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.