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Derivatives of Function in Parametric Form

Last updated date: 23rd May 2024
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Parametric Derivative of a Function. 

Sometimes, the relationship between two variables becomes so complicated that we find it necessary to introduce a third variable to reduce the complication and make it easy to handle. This third variable is called parameter in mathematics and the function is said to be in a parametric form. So instead of defining a function y(x) explicitly, both x and y are defined in terms of a third variable. Basically, it is a derivative of a dependent variable with reference to another dependent variable, and both the dependent variable depends on an independent variable. Therefore, there are two equations instead of only one equation. One equation relates x with the parameter and one equation relates y to the parameter.   

Derivation in a Parametric Form

It is extremely important to first understand the behavior of a parametric function before we jump into any other discussion. So let us start with an example:

We usually define acceleration as:

a = \[\frac{dy}{dt}\]                                   

But there is an alternative definition of acceleration that gives us:

 a = v\[\frac{dy}{dx}\]                                

The function v and x i.e., velocity and position respectively are expressed in terms of time that is the parameter here. So we can say that the velocity is equal to v(t) and the position is equal to x(t). So how shall we compute the derivative dvdx using the method of derivation? Let us find out. 

If x is equal to f(t) and y is equal to g(t) and they are the two different functions of a parameter t so that y can be defined as a function of x. Then: 

\[\frac{dy}{dx}\] = \[\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\], given that \[\frac{dx}{dt}\] ≠ 0 


\[\frac{dy}{dx}\] =  \[\frac{g’(t)}{f’(t)}\] provided that f’(t) ≠ 0  

It is very clear that this is the first derivative of the function y with reference to x when they are represented in a parametric form. Therefore, we can calculate the second derivative as: 

\[\frac{d²y}{dx²}\] = \[\frac{d}{dx}\](\[\frac{dy}{dx}\])

We can apply the first-order parametric differentiation again, considering \[\frac{dy}{dx}\] as a parametric function t: 

\[\frac{d²y}{dx²}\] =  \[\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}\]

We can calculate the higher-order derivative in the same way.  The only thing that we have to remember is that whenever we calculate a derivative, it will become the function of t.  

Solved Example

Question 1) Solve x = t² and y = t³

Solution 1) x\[_{t}^{‘}\] = (t²)’ = 2t,  y\[_{t}^{‘}\] = (t³)’ = 3t² 


\[\frac{dy}{dx}\] = y’\[_{x}\] = \[\frac{{y}'_{t}}{{x}’_{t}}\] = \[\frac{3t^{2}}{2t}\] = \[\frac{3t}{2}\](t ≠ 0).  

Question 2) x = 2t + 1, y = 4t - 3

Solution 2) x\[_{t}^{‘}\] = (2t + 1) = 2,  y\[_{t}^{‘}\] = (4t - 3)’ = 4 


\[\frac{dy}{dx}\] =  y’\[_{x}\] = \[\frac{{y}'_{t}}{{x}'_{t}}\] = \[\frac{4}{2}\] = 2

Question 3) x = e\[^{2t}\], y = e\[^{3t}\]

Solution 3) x’\[_{t}\] = (e\[^{2t}\])’ = 2e\[^{2t}\] , y’\[_{x}\] = (e\[^{3t}\])’ = 3e\[^{3t}\]


\[\frac{dy}{dx}\] = y’\[_{x}\] = \[\frac{{y}'_{t}}{{x}'_{t}}\] = \[\frac{3e^{3t}}{2e^{2t}}\] = \[\frac{3}{2}\] e\[^{3t-2t}\] = \[\frac{3}{2}\]e\[^{t}\]

Question 4) x = at, y = bt²

Solution 4) \[x_{{t}'}\] = (at)' = a, y\[_{t}^{‘}\] = (bt²)’ = 2bt   


\[\frac{dy}{dx}\] = y'\[_{x}\] = \[\frac{{y}'_{t}}{{x}'_{t}}\] = \[\frac{2bt}{a}\]

Question 5) x = sin²t, y = cos²t 

Solution 5)  x\[_{t}^{‘}\]  = (sin²t)' = 2sint . cos t = sin2t, 

y\[_{t}^{‘}\] = (cos²t )' = 2cost . (-sint) = -2sint cost = -sin2t


\[\frac{dy}{dx}\] = y'\[_{x}\] = \[\frac{{y}'_{t}}{{x}'_{t}}\] = \[\frac{-sin2t}{-sin2t}\] = -1. where, t ≠ \[\frac{πn}{2}\], n ∈ Z

FAQs on Derivatives of Function in Parametric Form

Question 1. How is the 2nd Derivative Test?

Answer. The second derivative is useful in the determination of local extrema

 of a function on the basis of a few conditions. Therefore, if a function shows a critical point for which  f′(x) = 0. Also, if the second derivative turns out to be positive at this point, then it is certain that f holds a local minimum. We call this technique the Second Derivative Test for Local Extrema.

Question 2. What Does the Second Derivative Tell us?

Answer. The second derivative of a function f is useful as it measures the concavity of a graph. A function that has a positive second derivative will be concave up. It means that in the graph of the function the tangent line will lie below it.   

Question 3. What is the Purpose of a Second Derivative?

Answer. The main purpose of the second derivative is that it can measure the instant change of a quantity that itself is changing. It is very useful in the analysis of the concavity or curvature of the graph of a function.

Question 4. How is the First Derivative Test Useful?

Answer. The first derivative test is useful because it tests a function's monotonic properties to see if the function is increasing or decreasing. It focuses on the exact point in its domain. If the function fluctuates from increasing to decreasing at the point, so it will be clear that the function will gain the highest value at that point.

Question 5. What is the Difference Between 1st and 2nd Order Derivative Test?

Answer. The major difference between the first-order and the second-order derivative test is that the first derivative test examines whether a function has a local maximum, or a local minimum, or neither of them. On the other hand, the second-order derivative test loses on yielding a proper conclusion when y is 0 at a critical value.