Question

If $x=4t$ and $y=2{{t}^{2}}$, then $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$at $x=\dfrac{1}{2}$ is
$\begin{aligned} & \left( \text{a} \right)\text{ }\dfrac{1}{2} \\\ & \left( b \right)\text{ }\dfrac{1}{4} \\\ & \left( c \right)\text{ 2} \\\ & \left( d \right)\text{ 4} \\\ \end{aligned}$

Answer 1

- Hint:Consider ‘x’ and ‘y’ separately, then differentiate them with respect to ‘t’ separately. Then apply the parametric form of derivation, i.e., $\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$. Then substitute the given values and find out the value of ‘t’. Later differentiate again to get a second order derivative and solve further.

Complete step-by-step solution -
As per the given information, $x=4t$, $y=2{{t}^{2}}$.
For this first let us find $\dfrac{dy}{dt}\And \dfrac{dx}{dt}.$
First we will derive $'y'$ with respect to $'t'$, we get
$\dfrac{dy}{dt}=\dfrac{d}{dt}(2{{t}^{2}})$
Differentiating on both the sides, we get
$\dfrac{dy}{dt}=2.2t=4t............(i)$
Now we will derive $'x'$ with respect to $'t'$, we get
$\dfrac{dx}{dt}=\dfrac{d}{dt}(4t)$
By differentiating on both sides, we get
$\dfrac{dx}{dt}=4(1)=4..............(ii)$
Now dividing equation (i) by equation (ii), we get
$\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{4t}{4}$
Cancelling the like terms, we get
$\Rightarrow \dfrac{dy}{dx}=t.......(iii)$
Now given $x=4t$, so
$\Rightarrow t=\dfrac{x}{4}$
Substitute this in equation (iii), we get
$\dfrac{dy}{dx}=\dfrac{x}{4}$
Now, we will again differentiate both sides with respect to ‘x’, we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{x}{4} \right)$
Taking out the constant term, we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{4}.\dfrac{d}{dx}(x)=\dfrac{1}{4}(1)$
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{4}$
And as the second derivative is a constant, which is independent of x-value, so it will be the same value at any point, i.e.,
${{\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{x=\dfrac{1}{2}}}=\dfrac{1}{4}$
Hence the correct answer is option (b).

Note: The given equations are of parametric form. Here we can also find the value of $'t'$from the given value of $'x'$, i.e., $t=\dfrac{x}{4}$.
Now substituting the value of $'t'$in the given value of $'y'$, i.e., $y=2{{\left( \dfrac{x}{4} \right)}^{2}}$. Here we can observe that $'y'$is independent of $'t'$. Now we can differentiate $'y'$directly with respect to $'x'$ and find the required values.