Higher Order Derivatives of Parametric Equations

Here you will learn higher order derivatives of parametric equations with examples.

Let’s begin –

Higher Order Derivatives of Parametric Equations

We know that the differentiation of parametric equations of type x = at and y = 2at is given by formula

\(dy\over dx\) = \(dy/dt\over dx/dt\)

where t is the parameter

Now, by again differentiating the above equation we obtain 2nd order derivative given below:

\(d^2y\over dx^2\) = \(d\over dx\)(\(dy\over dx\))

Example : find \(d^2y\over dx^2\) , if x = \(at^2\) , y = 2at.

Solution : We have, 

x = \(at^2\) , y = 2at.

Differentiating both sides with respect to t,

\(\implies\) \(dx\over dt\) = 2at and \(dy\over dt\) = 2a           …….(i)

\(\therefore\) \(dy\over dx\) = \(dy/dt\over dx/dt\) = \(2a\over 2at\) = \(1\over t\)

Differentiating both sides with respect to x, we get

\(d^2y\over dx^2\) = \(d\over dx\)(\(1\over t\))

\(\implies\) \(d^2y\over dx^2\) = \(-1\over t^2\)\(dt\over dx\)

from (i), [\(dx\over dt\) = 2at \(\therefore\)  \(dt\over dx\) = \(1\over 2at\)]

\(\implies\) \(d^2y\over dx^2\)  = -\(1\over 2at^3\)

Example : If x = \(acos^3\theta\) , y = \(asin^3\theta\) , find \(d^2y\over dx^2\). Also find its value at \(\theta\) = \(\pi\over 6\).

Solution : We have, 

x = \(acos^3\theta\) and y = \(asin^3\theta\)

Differentiating both sides with respect to \(\theta\), 

\(\therefore\) \(dx\over d\theta\) = \(-3acos^2\theta sin\theta\) and \(dy\over d\theta\) = \(3asin^2\theta cos\theta\)               …….(i)

 So, \(dy\over dx\) = \(dy/d\theta\over dx/d\theta\) = \(3asin^2\theta cos\theta\over {-3acos^2\theta sin\theta}\) = \(-tan\theta\)

Differentiating both sides with respect to x, we obtain

\(d^2y\over dx^2\) = \(d\over dx\)(\(-tan\theta\)) 

= \(sec^2\theta\)\(d\theta\over dx\) 

from (i), [\(dx\over d\theta\) = \(-3acos^2\theta sin\theta\) \(\therefore\)  \(d\theta\over dx\) = \(1\over {-3acos^2\theta sin\theta}\)]

\(d^2y\over dx^2\) = -\(sec^2\theta\)\(\times\)\(1\over {-3acos^2\theta sin\theta}\)

\(\implies\) \(d^2y\over dx^2\) = \(1\over 3a\) \(sec^4\theta\)\(cosec\theta\)

\(\therefore\) At \(\theta\) = \(\pi\over 6\) , \(d^2y\over dx^2\)  = \(1\over 3a\) \(sec^4{\pi\over 4}\)\(cosec{\pi\over 6}\)

= \(1\over 3a\) \(\times\) \((2\over \sqrt{3})^4\) \(\times\) 2 = \(32\over 27a\) 

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