Derivatives of a function in parametric form relate to situations where you can either express a function implicitly or explicitly. It will always be defined using a third variable. This is a parametric form, which is when a function y(x) is represented by a third variable called the parameter. A relationship between x and y may be expressed as x = f(t) and y = g(t), which is a parametric form representation with t as the parameter. In this post, we’ll look at how to use parametric differentiation to differentiate these functions.
It is critical to first comprehend the behavior of a parametric function before moving on to the next topic.
Let’s have a look at an example:
Acceleration is commonly defined as:
The functions v and x, which represent velocity and location, are represented in terms of time, which is the parameter in this case. As a result, we may state that the velocity is v(t) and the location is x. (t). So, using the derivation approach, how do we compute the derivative dv/dx? Let’s see what happens.
If x equals f(t) and y equals g(t), and the two functions of a parameter t are distinct, then y can be described as a function of x. Then:
When they are given in parametric form, it is obvious that this is the first derivative of the function y with regard to x.
As a result, the second derivative may be calculated as follows:
We may use first-order parametric differentiation once more, taking into account
in the form of a parametric function t:
In the same method, we may compute the higher-order derivative. The only thing we need to keep in mind is that every time we calculate a derivative, it becomes a function of t. For your convenience, an online differentiation calculator is also accessible.
When a connection of a third variable is used to build a relationship between two variables, this third variable is referred to as a parameter.
Suppose x = f(t), y = g(t)
This relationship between x and y is said to be of parametric form with t as a parameter.
To find the derivative of a function in this form, we use the chain rule.
i.e dy/dx = dy/dt * dt/dx
=> dy/dx = (dy/dt)/(dx/dt), where dx/dt ≠ 0
=> dy/dx = [d{g(t)}/dt]/ [d{f(t)}/dt]
=> dy/dx = g’(t)/f’(t)
Here d{g(t)}/dt] = g’(t), [d{f(t)}/dt = f’(t) and f’(t) ≠ 0
Problem: Find the derivative of the function x = sin t, y = cos 2t
Solution:
The given equations are: x = sin t, y = b cos 2t
Now, dx/dt = d(sin t)/dt
=> dx/dt = cos t
and dy/dt = d(cos 2t)/dt
=> dy/dt = -sin 2t * d(2t)/dt
=> dy/dt = -2 sin 2t
Now, dy/dx = (dy/dt)/(dx/dt)
=> dy/dx = (-2 sin 2t)/cos t
=> dy/dx = (-2 * 2 * sin t * cos t)/cos t
=> dy/dx = -4 sin t
This is the required solution.
We’ve covered everything there is to know about parametric functions. To get a better understanding of the derivative of a function, derivative applications, and other related topics. Join the online Unacademy community for an additional in-depth understanding of this and other key topics.
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