# 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) \[1\]

Last updated date: 23rd Mar 2023

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Answer

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

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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