Question

If $x=\frac{2at}{1+t^3}$ and $y=\frac{2a{t}^2}{{\left(1+t^3\right)}^2}$, then $\frac{dy}{dx}$ is equal to:

• A. $ax$
• B. $a^2{x}^2$
• C. $\frac{x}{a}$
• D. $\frac{x}{2a}$

Answer 1

The correct option is C

$\frac{x}{a}$

Explanation for the correct option:

Step -1: Differentiate $x$ w.r.t $t$:

$x=\frac{2at}{1+t^3}$, so

$\begin{array}{rcl}\frac{dx}{dt}& =& \frac{d}{dt}\left(\frac{2at}{1+t^3}\right)\\ & =& 2a\frac{d}{dt}\left(\frac{t}{1+t^3}\right)\\ & =& 2a\left[\frac{\left(1+t^3\right){\displaystyle \frac{d}{dt}}\left(t\right)-t{\displaystyle \frac{d}{dt}}\left(1+t^3\right)}{{\left(1+t^3\right)}^2}\right]\\ & =& 2a\left(\frac{1-2t^3}{{\left(1+t^3\right)}^2}\right)......\left(1\right)\end{array}$

Step- 2: Differentiate $y$ w.r.t $t$.:

$y=\frac{2a{t}^2}{{\left(1+t^3\right)}^2}$, so

$\begin{array}{rcl}\frac{dy}{dt}& =& \frac{d}{dt}\left(\frac{2a{t}^2}{{\left(1+t^3\right)}^2}\right)\\ & =& 2a\frac{d}{dt}\left(\frac{t^2}{{\left(1+t^3\right)}^2}\right)\\ & =& 2a\left[\frac{{\left(1+t^3\right)}^2{\displaystyle \frac{d}{dt}}\left(t^2\right)-t^2{\displaystyle \frac{d}{dt}}{\left(1+t^3\right)}^2}{{\left(1+t^3\right)}^4}\right]\\ & =& 2a\left[\frac{{\left(1+t^3\right)}^{2}2t-t^2·2{\displaystyle \left(1+t^3\right)\frac{d}{dt}}\left(1+t^3\right)}{{\left(1+t^3\right)}^4}\right]\\ & =& 2a\left[\frac{2t{\left(1+t^3\right)}^2-2t^2\left(1+t^3\right)\left(3t^2\right)}{{\left(1+t^3\right)}^4}\right]\\ & =& 2a\times 2t\left(1+t^3\right)\left[\frac{1+t^3-3t^3}{{\left(1+t^3\right)}^4}\right]\\ & =& 4at\left[\frac{1-2t^3}{{\left(1+t^3\right)}^3}\right]......\left(2\right)\end{array}$

Step- 3: Find $\frac{dy}{dx}$:

Dividing $\left(2\right)\mathrm{by}\left(1\right)$, we get

$\begin{array}{rcl}\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}& =& \frac{4at\left[\frac{1-2t^3}{{\left(1+t^3\right)}^3}\right]}{2a\left(\frac{1-2t^3}{{\left(1+t^3\right)}^2}\right)}\\ & \Rightarrow & \frac{dy}{dx}=\frac{2t}{\left(1+t^3\right)}\\ & \Rightarrow & \frac{dy}{dx}=\frac{x}{a}\end{array}$

Hence, option C is correct.