[Maths Class Notes] on Derivatives of Function in Parametric Form Pdf for Exam

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 

Or, 

[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 

Therefore, 

[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 

Therefore, 

[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}]

Therefore, 

[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   

Therefore, 

[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

Therefore, 

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