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.
Question 1 :
Differentiate the following
x = a cos^{3}t, y = a sin^{3}t
Solution :
x = a cos^{3}t dx/dt = a(3cos^{2}t)(-sint) = -3a cos^{2}t sint----(1) |
y = a sin^{3}t dy/dt = a(3sin^{2}t)(cost) = 3a sin^{2}t cost ----(2) |
Divide (2) by (1)
dy/dx = (3asin^{2}t cost) / (-3acos^{2}t 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-t^{2})/(1+t^{2}) ; y = 2t/(1+t^{2})
Solution :
x = (1-t^{2})/(1+t^{2})
Differentiate with respect to "t"
dx/dt = [(1+t^{2})(-2t) - (1-t^{2})(2t)]/(1+t^{2})^{2}
= [-2t - 2t^{3} - (2t - 2t^{3})]/(1+t^{2})^{2}
= [-2t - 2t^{3} - 2t + 2t^{3})]/(1+t^{2})^{2}
dx/dt = -4t/(1+t^{2})^{2 } ------(1)
y = 2t/(1+t^{2})
Differentiate with respect to "t"
dy/dt = [(1+t^{2})(2) - (2t)(2t)]/(1+t^{2})^{2}
= [2+2t^{2} - (4t^{2})]/(1+t^{2})^{2}
dy/dt = (2 - 2t^{2})/(1+t^{2})^{2} ------(2)
(2)/(1)
(dy/dt)/(dx/dt) = [2(1-t^{2})/(1+t^{2})^{2}] / [-4t/(1+t^{2})^{2}]
= 2(1-t^{2})/(-4t)
= (t^{2 }- 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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