Question

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 1

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 \[t=1,\dfrac{1}{3}\].