Question

For curve $y=\left(\frac{x^2}{2}\right)-2x+5$, the rate of change of $y$ is twice the rate of change of $x$, then the value of $x$ is

Answer 1

Solve for the required value

Given, $y=\left(\frac{x^2}{2}\right)-2x+5$ and $\frac{dy}{dt}=2\frac{dx}{dt}$

Applying differentiation on both sides of the equation $y=\left(\frac{x^2}{2}\right)-2x+5$, we get,

$\begin{array}{rcl}\frac{d}{dt}\left(y\right)& =& \frac{d}{dt}\left(\left(\frac{x^2}{2}\right)-2x+5\right)\\ & \Rightarrow & \frac{dy}{dt}=\frac{1}{2}\frac{d}{dt}\left(x^2\right)-2\frac{dx}{dt}+0\left[\because \frac{d}{dt}\left(5\right)=0\right]\\ & =& \frac{1}{2}\times \left(2x\right)\frac{dx}{dt}-2\frac{dx}{dt}\left[\because \frac{d}{dt}\left(x^n\right)=n.x^{\left(n-1\right)}\right]\\ & \Rightarrow & \frac{dy}{dt}=\left(x-2\right)\frac{dx}{dt}\\ & \Rightarrow & 2\frac{dx}{dt}=\left(x-2\right)\frac{dx}{dt}\left[\because \frac{dy}{dt}=2\frac{dx}{dt}\right]\\ & \Rightarrow & 2=x-2\\ & \Rightarrow & x=4\end{array}$

Hence, the value of $x$ is $4$