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) $$1$$

Question

Vedantu Content Team · Accepted Answer

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 $$t=1,\dfrac{1}{3}$$.

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) \[1\]

NCERT Book Solutions

Reference Book Solutions

Question Papers

Exams

Quick Links

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) \[1\]

Book your FREE Online counselling session

Your FREE Online counselling session has been booked!

Please try again later