## DIFFERENTITATION 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.

Example 1 :

Find  dy/dx for the below parametric functions.

x = a cos θ and 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

Example 2 :

Find dy/dx by using the parametric functions

x = a sec3 θ  and  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

Question 1 :

Differentiate the following

x = a cos3t, y = a sin3t

Solution :

x = a cos3t

dx/dt  =  a(3cos2t)(-sint)

dx/dt  =  -3a cos2t sint----(1)

y = a sin3t

dy/dt  =  a(3sin2t)(cost)

=  3a sin2t cost ----(2)

Divide (2) by (1)

dy/dx  =  (3asin2t cost) / (-3acos2t sint)

=  -sint/cost

dy/dx  =  -tant Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here

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