# 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
${y_2} = – \frac{{{a^2}}}{{{y^3}}}$