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

& \dfrac{2t}{1+3{{t}^{2}}}=\dfrac{1}{2} \\\ & \Rightarrow 4t=1+3{{t}^{2}} \\\ \end{aligned}$$ $\Rightarrow$ $$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}$$.