## Differentiation Of Parametric Function

Differentiation of parametric function is another interesting method in the topic differentiation. In this method we will have two functions known as x and y. Each function will be defined using another third variable. To understand this topic more let us see some examples. Example 1:

Find  dy/dx for the below parametric functions.

x = a cos θ y = b sin θ

Solution:

Here the two variables x and y are defined using third variable θ. By using the function x we can only find dx/dθ and by using the function y we can only find dy/dθ. By dividing these two derivatives we can find dy/dx.

x = a cos θ                       y = b sin θ

dx/dθ = a (-sin θ)             dy/dθ = b cos θ

dx/dθ = - a sin θ

Now we can find dy/dx

dy/dx = (dy/dθ)/(dx/dθ)

dy/dx = (b cos θ)/(- a sin θ)

dy/dx = -b/a (cos θ/ sin θ)

dy/dx = -(b/a) cot θ          differentiation of parametric function      differentiation of parametric function

Example 2:

Find dy/dx by using the parametric functions

x = a sec³ θ    y = b tan³ θ

Solution:

Here the two variables x and y are defined using third variable θ. By using the function x we can only find dx/dθ and by using the function y we can only find dy/dθ. By dividing these two derivatives we can find dy/dx.

x = a sec³ θ

dx/dθ = 3a sec² θ(sec θ tan θ)

= 3a sec³ θ tan θ

y = b tan³ θ

dy/dθ = 3b tan² θ  sec² θ

Now we have divide those two derivatives to find dy/dx

dy/dx = (dy/dθ)/(dx/dθ)

= (3b tan² θ  sec² θ)/(3a sec³ θ tan θ)

= (b/a) (tan θ/sec θ)

= (b/a) (sin θ/cos θ)/(1/cos θ)

= (b/a) (sin θ/cos θ) x (cos θ/1)

= (b/a) sin θ

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Math is not only solving problems and finding solutions and it is also doing many things in our day to day life.  They are: 