How to Solve Questions Based on the Parametric Form?

[Click Here for Sample Questions]

Any question based on this topic can be solved in three steps:

  1. The variables are written as a function of the parameter, t.
  2. Then differentiate the variables in terms of dy/dt and dx = dx/dt.
  3. Use the final formula to determine the values.

Discover about the Chapter video:

Continuity and Differentiability Detailed Video Explanation:

Read more: Class 12 Mathematics Chapter 5 Continuity and Differentiability


Application of Parametric Equation

[Click Here for Sample Questions]

Parametric equation is used widely in the branch of science. Two widely used areas are:

  1. Circular motion and curves: Here the variables are expressed in terms of sine, and cosine. 
  2. Projectile motion: Projectile motion has a vertical parameter, and a horizontal parameter. Both of them are independent of each other and are represented in terms of sin and cos.

Also Read:


Things to Remember

  • A mathematical function is an expression, law, or rule, used to establish a relation between two variables.
  • When the use of the function to correlate an independent and a dependent variable becomes complex, a third variable known as a parameter is used. 
  • The derivative of a function of parametric form is derived in two parts: the first derivative and the second derivative.
  • First derivative can be simply written as, \(\frac{dy}{dx}\)(t)
  • Second derivative can be written as,  \( \)

Sample Questions

Ques: Determine dy/dx value for a given set of variables having values, x = sin (t) and y = t3 (2 marks)

Ans: dy/dt = 3t

dx/dt = cos(t)

dy/dx = dy/dt X dt/dx

dy/dx = 3t/cos(t)

Ques: Determine dy/dx value for a given set of variables having values, x = 4/t4 and y = 3t4 + 5 (3 marks)

Ans: dy/dt = 12t3

dx/dt = 16/t5

dy/dx = dy/dt X dt/dx

dy/dx = 12t8/16 = 3t8/4

Ques: Determine dy/dx value for a given set of variables having values, x = t3 and y = t2 (3 marks)

Ans: dy/dt = 2t

dx/dt = 3t2

dy/dx = dy/dt X dt/dx

dy/dx = 2t/3t

Ques: Determine dy/dx value for a given set of variables having values, x = 2 and y = 4 (2 marks)

Ans: The differentiation value for constants is 1. 

Thus, dy/dx = 1

Ques: Determine dy/dx value for a given set of variables having values, x = e3t and y = e2t (3 marks)

Ans: dy/dt = 2e2t

dx/dt = 3e3t

dy/dx = dy/dt X dt/dx

dy/dx = 2e2t / 3e3t = (2/3) e-t 

Ques: Calculate the parameter equation for a given set of variables having values, x = at3 and y = bt2 (3 marks)

Ans: dy/dt = 2bt

dx/dt = 3at2

dy/dx = dy/dt X dt/dx

dy/dx = 2bt / 3at2 = (2b/3a) t-1

Ques: Determine the parameter equation value for a given set of variables having values, x = 2 and y = 4 (3 marks)

Ans: dy/dt = 2bt

dx/dt = 3at2

dy/dx = dy/dt X dt/dx

dy/dx = 2bt / 3at2 = (2b/3a) t-1 

Ques: Determine the parameter equation value for a given set of variables having values, x = sin t and y = cos t (3 marks)

Ans: dy/dt = sin t

dx/dt = sin t

dy/dx = dy/dt X dt/dx

dy/dx = sin t / sin t = 1

Ques: Determine the parameter equation value for a given set of variables having values, x = sin2 (t) and y = cos2 (t) (3 marks)

Ans: dy/dt = -2sin cos t

dx/dt = 2sin cos t

dy/dx = dy/dt X dt/dx

dy/dx = -2sin cos t / 2sin cos t = -1

Ques: Determine the parameter equation value for a given set of variables having values, x = sin2 t and y = cos t (3 marks)

Ans: dy/dt = -2sin cos t

dx/dt = sin t

dy/dx = dy/dt X dt/dx

dy/dx = -2sin cos t / sin t = -2 cos t

For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates


Check-Out: 

CBSE CLASS XII Related Questions

1.

 If \(\frac{d}{dx}f(x) = 4x^3-\frac{3}{x^4}\) such that \(f(2)=0\), then \(f(x)\) is

    • \(x^4+\frac{1}{x^3}-\frac{129}{8}\)

    • \(x^3+\frac{1}{x^4}+\frac{129}{8}\)

    • \(x^4+\frac{1}{x^3}+\frac{129}{8}\)

    • \(x^3+\frac{1}{x^4}-\frac{129}{8}\)

    2.
    By using the properties of definite integrals, evaluate the integral: \(∫_0^π log(1+cosx)dx\)

        3.

        Let A=\(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\),show that(aI+bA)n=anI+nan-1bA,where I is the identity matrix of order 2 and n∈N

            4.
            For what values of x,\(\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\)\(\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 & 1 \\1&0&2 \end{bmatrix}\)\(\begin{bmatrix} 0 \\2\\x\end{bmatrix}\)=O?

                5.
                Find the vector and the cartesian equations of the lines that pass through the origin and(5,-2,3).

                    6.
                    Find the following integral: \(\int (ax^2+bx+c)dx\)

                        CBSE CLASS XII Previous Year Papers

                        Comments



                        No Comments To Show