Chapter 4.6, Problem 41E

To determine

To calculate: The shortest possible length of the line segment that is cut off by the first quadrant and is tangent to the curve $y=\frac{3}{x}$ at some point.

Answer to Problem 41E

The shortest distance is $2\sqrt{6}$

Explanation of Solution

Given information:

The line segment that is cut off by the first quadrant and is tangent to the curve $y=\frac{3}{x}$ at some point

Formula used:

Point-Slope form of the equation of a line is

  $y-y_1=m\left(x-x_1\right)$

Where $m$ is the slope and $\left(x_1,y_1\right)$ is the point in the line.

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

• If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
• If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Calculation:

As per the given problem

The line segment that is cut off by the first quadrant and is tangent to the curve $y=\frac{3}{x}$ at some point

Let $x=a$ be where the line segment is tangent to the curve,

The curve

  $y=\frac{3}{x}$

Substitute $x=a$ to get,

  $y=\frac{3}{a}$

For slope, differentiate equation with respect to $x$

  $y^{\prime }=-\frac{3}{x^2}$

Let us denote the slope of tangent as $M\left(x\right)$

  $M\left(x\right)=-\frac{3}{x^2}$

An equation of the line that passing through $\left(a,\frac{3}{a}\right)$ with slope $M\left(x\right)=-\frac{3}{x^2}$ is

Recall that,

Point-Slope form of the equation of a line is

  $y-y_1=m\left(x-x_1\right)$

Where $m$ is the slope and $\left(x_1,y_1\right)$ is the point on the line

  $\begin{array}{l}y-\frac{3}{a}=-\frac{3}{a^2}\left(x-a\right)\\ \Rightarrow y=-\frac{3x}{a^2}+\frac{3}{a}+\frac{3}{a}\\ \Rightarrow y=-\frac{3x}{a^2}+\frac{6}{a}\left(1\right)\end{array}$

Substitute $x=0$ in equation $\left(1\right)$ to get $y-\text{intercept}$

  $y=\frac{6}{a}$

Substitute $y=0$ in equation $\left(1\right)$ to get $x-\text{intercept}$

  $\begin{array}{l}\Rightarrow 0=-\frac{3x}{a^2}+\frac{6}{a}\\ \Rightarrow \frac{3x}{a^2}=\frac{6}{a}\\ \Rightarrow x=2a\end{array}$

The distance of the line segment that has end points in at the intercepts.

  $\begin{array}{l}d=\sqrt{{\left(2a-0\right)}^2+{\left(\frac{6}{a}-0\right)}^2}\\ =\sqrt{4a^2+\frac{36}{a^2}}\end{array}$

Square both on both sides, to get

  $d^2=4a^2+\frac{36}{a^2}$

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

• If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
• If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Differentiate with respect to $a$

  $\begin{array}{l}\frac{d}{da}{d}^2\left(a\right)=\frac{d}{da}\left\{4a^2+\frac{36}{a^2}\right\}\\ =8a-\frac{72}{a^3}\mathrm{......}\left(2\right)\end{array}$

Solve for $\frac{d}{da}{d}^2=0$ , and simplified

  $\begin{array}{l}8a-\frac{72}{a^3}=0\\ \Rightarrow 8a=\frac{72}{a^3}\\ \Rightarrow a^4=9\\ \Rightarrow a^2=3\\ \Rightarrow a=\sqrt{3}\end{array}$

Differentiate equation $\left(2\right)$ with respect to $a$

  $\begin{array}{l}\frac{d^2}{d{a}^2}{d}^2\left(a\right)=\frac{d}{da}\left\{8a-\frac{72}{a^3}\right\}\\ =8+\frac{216}{a^4}\end{array}$

Therefore,

  $\frac{d^2}{d{a}^2}{d}^2\left(a\right)=8+\frac{216}{a^4}\ge 0$ at $a=\sqrt{3}$ , hence there is local minimum at $a=\sqrt{3}$

Now put $a=\sqrt{3}$ in $d=\sqrt{4a^2+\frac{36}{a^2}}$ and solve

  $\begin{array}{l}d=\sqrt{4{\left(\sqrt{3}\right)}^2+\frac{36}{{\left(\sqrt{3}\right)}^2}}\\ =\sqrt{12+12}\\ =2\sqrt{6}\end{array}$

Conclusion:

Thus the shortest distance is $2\sqrt{6}$