DIFFERENTIATION IN PARAMETRIC FORM

About "Differentiation in Parametric Form"

Differentiation in Parametric Form :

Here we are going to see some example problems to understand how to differentiate equations 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.

Differentiation in Parametric Form - Examples

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

After having gone through the stuff given above, we hope that the students would have understood, "Differentiation in Parametric Form"

Apart from the stuff given in "Differentiation in Parametric Form", if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.