Question

find an equation of the normal line to the curve of y = √x that is parallel to the line 6x + y = 1.

Explanation:

Step1: Rewrite the given line in slope - intercept form

The line $6x + y=1$ can be rewritten as $y=-6x + 1$. Its slope $m=-6$.

Step2: Find the derivative of the curve $y = \sqrt{x}=x^{\frac{1}{2}}$

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $y^\prime=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}$.

Step3: Find the slope of the normal line

The slope of the normal line $m_n$ to the curve at a point is related to the slope of the tangent line $m_t$ by $m_n=-\frac{1}{m_t}$. Since the normal line is parallel to $y=-6x + 1$ (so $m_n=-6$), then the slope of the tangent line $m_t=\frac{1}{6}$.

Step4: Solve for $x$ on the curve

Set $\frac{1}{2\sqrt{x}}=\frac{1}{6}$. Cross - multiply to get $2\sqrt{x}=6$, then $\sqrt{x}=3$, and $x = 9$.

Step5: Find the $y$ - value on the curve

When $x = 9$, $y=\sqrt{9}=3$.

Step6: Write the equation of the normal line

Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(9,3)$ and $m=-6$, we have $y - 3=-6(x - 9)$. Expand to get $y-3=-6x + 54$, and then $y=-6x+57$.

Answer:

$y=-6x + 57$