Second Derivative of the Parametric Equation

Parametric Function
A function in which $$x$$ and $$y$$ are expressed as a function of a third variable is called a parametric function. For example, the function defined by the equations $$x = a{t^2}$$ and $$y = 2at$$ is a parametric function.

Now we shall give an example to find the second derivative of the parametric function.

Example: If the parametric function $$x = a\cos \theta $$, $$y = a\sin \theta $$, then show that
\[{y_2} = – \frac{{{a^2}}}{{{y^2}}}\]

We have the given parametric function
\[\begin{gathered} x = a\cos \theta \,\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ y = a\sin \theta \,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right) \\ \end{gathered} \]

Differentiating both sides of equation (i) with respect to $$\theta $$, we have
\[\begin{gathered} \frac{{dx}}{{d\theta }} = – a\sin \theta \\ \Rightarrow \frac{{d\theta }}{{dx}} = – \frac{1}{{a\sin \theta }} \\ \end{gathered} \]

Differentiating both sides of equation (ii) with respect to $$\theta $$, we have
\[\frac{{dy}}{{d\theta }} = a\cos \theta \]

Using the chain rule of differentiation , we have
\[\frac{{dy}}{{dx}} = \frac{{dy}}{{d\theta }} \times \frac{{d\theta }}{{dx}}\]

Putting the values of $$\frac{{d\theta }}{{dx}}$$ and $$\frac{{dy}}{{d\theta }}$$ in the above chain rule formula, we have
\[\frac{{dy}}{{dx}} = a\cos \theta \times \left( { – \frac{1}{{a\sin \theta }}} \right) = – \cot \theta \]

Again, differentiating both sides with respect to $$\theta $$, we have
\[\begin{gathered} \frac{{{d^2}y}}{{d{x^2}}} = – \frac{d}{{dx}}\cot \theta \\ \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = – {\csc ^2}\theta \frac{{d\theta }}{{dx}} \\ \end{gathered} \]

Using the values of $$\frac{{d\theta }}{{dx}}$$, we get
\[\begin{gathered} \frac{{{d^2}y}}{{d{x^2}}} = – {\csc ^2}\theta \left( { – \frac{1}{{a\sin \theta }}} \right) \\ \Rightarrow {y_2} = – \frac{1}{{{{\sin }^2}\theta }}\left( { – \frac{1}{{a\sin \theta }}} \right) \\ \Rightarrow {y_2} = – \frac{1}{{a{{\sin }^3}\theta }} = – \frac{{{a^2}}}{{{a^3}{{\sin }^3}\theta }} \\ \Rightarrow {y_2} = – \frac{{{a^2}}}{{{{\left( {a\sin \theta } \right)}^3}}} \\ \end{gathered} \]

Putting the value of $$a\sin \theta $$ in the above result, we have
\[{y_2} = – \frac{{{a^2}}}{{{y^3}}}\]