Single Variable Calculus, Chapter 2, 2.1, Section 2.1, Problem 4

Given the curve $y= \sqrt{x-2}$ and the point $J(3, 1)$ that lies on the curve.


a. Use your calculator to find the slope of the secant line $JK$ (correct to six decimal places) for the following values of $x$ if $K$ is the point $(x,\sqrt{x-2})$.


$
\begin{equation}
\begin{aligned}

&\text{ (i) } 2.5 && \text{ (v) } 3.5 \\
&\text{ (ii) } 2.9 && \text{(vi) } 3.5 \\
&\text{ (iii) }2.99 && \text{(vii) } 3.01\\
&\text{ (iv) } 2.999 && \text{(viii) } 3.001
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

\begin{array}{|c|c|c|c|c|c|}
\hline\\
& K_x(x) & K_y(\sqrt{x - 2}) & J_x & J_y & \text{Slope of secant line} JK = \left(\frac{J_y - K_y}{J_x - K_x}\right) \\
\hline\\
(i) & 2.5 & \frac{\sqrt{2}}{2} & 3 & 1 & 0.585786 \\
\hline\\
(ii) & 2.9 & \frac{3 \sqrt{10}}{10} & 3 & 1 & 0.513167 \\
\hline\\
(iii) & 2.99 & \frac{3\sqrt{11}}{10} & 3 & 1 & 0.501256 \\
\hline\\
(iv) & 2.999 & 0.99 & 3 & 1 & 0.500125 \\
\hline\\
(v) & 3.5 & \frac{\sqrt{6}}{2} & 3 & 1 & 0.449489 \\
\hline\\
(vi) & 3.1 & \frac{\sqrt{110}}{10} & 3 & 1 & 0.488088 \\
\hline\\
(vii) & 3.01 & \frac{\sqrt{101}}{10} & 3 & 1 & 0.498756 \\
\hline\\
(viii) & 3.001 & 1.0005 & 3 & 1 & 0.499875\\
\hline
\end{array}

\end{aligned}
\end{equation}
$


For example,
@ x = 2.5,

$
\begin{equation}
\begin{aligned}
J_y \sqrt{x-2} & = \sqrt{2.5-2}\\
& = \frac{\sqrt{2}}{2}
\end{aligned}
\end{equation}
$

@ x = 2.9,

$
\begin{equation}
\begin{aligned}
J_y \sqrt{x-2} & = \sqrt{2.9-2}\\
& = \frac{3\sqrt{10}}{10}
\end{aligned}
\end{equation}
$



b. Guess the value of the slope of the tangent line to the curve at $J(3,1)$ using the results in part (a).

Based from the values we obtain from the table, the slope of the tangent line seems to have a value of $\displaystyle \frac{1}{2}$

c. Find an equation of the tangent line to the curve at $J(3,1)$ using the slope in part (b).

Using point slope form:


$
\begin{equation}
\begin{aligned}

y - y_1 =& m(x - x_1)\\
y - 1 =& \frac{1}{2} (x-3)\\
2(y- 1) =& x - 3\\
2y - 2 =& x -3\\
2y =& x - 3 + 2\\
y =& \frac{x - 1}{2}


\end{aligned}
\end{equation}
$


d. Sketch the curve, two of the secant lines, and the tangent line.