DIFFERENTIATION IN PARAMETRIC FORM

Consider the equations x = f(t), y = g(t).

These equations give a functional relationship between the variables x and y. Given the value of t in some domain [a, b], we can find x and y.

If two variables x and y are defined separately as a function of an intermediating (auxiliary) variable t, then the specification of a functional relationship between x and y is described as parametric and the auxiliary variable is known as parameter.

The operation of finding the direct connection between x and y without the presence of the auxiliary variable t is called elimination of the parameter. The question as to why should we deal with parametric equations is that two or more variables are reduced to a single variable, t.

Question 1 :

Differentiate the following

x = a cos³t, y = a sin³t

Solution :

Divide (2) by (1)

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

  =  -sint/cost

 dy/dx  =  -tant

Question 2 :

Differentiate the following

x = a (cos t + t sin t) ; y = a (sin t - t cos t)

Solution :

x = a (cos t + t sin t) ; y = a (sin t - t cos t)

Differentiate with respect to "t"

x = a (cos t + t sin t)

dx/dt  =  a(-sin t + t(cos t) + sin t (1))

  =  a(-sin t + t(cos t) + sin t)

dx/dt  =  a t cos t  -----(1)

y = a (sin t - t cos t)

dy/dt  =  a(cos t - [t(-sin t) + cos t(1)])

  =  a(cos t + t sin t - cos t)

dy/dt  =  a t sin t  -----(2)

(2) / (1)

(dy/dt) / (dx/dt)  =  (a t sin t) / (a t cos t )

  =  tan t

Question 3 :

Differentiate the following

x = (1-t²)/(1+t²) ; y = 2t/(1+t²)

Solution :

x = (1-t²)/(1+t²)

Differentiate with respect to "t"

dx/dt  =  [(1+t²)(-2t) - (1-t²)(2t)]/(1+t²)²

 =  [-2t - 2t³ - (2t - 2t³)]/(1+t²)²

 =  [-2t - 2t³ - 2t + 2t³)]/(1+t²)²

dx/dt  =  -4t/(1+t²)²    ------(1)

y = 2t/(1+t²)

Differentiate with respect to "t"

dy/dt  =  [(1+t²)(2) - (2t)(2t)]/(1+t²)²

  =  [2+2t² - (4t²)]/(1+t²)²

dy/dt  =  (2 - 2t²)/(1+t²)²  ------(2)

(2)/(1)

(dy/dt)/(dx/dt)  =  [2(1-t²)/(1+t²)²] / [-4t/(1+t²)²]

=  2(1-t²)/(-4t)

=  (t² - 1)/2t

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.