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 :

x = a cos3t

dx/dt  =  a(3cos2t)(-sint)

  =  -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

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)  ------(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"

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