Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 10

a.) Determine the slope of the tangent to the curve $\displaystyle y = \frac{1}{\sqrt{x}}$ at the point where $x = a$

Using the equation

$\displaystyle m = \lim \limits_{h \to 0} \frac{f(a + h) - f(a)}{h}$

Let $\displaystyle f(x) = \frac{1}{\sqrt{x}}$. So the slope of the tangent to the curve at the point where $x = a$ is


$
\begin{equation}
\begin{aligned}

\displaystyle m =& \lim \limits_{h \to 0} \frac{f(a + h) - f(a)}{h}
&& \\
\\
\displaystyle m =& \lim \limits_{h \to 0} \frac{\frac{1}{\sqrt{a + h}} - \frac{1}{\sqrt{a}}}{h}
&& \text{Substitute value of $a$}\\
\\
\displaystyle m =& \lim \limits_{h \to 0} \frac{\sqrt{a} - \sqrt{a + h}}{(h)(\sqrt{a}) (\sqrt{a + h})}
&& \text{Get the LCD on the numerator and simplify}\\
\\
\displaystyle m =& \lim \limits_{h \to 0} \frac{\sqrt{a} - \sqrt{a + h}}{(h)(\sqrt{a}) (\sqrt{a + h})} \cdot
\frac{\sqrt{a} + \sqrt{a + h}}{\sqrt{a} + \sqrt{a + h}}
&& \text{Multiply both numerator and denominator by $(\sqrt{a} + \sqrt{a + h})$}\\
\\
\displaystyle m =& \frac{a - (a+ h)}{(h)(\sqrt{a})(\sqrt{a + h})(\sqrt{a} + \sqrt{a + h})}
&& \text{Combine like terms}\\
\\
\displaystyle m =& \frac{-\cancel{h}}{\cancel{(h)}(\sqrt{a})(\sqrt{a + h})(\sqrt{a} + \sqrt{a + h})}
&& \text{Cancel out like terms}\\
\\
\displaystyle m =& \frac{-1}{(\sqrt{a}) (\sqrt{a + h})(\sqrt{a} + \sqrt{a + h})} = \frac{-1}{(\sqrt{a}) (\sqrt{a + 0}) (\sqrt{a} + \sqrt{a + 0})}
&& \text{Evaluate the limit}\\
\\
\displaystyle m =& \frac{-1}{2a \sqrt{a}}
&& \text{Slope of the tangent}

\end{aligned}
\end{equation}
$


b.) Determine the equations of the tangent lines at the points $(1, 1)$ and $\displaystyle\left(4, \frac{1}{2}\right)$

Solving for the equation of the tangent line at $(1, 1)$

Using the equation of slope of the tangent in part (a), we have


$
\begin{equation}
\begin{aligned}

a =& 1
&& \text{So the slope is }\\
\\
m =& \frac{-1}{2a \sqrt{a}}
&& \\
\\
m =& \frac{-1}{2(1) \sqrt{1}}
&& \text{Substitute value of $a$}\\
\\
m =& \frac{-1}{2}
&& \text{Slope of the tangent line at $(1, 1)$}\\
\end{aligned}
\end{equation}
$

Using point slope form

$
\begin{equation}
\begin{aligned}

y - y_1 =& m ( x - x_1)
&& \\
\\
y - 1 =& \frac{-1}{2}(x - 1)
&& \text{Substitute value of $x, y$ and $m$}\\
\\
y - 1 =& \frac{- x + 1}{2} + 1
&& \text{Get the LCD}\\
\\
y =& \frac{- x + 1 + 2}{2}
&& \text{Combine like terms}
\\
y =& \frac{-x + 3}{2}
\end{aligned}
\end{equation}
$

Therefore,
The equation of the tangent line at $(1,1)$ is $ y = \displaystyle \frac{-x + 3}{2}$

Solving for the equation of the tangent line at $\displaystyle \left(4, \frac{1}{2}\right)$

Using the equation of slope of the tangent in part (a), we have $a = 4$. So the slope is


$
\begin{equation}
\begin{aligned}

m =& \frac{-1}{2a \sqrt{a}}
&& \\
\\
m =& \frac{-1}{2(4)\sqrt{4}}
&& \text{Substitute the value of $x, y$ and $m$}\\
\\
m =& \frac{-1}{16}
&& \text{Slope of the tangent line at $\left(4, \frac{1}{2}\right)$}

\end{aligned}
\end{equation}
$


Using point slope form



$
\begin{equation}
\begin{aligned}

y - y_1 =& m ( x - x_1)
&& \\
\\
y - \frac{1}{2} =& \frac{-1}{16} (x - 4 )
&& \text{Substitute value of $x, y$ and $m$}\\
\\
y =& \frac{-x + 4}{16} + \frac{1}{2}
&& \text{Get the LCD}\\
\\
y =& \frac{-x + 4 + 8}{16}
&& \text{Combine like terms}
\\
y =& \displaystyle \frac{-x + 12}{16}
\end{aligned}
\end{equation}
$


Therefore,
The equation of the tangent line at $\displaystyle\left(4, \frac{1}{2}\right)$ is $ y = \displaystyle \frac{-x + 12}{16}$


c.) Graph the curve and both tangent lines on a common screen.