Question

How do you find the second derivative of parametric function?

Answer 1

Hint : In order to determine to determine the second derivative of parametric function or equations , we have to use the chain rule twice and with the help of the result obtained for first derivative , you will see that the second derivative is equal to division of the derivative with respect to $t$ of the first derivative with the derivative of $x$ with respect to t.

Complete step-by-step answer :
To find out the second derivative of a parametric function or equation lets understand what are parametric equations.
Parametric equations are equations in which the dependent variable of derivative i.e. $x\,and\,y$ are dependent on some other independent third variable $(t)$ .
$x = x\left( t \right)$ and $y = y\left( t \right)$
Now the first derivative of dependent variable $y$ with respect to the another dependent variable $x$ comes to be
$\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}} \\\ \dfrac{{dy}}{{dx}} = \dfrac{{y'(t)}}{{x'(t)}} \;$
Here, $\dfrac{{dx}}{{dt}} = x'(t)$ denotes the derivative of parametric equation $x$ with respect to $t$ and similarly $\dfrac{{dy}}{{dt}} = y'(t)$ denotes the derivative of parametric equation $y$ with respect to $t$ .
Now to calculate the second derivative of parametric equations, we have to use the chain rule twice.
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left( {\dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}} \right) \\\ \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dt}}\left( {\dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}} \right)\dfrac{{dt}}{{dx}} \\\ \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{\dfrac{d}{{dt}}\left( {\dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}} \right)}}{{\dfrac{{dx}}{{dt}}}} \;$
Therefore, to find out the second derivative of the parametric function, find out the derivative with respect to $t$ of the first derivative and after that divide it by the derivative of $x$ with respect to t.

Note : 1. Calculus consists of two important concepts one is differentiation and other is integration.
2.What is Differentiation?
It is a method by which we can find the derivative of the function .It is a process through which we can find the instantaneous rate of change in a function based on one of its variables.
Let y = f(x) be a function of x. So the rate of change of $y$ per unit change in $x$ is given by:
$\dfrac{{dy}}{{dx}}$ .
1.Don’t forget to cross-check your answer at least once.
2.Differentiation is basically the inverse of integration.
3. $x$ and $y$ are the dependent variables of the derivative and both $x$ and $y$ are dependent on some independent variable $t$ .