Chapter 4.6, Problem 16E

To determine

To calculate: Thecoordinatesof the point on the curve $y=\tan x$ that is closest to the point $\left(1,1\right)$ correct to two decimal places.

Answer to Problem 16E

Thecoordinates of the point on the e the curve $y=\tan x$ that is closest to the point $\left(1,1\right)$ is $\left(0.82,1.07\right)$

Explanation of Solution

Given information:

The curve $y=\tan x$ and a point $\left(1,1\right)$

Formula used:

Let $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are two given points, the distance $d$ between these two points is given by the formula

  $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

And

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.

  $f^{\prime }(a)=0$ 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.

3. 1. if $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
   2. If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Calculation:

As per the given problem

The curve

  $y=\tan x\mathrm{......}\left(1\right)$

Let $\left(x,y\right)$ the point on the curvethat is closest from the point $\left(1,0\right)$

Recall that, let $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are two points, the distance $d$ between these two points is given by the formula

  $\begin{array}{l}D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ \Rightarrow D=\sqrt{{\left(x-1\right)}^2+{\left(y-1\right)}^2}\\ =\sqrt{x^2+1^2-2\times x\times 1+y^2+1^2-2\times y\times 1}\left\{\text{Use }{\left(a+b\right)}^2=a^2+b^{{}^2}+2ab\right\}\end{array}$

Substitute the value $y=\tan x$ and simplified

  $\begin{array}{l}D=\sqrt{x^2+1^2-2\times x\times 1+y^2+1^2-2\times y\times 1}\\ D=\sqrt{x^2+1-2x+{\tan }^{2}x+1-2\tan x}\\ D=\sqrt{x^2+1-2x+{\sec }^{2}x-2\tan x}\left\{{\text{Use sec}}^{2}x=1+{\tan }^{2}x\right\}\end{array}$

Square both sides, to get

  $D^2(x)=x^2+1-2x+{\sec }^{2}x-2\tan x$

Recall that,

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.

Differentiate on both sides,

  $\begin{array}{l}\frac{d}{dx}{D}^2\left(x\right)=\frac{d}{dx}\left(x^2+1-2x+{\sec }^{2}x-2\tan x\right)\\ =2x-2+2\sec x\times \sec x\tan x-2{\sec }^{2}x\\ =2x-2+2{\sec }^{2}x\tan x-2{\sec }^{2}x\\ \end{array}$

Solve for $\frac{d}{dx}{D}^2\left(x\right)=0$ ,

  $2x-2+2{\sec }^{2}x\tan x-2{\sec }^{2}x=0$

Solve by graphically, to get

  

  $x=0.82$

Substitute $x=0.82$ in equation $(1)$ and simplified

  $\begin{array}{l}y=\tan (0.82)\\ =1.07\end{array}$

Therefore,

  $x=0.82,y=1.07$

Thus thecoordinates of the point on the e the curve $y=\tan x$ that is closest to the point $\left(1,1\right)$ is $\left(0.82,1.07\right)$