Chapter 2.6, Problem 10E

(a)

To determine

To find: The slope of the tangent line to the curve at the given point.

Answer to Problem 10E

The slope of the tangent line to the curve $y=\frac{1}{\sqrt{x}}$ at the point where $x=a$ is $m=-\frac{1}{2a^{\frac{3}{2}}}\text{ or }m=-\frac{1}{2}{a}^{-\frac{3}{2}}$, $\left[a>0\right]$.

Explanation of Solution

Given:

The equation of the curve is $y=\frac{1}{\sqrt{x}}$.

The curve passing through the points (1, 1) and $\left(4,\frac{1}{2}\right)$.

Formula used:

The slope of the tangent curve $y=f\left(x\right)$ at the point $P\left(a,f\left(a\right)\right)$ is,

$m=\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ (1)

Difference of square formula: $\left(a^2-b^2\right)=\left(a+b\right)\left(a-b\right)$

Calculation:

Obtain the slope of the tangent to the curve at the point $x=a$.

Since $f\left(x\right)=\frac{1}{\sqrt{x}}$, substitute $f\left(a\right)=\frac{1}{\sqrt{a}}$ in equation (1),

$\begin{array}{c}m=\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}\\ =\underset{x\to a}{\lim }\frac{\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{a}}}{x-a}\\ =\underset{x\to a}{\lim }\frac{\frac{\sqrt{a}-\sqrt{x}}{\sqrt{ax}}}{x-a}\\ =\underset{x\to a}{\lim }\frac{\sqrt{a}-\sqrt{x}}{\left(x-a\right)\sqrt{ax}}\end{array}$

Multiply both the numerator and the denominator by the conjugate of the numerator.

$m=\underset{x\to a}{\lim }\frac{\sqrt{a}-\sqrt{x}}{\left(x-a\right)\sqrt{ax}}\times \frac{\sqrt{a}+\sqrt{x}}{\sqrt{a}+\sqrt{x}}$

Apply the difference of squares formula,

$\begin{array}{c}m=\underset{x\to a}{\lim }\frac{{\left(\sqrt{a}\right)}^2-{\left(\sqrt{x}\right)}^2}{\left(x-a\right)\left(\sqrt{a}+\sqrt{x}\right)\sqrt{ax}}\\ =\underset{x\to a}{\lim }\frac{a-x}{\left(x-a\right)\left(\sqrt{a}+\sqrt{x}\right)\sqrt{ax}}\\ =\underset{x\to a}{\lim }\frac{-\left(x-a\right)}{\left(x-a\right)\left(\sqrt{a}+\sqrt{x}\right)\sqrt{ax}}\end{array}$

Since the limit x approaches to a but not equal to a, cancel the common term $x-a\left(\ne 0\right)$ from both the numerator and the denominator,

$\begin{array}{c}m=\underset{x\to a}{\lim }\frac{-1}{\left(\sqrt{a}+\sqrt{x}\right)\sqrt{ax}}\\ =\frac{-1}{\left(\sqrt{a}+\sqrt{a}\right)\sqrt{a\left(a\right)}}\\ =\frac{-1}{\left(2\sqrt{a}\right)\sqrt{a^2}}\end{array}$

Perform the mathematical operations and compute the value of the function as shown below.

$\begin{array}{c}m=\frac{-1}{2\sqrt{a}\left(a\right)}\\ =\frac{-1}{2{\left(a\right)}^{\frac{1}{2}}\left(a\right)}\\ =-\frac{1}{2{\left(a\right)}^{\frac{3}{2}}}\\ =-\frac{1}{2}{a}^{-\frac{3}{2}}\end{array}$

Thus, the slope of the tangent line to the curve at the point $x=a$ is $m=-\frac{1}{2a^{\frac{3}{2}}}\text{ or }m=-\frac{1}{2}{a}^{-\frac{3}{2}}$, $\left[a>0\right]$.

(b)

To determine

To find: The equation of the tangent lines to the curve at the given points.

Answer to Problem 10E

The equation of the tangent lines to the curve $y=\frac{1}{\sqrt{x}}$ at the points (1, 1) and $\left(4,\frac{1}{2}\right)$ are $y=-\frac{1}{2}x+\frac{3}{2}$ and $y=-\frac{1}{16}x+\frac{3}{4}$ respectively.

Explanation of Solution

Formula used:

The equation of the tangent line to the curve $y=f\left(x\right)$ at the point is,

$y-f\left(a\right)=f^{\prime }\left(a\right)\left(x-a\right)$ (2)

Calculation:

Obtain the equation of the tangent line at the point (1, 1).

Since the tangent line to the curve $y=f\left(x\right)$ at $\left(a,f\left(a\right)\right)$ is the line through the point $\left(a,f\left(a\right)\right)$ whose slope is equal to the derivative of a, $f^{\prime }\left(a\right)=-\frac{1}{2a^{\frac{3}{2}}}$.

At the point (1, 1), consider $a=1$ and $f\left(a\right)=1$,

$\begin{array}{c}{f}^{\prime }\left(1\right)=-\frac{1}{2{\left(1\right)}^{\frac{3}{2}}}\\ =-\frac{1}{2\left(1\right)}\\ =-\frac{1}{2}\end{array}$

Substitute $a=1$, $f\left(a\right)=1$ and $f^{\prime }\left(a\right)=-\frac{1}{2}$ in equation (2),

$\begin{array}{c}\left(y-f\left(a\right)\right)=f^{\prime }\left(a\right)\left(x-a\right)\\ y-1=-\frac{1}{2}\left(x-1\right)\\ y-1=-\frac{1}{2}x+\frac{1}{2}\end{array}$

Isolate y as shown below:

$\begin{array}{c}y=-\frac{1}{2}x+\frac{1}{2}+1\\ =-\frac{1}{2}x+\frac{3}{2}\end{array}$

Thus, the equation of the tangent line is $y=-\frac{1}{2}x+\frac{3}{2}$.

Obtain the equation of the tangent line at the point $\left(4,\frac{1}{2}\right)$.

Since the tangent line to the curve $y=f\left(x\right)$ at $\left(a,f\left(a\right)\right)$ is the line through the point $\left(a,f\left(a\right)\right)$ whose slope is equal to the derivative of a, $f^{\prime }\left(a\right)=-\frac{1}{2a^{\frac{3}{2}}}$.

At the point $\left(4,\frac{1}{2}\right)$, take $a=4$ and $f\left(a\right)=\frac{1}{2}$

$\begin{array}{c}{f}^{\prime }\left(4\right)=-\frac{1}{2{\left(4\right)}^{\frac{3}{2}}}\\ =-\frac{1}{2\left(4\right)\sqrt{4}}\\ =-\frac{1}{2\left(4\right)\left(2\right)}\\ =-\frac{1}{16}\end{array}$

Substitute $a=2$, $f\left(a\right)=3$ and $f^{\prime }\left(a\right)=-\frac{1}{16}$ in equation (2).

$\begin{array}{c}\left(y-f\left(a\right)\right)=f^{\prime }\left(a\right)\left(x-a\right)\\ y-\frac{1}{2}=-\frac{1}{16}\left(x-4\right)\\ y-\frac{1}{2}=-\frac{1}{16}x+\frac{1}{4}\end{array}$

Isolate y as shown below.

$\begin{array}{c}y=-\frac{1}{16}x+\frac{1}{4}+\frac{1}{2}\\ =-\frac{1}{16}x+\frac{1+2}{4}\\ =-\frac{1}{16}x+\frac{3}{4}\end{array}$

Thus, the equation of the tangent line is $y=-\frac{1}{16}x+\frac{3}{4}$.

(c)

To determine

To sketch: The graph of the curve and the tangent lines.

Explanation of Solution

Given:

The equation of the curve is $y=\frac{1}{\sqrt{x}}$.

The equation of the tangent lines are $y=-\frac{1}{2}x+\frac{3}{2}$ and $y=-\frac{1}{16}x+\frac{3}{4}$.

Graph:

Use the online graphing calculator to draw the graph of the functions as shown below in Figure 1.

From Figure 1, it is noticed that the two lines $y=-\frac{1}{2}x+\frac{3}{2}$ and $y=-\frac{1}{16}x+\frac{3}{4}$ touches the curve $y=\frac{1}{\sqrt{x}}$ at a unique point. That is, the two lines are tangent to the curve $y=\frac{1}{\sqrt{x}}$.