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# A curve parametrically given by $x=t+{{t}^{3}}$and $y={{t}^{2}}$, where $t\in R$. For what value(s) of $t$ is $\dfrac{dy}{dx}=\dfrac{1}{2}$?(A) $\dfrac{1}{3}$ (B) $2$(C) $3$(D) $1$  Answer Verified
Hint: If $x=f(t)$ and $y=g(t)$ then $\dfrac{dy}{dx}=\dfrac{dy}{dt}\div \dfrac{dx}{dt}$.

Complete step-by-step answer:
The given equation of the curve is $x=t+{{t}^{3}}$and $y={{t}^{2}}$.
We can clearly see that $y$ and $x$ are not given in terms of each other but in terms of another parameter $t$ . Hence, $\dfrac{dy}{dx}$ cannot be directly calculated.
So, we can write $\dfrac{dy}{dx}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}.........$equation$(1)$
Now, to find $\dfrac{dy}{dx}$, we need to find the values of $\dfrac{dy}{dt}$ and $\dfrac{dt}{dx}$.
Now, we have $y={{t}^{2}}$
We will differentiate $y$ with respect to $t$.
On differentiating $y$ with respect to $t$, we get ,
$\Rightarrow$ $\dfrac{dy}{dt}=\dfrac{d}{dt}({{t}^{2}})=2t$

Now, we will differentiate $x$ with respect to $t$.
On differentiating $x$ with respect to $t$, we get,
$\Rightarrow$ $\dfrac{dx}{dt}=\dfrac{d}{dt}(t+{{t}^{3}})=1+3{{t}^{2}}$

Now, we know inverse function theorem of differentiation says that if $x=f(t)$ and $\dfrac{dx}{dt}=x'$ then, $\dfrac{dt}{dx}=t'=\dfrac{1}{x'}$ .
So, we can write $\dfrac{dt}{dx}=\dfrac{1}{\dfrac{dx}{dt}}=\dfrac{1}{1+3{{t}^{2}}}$
Now, to find the value of $\dfrac{dy}{dx}$ , we will substitute the values of $\dfrac{dy}{dt}$ and $\dfrac{dt}{dx}$in equation$(1)$.
On substituting the values of $\dfrac{dy}{dt}$ and $\dfrac{dt}{dx}$in equation$(1)$, we get ,
$\Rightarrow$ $\dfrac{dy}{dx}=2t\times \dfrac{1}{1+3{{t}^{2}}}$
$=\dfrac{2t}{1+3{{t}^{2}}}$
Now, it is given that the value of $\dfrac{dy}{dx}$ is equal to $\dfrac{1}{2}$.
So, we can write
\begin{align} & \dfrac{2t}{1+3{{t}^{2}}}=\dfrac{1}{2} \\ & \Rightarrow 4t=1+3{{t}^{2}} \\ \end{align}
$\Rightarrow$ $3{{t}^{2}}-4t+1=0$
Clearly, it is a quadratic equation in $t$.
Now, we will solve this quadratic equation by factorisation method.
$3{{t}^{2}}-4t+1=0$
$\Rightarrow 3{{t}^{2}}-3t-t+1=0$
$\Rightarrow 3t(t-1)-1(t-1)=0$
$\Rightarrow (3t-1)(t-1)=0$
$\Rightarrow t=\dfrac{1}{3}$or $t=1$
Hence , the values of $t$ for $\dfrac{dy}{dx}$ to be equal to $\dfrac{1}{2}$ are $\dfrac{1}{3}$ and $1$.
Answers are options (D), (A).

Note: $3{{t}^{2}}-4t+1=0$ can alternatively be solved using the quadratic formula.
We know, for a quadratic equation given by $a{{x}^{2}}+bx+c=0$, the values of $x$ satisfying the equation are known as the roots of the equation and are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
So , $t=\dfrac{-(-4)\pm \sqrt{{{(-4)}^{2}}-4(3)(1)}}{2(3)}$
$\Rightarrow t=\dfrac{4\pm \sqrt{16-12}}{6}$
$\Rightarrow t=\dfrac{4\pm 2}{6}$
$\Rightarrow t=\dfrac{6}{6},\dfrac{2}{6}$
$\Rightarrow t=1,\dfrac{1}{3}$
Hence, the values of $t$ satisfying the equation $3{{t}^{2}}-4t+1=0$ are $t=1,\dfrac{1}{3}$.
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