Question

Find an equation of the normal line to the curve \(y = \sqrt x \) that is parallel to the line \(2x + y = 1\).

Short answer

The equation of the normal line to the curve is \(y = - 2x + 3\).

Step-by-step solution

 Normal line to a curve

The normal line to the curve \(C\) at a point \(P\) is the line through \(P\) which is perpendicular to the tangent line at \(P\).

Determine the equation of the normal line to the curve

The slope of \(y = \sqrt x \) as shown below:

\(\begin{align}y' &= \frac{d}{{dx}}\left( {\sqrt x } \right)\\ &= \frac{d}{{dx}}\left( {{x^{\frac{1}{2}}}} \right)\\ &= \frac{1}{2}{x^{ - \frac{1}{2}}}\\ &= \frac{1}{{2\sqrt x }}\end{align}\)

Rewrite the equation as shown below:

\(\begin{align}2x + y &= 1\\y &= - 2x + 1\end{align}\)

Evaluate the slope of the line as shown below

\(\begin{align}y' &= \frac{d}{{dx}}\left( { - 2x + 1} \right)\\ &= - 2\frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( 1 \right)\\ &= - 2 + 0\\ &= - 2\end{align}\)

Therefore, the desired normal line should have a slope \( - 2\). Therefore, the slope of the tangent lines to the curve must be \(\frac{1}{2}\).

When \(\frac{1}{{2\sqrt x }} = \frac{1}{2}\) then,

\(\begin{align}\frac{1}{{2\sqrt x }} &= \frac{1}{2}\\\sqrt x &= 1\\x &= 1\end{align}\)

Substitute \(x = 1\) in the equation \(y = \sqrt x \) as shown below:

\(\begin{align}y &= \sqrt 1 \\ &= 1\end{align}\)

The point on the curve is \(\left( {1,1} \right)\).

The equation of the normal line at \(\left( {1,1} \right)\) as shown below:

\(\begin{align}y - 1 &= - 2\left( {x - 1} \right)\\y &= - 2x + 2 + 1\\y &= - 2x + 3\end{align}\)

Thus, the equation of the normal line to the curve is \(y = - 2x + 3\).