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 $${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{1}{2}}$$?
(A) $${\displaystyle \frac{1}{3}}$$
(B) $$2$$
(C) $$3$$
(D) $$1$$

Answer 1

Hint: If $$x=f(t)$$ and $$y=g(t)$$ then $${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{dy}{dt}}÷{\displaystyle \frac{dx}{dt}}$$.

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, $${\displaystyle \frac{dy}{dx}}$$ cannot be directly calculated.
So, we can write $${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{dy}{dt}}\cdot {\displaystyle \frac{dt}{dx}}.........$$equation$$(1)$$
Now, to find $${\displaystyle \frac{dy}{dx}}$$, we need to find the values of $${\displaystyle \frac{dy}{dt}}$$ and $${\displaystyle \frac{dt}{dx}}$$.
Now, we have $$y=t^2$$
 We will differentiate $$y$$ with respect to $$t$$.
On differentiating $$y$$ with respect to $$t$$, we get ,
$\Rightarrow$ $${\displaystyle \frac{dy}{dt}}={\displaystyle \frac{d}{dt}}(t^2)=2t$$

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

Now, we know inverse function theorem of differentiation says that if $$x=f(t)$$ and $${\displaystyle \frac{dx}{dt}}=x^{\prime }$$ then, $${\displaystyle \frac{dt}{dx}}=t^{\prime }={\displaystyle \frac{1}{x^{\prime }}}$$ .
So, we can write $${\displaystyle \frac{dt}{dx}}={\displaystyle \frac{1}{{\displaystyle \frac{dx}{dt}}}}={\displaystyle \frac{1}{1+3t^2}}$$
Now, to find the value of $${\displaystyle \frac{dy}{dx}}$$ , we will substitute the values of $${\displaystyle \frac{dy}{dt}}$$ and $${\displaystyle \frac{dt}{dx}}$$in equation$$(1)$$.
On substituting the values of $${\displaystyle \frac{dy}{dt}}$$ and $${\displaystyle \frac{dt}{dx}}$$in equation$$(1)$$, we get ,$$$$
$\Rightarrow$ $${\displaystyle \frac{dy}{dx}}=2t\times {\displaystyle \frac{1}{1+3t^2}}$$
$$={\displaystyle \frac{2t}{1+3t^2}}$$
Now, it is given that the value of $${\displaystyle \frac{dy}{dx}}$$ is equal to $${\displaystyle \frac{1}{2}}$$.
So, we can write
$$\begin{array}{rl}& {\displaystyle \frac{2t}{1+3t^2}}={\displaystyle \frac{1}{2}}\\ & \Rightarrow 4t=1+3t^2\end{array}$$
$\Rightarrow$ $$3t^2-4t+1=0$$
Clearly, it is a quadratic equation in $$t$$.
Now, we will solve this quadratic equation by factorisation method.
$$3t^2-4t+1=0$$
$$\Rightarrow 3t^2-3t-t+1=0$$
$$\Rightarrow 3t(t-1)-1(t-1)=0$$
$$\Rightarrow (3t-1)(t-1)=0$$
$$\Rightarrow t={\displaystyle \frac{1}{3}}$$or $$t=1$$
Hence , the values of $$t$$ for $${\displaystyle \frac{dy}{dx}}$$ to be equal to $${\displaystyle \frac{1}{2}}$$ are $${\displaystyle \frac{1}{3}}$$ and $$1$$.
Answers are options (D), (A).

Note: $$3t^2-4t+1=0$$ 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 $$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$$ .
So , $$t={\displaystyle \frac{-(-4)\pm \sqrt{{(-4)}^2-4(3)(1)}}{2(3)}}$$
$$\Rightarrow t={\displaystyle \frac{4\pm \sqrt{16-12}}{6}}$$
$$\Rightarrow t={\displaystyle \frac{4\pm 2}{6}}$$
$$\Rightarrow t={\displaystyle \frac{6}{6}},{\displaystyle \frac{2}{6}}$$
$$\Rightarrow t=1,{\displaystyle \frac{1}{3}}$$
Hence, the values of $$t$$ satisfying the equation $$3t^2-4t+1=0$$ are $$t=1,{\displaystyle \frac{1}{3}}$$.