Question

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

• A. $\frac{t\left(2-t^3\right)}{1-2t^3}$
• B. $\frac{t\left(2+t^3\right)}{1-2t^3}$
• C. $\frac{t\left(2-t^3\right)}{1+2t^3}$
• D. $\frac{t\left(2+t^3\right)}{1+2t^3}$

Answer 1

The correct option is A

$\frac{t\left(2-t^3\right)}{1-2t^3}$

Explanation for the correct option.

Step 1. Find the value of $\frac{dy}{dt}$.

Differentiate $y=\frac{3a{t}^2}{1+t^3}$ with respect to $t$.

$\begin{array}{rcl}\frac{dy}{dt}& =& \frac{d}{dt}\left(\frac{3a{t}^2}{1+t^3}\right)\\ & =& \frac{3a\times 2t(1+t^3)-3a{t}^2\left(3t^2\right)}{{\left(1+t^3\right)}^2}\\ & =& \frac{6at+6a{t}^4-9a{t}^4}{{(1+t^3)}^2}\\ & =& \frac{6at-3a{t}^4}{{(1+t^3)}^2}\\ & =& \frac{3at\left(2-t^3\right)}{{(1+t^3)}^2}\end{array}$

So, $\frac{dy}{dt}=\begin{array}{rcl}& & \frac{3at\left(2-t^3\right)}{{(1+t^3)}^2}\end{array}...\left(1\right)$.

Step 2. Find the value of $\frac{dt}{dx}$.

Differentiate $x=\frac{3at}{1+t^3}$ with respect to $t$.

$\begin{array}{rcl}\frac{dx}{dt}& =& \frac{d}{dt}\left(\frac{3at}{1+t^3}\right)\\ & =& \frac{3a\times 1(1+t^3)-3at\left(3t^2\right)}{{\left(1+t^3\right)}^2}\\ & =& \frac{3a+3a{t}^3-9a{t}^3}{{(1+t^3)}^2}\\ & =& \frac{3a-6a{t}^3}{{(1+t^3)}^2}\\ & =& \frac{3a\left(1-2t^3\right)}{{(1+t^3)}^2}\end{array}$

So, $\frac{dx}{dt}=\begin{array}{rcl}& & \frac{3a\left(1-2t^3\right)}{{(1+t^3)}^2}\end{array}$ and thus:

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

Step 3. Find the value of $\frac{dy}{dx}$.

The term $\frac{dy}{dx}$ is given as:

$\frac{dy}{dx}=\frac{dy}{dt}·\frac{dt}{dx}....\left(3\right)$

Now, using equation $1$ and $2$, substitute $\frac{dy}{dt}=\begin{array}{rcl}& & \frac{3at\left(2-t^3\right)}{{(1+t^3)}^2}\end{array}$ and $\frac{dt}{dx}=\begin{array}{rcl}& & \frac{{(1+t^3)}^2}{3a\left(1-2t^3\right)}\end{array}$ in equation $3$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{3at\left(2-t^3\right)}{{(1+t^3)}^2}·\frac{{(1+t^3)}^2}{3a\left(1-2t^3\right)}\\ & =& \frac{t(2-t^3)}{(1-2t^3)}\end{array}$

So, $\frac{dy}{dx}=\begin{array}{rcl}& & \frac{t(2-t^3)}{(1-2t^3)}\end{array}$.

Hence, the correct option is A.