Derivatives Of A Function In Parametric Form

There are instances when rather than defining a function explicitly or implicitly we define it using a third variable. This representation when a function y(x)  is represented via a third variable which is known as the parameter is a parametric form. A relation between x and y  expressible in the form x = f(t) and  y = g(t) is a parametric form representation with parameter as t. Now we will concentrate on how to differentiate these functions using parametric differentiation.

Derivatives - Function In Parametric Form

As we already know by the chain rule,

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

This chain rule can also be rewritten as,

\( \frac {dy}{dx} = \frac {\frac {dy}{dt}}{\frac {dx}{dt}} \) (Where \( \frac {dx}{dt} \ne 0 \) )

Also as mentioned earlier  y’ = g'(t) = \( \frac {dy}{dt} \space and \space x’ = f'(t) = \frac {dx}{dt} \)

Thus we can say that,

\( \frac{dy}{dx} = \frac{g'(t)}{f'(t)} \) ( where f'(t) \( \ne 0 \) )

Rules for solving problems on derivatives of functions expressed in parametric form:

Step i) First of all we write the given functions x and y in terms of the parameter t.

Step ii) Using differentiation find out \( \frac{dy}{dt} \space  and \space \frac {dx}{dt} \) .

Step iii) Then by using the formula used for solving functions in parametric form i.e.\( \frac {dy}{dx} = \frac {\frac {dy}{dt}}{\frac {dx}{dt}} \)

Step iv) Lastly substituting the values of \( \frac {dy}{dt} \space and \space \frac{dx}{dt} \)  and simplify to obtain the result.

Now let us go through some examples to get a deeper insight for solving functions of parametric form.

Example 1: Find the value of \( \frac {dy}{dx} \) for the following functions which are expressed in the parametric form.

i) x = sin t , y = \( t^2 \)

ii) x = \( \frac {3}{t^3} , y = 3t^4 + 5 \)

Solution 1:

i) Since this function is represented in parametric function format beforehand therefore we need to find out the value  \( \frac {dy}{dt} \space and \frac {dx}{dt} \).

Now, \( \frac {dy}{dt} = 2t \)

\( \frac {dx}{dt} = cost \)

Now with the help of chain rule we can write,

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

Now substituting the value of \( \frac {dy}{dt} \space  and \space \frac {dx}{dt} \) into the above equation we can find the derivative of  w.r.t x

\( \Rightarrow \frac {dy}{dx} = \frac {2t}{cost} \)

This is the required solution of the differentiation of the parametric equation.

ii) Now as these functions are already expressed in terms of therefore to find \( \frac {dy}{dx} \) , we find out \( \frac {dy}{dt}~and~\frac {dx}{dt} \) sepatately. So

\( \frac {dy}{dt} = 12t^3 \)

\( \frac {dx}{dt} = \frac {-9}{t^4} \)

We know,

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

\( \frac {dy}{dx} = \frac {12t^3}{\frac {-9}{t^4}} \)

\( \frac {dy}{dx} = \frac {-4t^7}{3} \)

This is the required solution of the differentiation of the parametric equation.

Example 2: Find the value of  \( \frac{dy}{dx} \)  for y = \( e^{sin t} \) and \( x = 3t^3 \)

Solution 2: The given functions are parametric in nature.

And we know, \( \frac {dy}{dx} = \frac {\frac{dy}{dt}}{\frac{dx}{dt}} \)

\( \frac {dy}{dt} = e^{sin t} × cost \)

\( \frac {dx}{dt} = 9t^2 \)

\( \Rightarrow \frac {dy}{dx} = \frac {e^{sint} × cos t}{9t^2} \)<

This is the required solution.

We are now thorough through the concept of parametric function. To learn more derivative of a function visit our app-BYJU’s the learning app.


Practise This Question

The outline of a half moon can be best described as a