Answer
$y=-2x+3$
Work Step by Step
Given a curve $y=\sqrt{x}$ and a line $2x+y=1$.
Find the slope of the line:
$2x+y=1$
$y=-2x+1$
$m_{line}=-2$
Find the values of $x$ such that the tangent line to the curve is perpendicular to the line, that is $\frac{dy}{dx}=-\frac{1}{m_{line}}$:
$\frac{d}{dx}(\sqrt{x})=-\frac{1}{-2}$
$\frac{d}{dx}(x^{1/2})=\frac{1}{2}$
$\frac{1}{2}x^{-1/2}=\frac{1}{2}$
$\frac{1}{2\sqrt{x}}=\frac{1}{2}$
$2\sqrt{x}=2$
$x=1$
Find the value of $y$ when $x=1$:
$y=\sqrt{1}=1$
Since the tangent line to the curve at $(1,1)$ is perpendicular to the line $2x+y=1$, it must be that the normal line to the curve at that point is parallel to the line.
Now, find the equation of the normal line:
$y-1=-m_{line}(x-1)$
$y-1=-2(x-1)$
$y-1=-2x+2$
$y=-2x+3$
Thus, the normal line has an equation $y=-2x+3$.
