Question

find equations of the tangent line and normal line to the curve at the specified point. $y = \frac{\sqrt{x}}{x + 6}, (4, 0.2)$ tangent $y = $ normal $y = $

Explanation:

Step1: Find the derivative using quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = \sqrt{x}=x^{\frac{1}{2}}$, so $u^\prime=\frac{1}{2}x^{-\frac{1}{2}}$, and $v=x + 6$, so $v^\prime = 1$. Then $y^\prime=\frac{\frac{1}{2}x^{-\frac{1}{2}}(x + 6)-x^{\frac{1}{2}}\times1}{(x + 6)^{2}}=\frac{\frac{x + 6}{2\sqrt{x}}-\sqrt{x}}{(x + 6)^{2}}=\frac{\frac{x + 6-2x}{2\sqrt{x}}}{(x + 6)^{2}}=\frac{6 - x}{2\sqrt{x}(x + 6)^{2}}$.

Step2: Evaluate the derivative at $x = 4$

Substitute $x = 4$ into $y^\prime$: $y^\prime|_{x = 4}=\frac{6-4}{2\sqrt{4}(4 + 6)^{2}}=\frac{2}{2\times2\times100}=\frac{1}{200}$. This is the slope $m_t$ of the tangent line.

Step3: Find the equation of the tangent line

Use the point - slope form of a line $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(4,0.2)$ and $m=\frac{1}{200}$. So $y-0.2=\frac{1}{200}(x - 4)$, which simplifies to $y=\frac{1}{200}x- \frac{4}{200}+0.2=\frac{1}{200}x-0.02 + 0.2=\frac{1}{200}x+0.18$.

Step4: Find the slope of the normal line

The slope $m_n$ of the normal line is the negative reciprocal of the slope of the tangent line. So $m_n=-200$.

Step5: Find the equation of the normal line

Use the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(4,0.2)$ and $m=-200$. Then $y-0.2=-200(x - 4)$, which simplifies to $y=-200x+800 + 0.2=-200x+800.2$.

Answer:

tangent: $y=\frac{1}{200}x + 0.18$
normal: $y=-200x+800.2$