Question

Find \(dy/dx\) and \({d^2}y/d{x^2}\).

26. \(x = 1 + {t^2},\;\;\;y = t - {t^3}\)

Short answer

The required solution is:

\(\frac{{dy}}{{dx}} = \frac{{1 - 3{t^2}}}{{2t}}\)

\(\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - 3{t^2} - 1}}{{4{t^3}}}\)

Step-by-step solution

Definition of a parametric equation

A parametric equation in mathematics specifies a set of numbers as functions of one or more independent variables known as parameters.

Calculate \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\)

Let us solve the given problem.

Given equations are

\(x = 1 + {t^2},\;\;\;y = t - {t^3}\)

Formula for \(\frac{{dy}}{{dx}}\) is: \(\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\).

Formula for \(\frac{{{d^2}y}}{{d{x^2}}}\) is:

\(\begin{aligned}{c}\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right)\\ = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}}\end{aligned}\)

Now, find the derivatives of \(x = 1 + {t^2},\;\;\;y = t - {t^3}\) with respect to t.

\(\begin{aligned}{l}\frac{{dx}}{{dt}} = 2t{\rm{ }}......\left( 1 \right)\\\frac{{dy}}{{dt}} = 1 - 3{t^2}{\rm{ }}......\left( 2 \right)\end{aligned}\)

Now, divide equation (2) by (1)

\(\begin{aligned}{l}\frac{{\frac{{dy}}{{dy}}}}{{\frac{{dx}}{{dt}}}} = \frac{{1 - 3{t^2}}}{{2t}}\\\frac{{dy}}{{dx}} = \frac{{1 - 3{t^2}}}{{2t}}\end{aligned}\)

Therefore, \(\frac{{dy}}{{dx}} = \frac{{1 - 3{t^2}}}{{2t}}\).

Calculating

\(\frac{{{{\rm{d}}^{\rm{2}}}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^{\rm{2}}}}}\)

Consider the given problem.

\(\begin{aligned}{c}\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right)\\ = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}}\\ = \frac{{\frac{d}{{dt}}\left( {\frac{{1 - 3{t^2}}}{{2t}}} \right)}}{{2t}}\\ = \frac{{\frac{{ - 3{t^2} - 1}}{{2{t^2}}}}}{{2t}}\\ = \frac{{ - 3{t^2} - 1}}{{4{t^3}}}\end{aligned}\)

Therefore, calculated value is \(\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - 3{t^2} - 1}}{{4{t^3}}}\).