Question:
A particle moves along the curve $y=x^{2}+2 x .$ At what point(s) on the curve are the $x$ and $y$ coordinates of the particle changing at the same rate?
Solution:
Here,
$y=x^{2}+2 x$
$\Rightarrow \frac{d y}{d t}=(2 x+2) \frac{d x}{d t}$
$\Rightarrow 2 x+2=1$ $\left[\because \frac{d y}{d t}=\frac{d x}{d t}\right]$
$\Rightarrow 2 x=-1$
$\Rightarrow x=\frac{-1}{2}$
Substituting $x=\frac{-1}{2}$ in $y=x^{2}+2 x$, we get
$y=\frac{-3}{4}$
Hence, the coordinates of the point are $\left(\frac{-1}{2}, \frac{-3}{4}\right)$.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.