Q 14. - Calculus

Question

Use Theorem 12.45 to show that the point 

$\left(\frac{ad-ab{y}_0-ac{z}_0+b^2{x}_0+c^2{x}_0}{a^2+b^2+c^2},\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\right)$

provides an absolute minimum for the function

$D(x,y)={\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2+{\left(\frac{d-ax-by}{c}-z_0\right)}^2$

Step-by-Step Solution

Answer

$\left(\frac{ad-ab{y}_0-ac{z}_0+b^2{x}_0+c^2{x}_0}{a^2+b^2+c^2},\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\right)$

Step 1: Given information

$D(x,y)={\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2+{\left(\frac{d-ax-by}{c}-z_0\right)}^2$

Step 2: Calculation

The goal is to figure out how to solve the system of equations.

$\left(\frac{ad-b{y}_0-ac{z}_0+b^2{x}_0+c^2{x}_0}{a^2+b^2+c^2},\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\right)$

gives an absolute minimum for the given function.

Differentiate $\left(1\right)$ partially with respect to $x$

$$\begin{gathered} \frac{\partial }{\partial x}D(x,y)=\frac{\partial }{\partial x}\left\{{\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2+{\left(\frac{d-ax-by}{c}-z_0\right)}^2\right\} \\ {D}_x(x,y)=\frac{\partial }{\partial x}{\left(x-x_0\right)}^2+\frac{\partial }{\partial x}{\left(y-y_0\right)}^2+\frac{\partial }{\partial x}{\left(\frac{d-ax-by}{c}-z_0\right)}^2 \\ =2\left(x-x_0\right)+0+2\left(\frac{d-ax-by}{c}-z_0\right)\frac{\partial }{\partial x}\left(\frac{d-ax-by}{c}-z_0\right) \\ =2\left(x-x_0\right)+2\left(\frac{d-ax-by}{c}-z_0\right)\left\{\frac{\partial }{\partial x}\left(\frac{d-ax-by}{c}\right)-\frac{\partial }{\partial x}{z}_0\right\} \\ =2\left(x-x_0\right)+2\left(\frac{d-ax-by}{c}-z_0\right)\left(-\frac{a}{c}-0\right) \\ =2\left(x-x_0\right)-2·\frac{a}{c}\left(\frac{d-ax-by}{c}-z_0\right)\cdots ..\left(2\right)\frac{\partial }{\partial x}D(x,y)=\frac{\partial }{\partial x}\left\{{\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2+{\left(\frac{d-ax-by}{c}-z_0\right)}^2\right\} \\ {D}_x(x,y)=\frac{\partial }{\partial x}{\left(x-x_0\right)}^2+\frac{\partial }{\partial x}{\left(y-y_0\right)}^2+\frac{\partial }{\partial x}{\left(\frac{d-ax-by}{c}-z_0\right)}^2 \\ =2\left(x-x_0\right)+0+2\left(\frac{d-ax-by}{c}-z_0\right)\frac{\partial }{\partial x}\left(\frac{d-ax-by}{c}-z_0\right) \\ =2\left(x-x_0\right)+2\left(\frac{d-ax-by}{c}-z_0\right)\left\{\frac{\partial }{\partial x}\left(\frac{d-ax-by}{c}\right)-\frac{\partial }{\partial x}{z}_0\right\} \\ =2\left(x-x_0\right)+2\left(\frac{d-ax-by}{c}-z_0\right)\left(-\frac{a}{c}-0\right) \\ =2\left(x-x_0\right)-2·\frac{a}{c}\left(\frac{d-ax-by}{c}-z_0\right)\cdots ..\left(2\right) \end{gathered}$$uncaught exception: Invalid chunk

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Differentiate $\left(1\right)$ partially with respect to $y$

$$\begin{gathered} \frac{\partial }{\partial y}D(x,y)=\frac{\partial }{\partial y}\left\{{\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2+{\left(\frac{d-ax-by}{c}-z_0\right)}^2\right\} \\ {D}_y(x,y)=\frac{\partial }{\partial y}{\left(x-x_0\right)}^2+\frac{\partial }{\partial y}{\left(y-y_0\right)}^2+\frac{\partial }{\partial y}{\left(\frac{d-ax-by}{c}-z_0\right)}^2 \\ =0+2\left(y-y_0\right)+2\left(\frac{d-ax-by}{c}-z_0\right)\frac{\partial }{\partial y}\left(\frac{d-ax-by}{c}-z_0\right) \\ =2\left(y-y_0\right)+2\left(\frac{d-ax-by}{c}-z_0\right)\left(\frac{\partial }{\partial y}\frac{d-ax-by}{c}-\frac{\partial }{\partial y}{z}_0\right) \\ =2\left(y-y_0\right)+2\left(\frac{d-ax-by}{c}-z_0\right)\left(-\frac{b}{c}-0\right) \\ =2\left(y-y_0\right)-2·\frac{b}{c}\left(\frac{d-ax-by}{c}-z_0\right)\cdots \cdots \left(3\right) \end{gathered}$$

The stationary point is given by

$D_x(x,y)=0$ and $D_y(x,y)=0$

Thus,

$$\begin{gathered} D_x(x,y)=0 \\ 2\left(x-x_0\right)-2·\frac{a}{c}\left(\frac{d-ax-by}{c}-z_0\right)=0 \\ 2x-2x_0-\frac{2ad}{c^2}+\frac{2a^2}{c^2}x+\frac{2aby}{c^2}+\frac{2a}{c}{z}_0=0 \\ 2x+\frac{2a^2}{c^2}x-2x_0-\frac{2ad}{c^2}+\frac{2aby}{c^2}+\frac{2a}{c}{z}_0=0 \\ \left(2+\frac{2a^2}{c^2}\right)x+\frac{2ab}{c^2}y-2x_0-\frac{2ad}{c^2}+\frac{2a}{c}{z}_0=0 \\ \left(2+\frac{2a^2}{c^2}\right)x+\frac{2ab}{c^2}y=2x_0+\frac{2ad}{c^2}-\frac{2a}{c}{z}_0\dots \dots \end{gathered}$$

And,

$$\begin{gathered} D_y(x,y)=0 \\ 2\left(y-y_0\right)-2·\frac{b}{c}\left(\frac{d-ax-by}{c}-z_0\right)=0 \\ 2y-2y_0-\frac{2bd}{c^2}+\frac{2ab}{c^2}x+\frac{2b^{2}y}{c^2}+\frac{2b}{c}{z}_0=0 \\ 2y+\frac{2b^2}{c^2}y-2y_0-\frac{2bd}{c^2}+\frac{2ab}{c^2}x+\frac{2b}{c}{z}_0=0 \\ \left(2+\frac{2b^2}{c^2}\right)y+\frac{2ab}{c^2}x-2y_0-\frac{2bd}{c^2}+\frac{2b}{c}{z}_0=0 \end{gathered}$$

$\frac{2ab}{c^2}x+\left(2+\frac{2b^2}{c^2}\right)y=2y_0+\frac{2bd}{c^2}-\frac{2b}{c}{z}_0$

Step 3: Calculation

Multiply $( 4 )$ by $\frac{2ab}{c^2}$ and multiply $\left(5\right)$ by $\left(2+\frac{2a^2}{c^2}\right)$ and then subtract

$\frac{2ab}{c^2}\left\{\left(2+\frac{2a^2}{c^2}\right)x+\frac{2ab}{c^2}y\right\}-\left(2+\frac{2a^2}{c^2}\right)\left\{\frac{2ab}{c^2}x+\left(2+\frac{2b^2}{c^2}\right)y\right\}$

$=\frac{2ab}{c^2}\left(2x_0+\frac{2ad}{c^2}-\frac{2a}{c}{z}_0\right)-\left(2+\frac{2a^2}{c^2}\right)\left(2y_0+\frac{2bd}{c^2}-\frac{2b}{c}{z}_0\right)$

$\frac{2ab}{c^2}\left(2+\frac{2a^2}{c^2}\right)x+\frac{2ab}{c^2}·\frac{2ab}{c^2}y-\left(2+\frac{2a^2}{c^2}\right)\frac{2ab}{c^2}x-\left(2+\frac{2a^2}{c^2}\right)\left(2+\frac{2b^2}{c^2}\right)y$

$=\frac{4ab}{c^2}{x}_0+\frac{4a^{2}bd}{c^4}-\frac{4a^{2}b}{c^3}{z}_0-4y_0-\frac{4bd}{c^2}$

$+\frac{4b}{c}{z}_0-\frac{4a^2}{c^2}{y}_0-\frac{4a^{2}bd}{c^4}+\frac{4a^{2}b}{c^3}{z}_0$

$\left\{\frac{4a^2{b}^2}{c^4}-\left(2+\frac{2a^2}{c^2}\right)\left(2+\frac{2b^2}{c^2}\right)\right\}y=\frac{4ab}{c^2}{x}_0-4y_0-\frac{4bd}{c^2}+\frac{4b}{c}{z}_0-\frac{4a^2}{c^2}{y}_0$

$\left(\frac{4a^2{b}^2}{c^4}-4-\frac{4b^2}{c^2}-\frac{4a^2}{c^2}-\frac{4a^2{b}^2}{c^4}\right)y=-\frac{4bd}{c^2}+\frac{4ab}{c^2}{x}_0+\frac{4b}{c}{z}_0-4y_0-\frac{4a^2}{c^2}{y}_0$$\frac{2ab}{c^2}\left(2+\frac{2a^2}{c^2}\right)x+\frac{2ab}{c^2}·\frac{2ab}{c^2}y-\left(2+\frac{2a^2}{c^2}\right)\frac{2ab}{c^2}x-\left(2+\frac{2a^2}{c^2}\right)\left(2+\frac{2b^2}{c^2}\right)y$

$$\begin{gathered} =\frac{4ab}{c^2}{x}_0+\frac{4a^{2}bd}{c^4}-\frac{4a^{2}b}{c^3}{z}_0-4y_0-\frac{4bd}{c^2} \\ +\frac{4b}{c}{z}_0-\frac{4a^2}{c^2}{y}_0-\frac{4a^{2}bd}{c^4}+\frac{4a^{2}b}{c^3}{z}_0 \\ \left\{\frac{4a^2{b}^2}{c^4}-\left(2+\frac{2a^2}{c^2}\right)\left(2+\frac{2b^2}{c^2}\right)\right\}y=\frac{4ab}{c^2}{x}_0-4y_0-\frac{4bd}{c^2}+\frac{4b}{c}{z}_0-\frac{4a^2}{c^2}{y}_0 \\ \left(\frac{4a^2{b}^2}{c^4}-4-\frac{4b^2}{c^2}-\frac{4a^2}{c^2}-\frac{4a^2{b}^2}{c^4}\right)y=-\frac{4bd}{c^2}+\frac{4ab}{c^2}{x}_0+\frac{4b}{c}{z}_0-4y_0-\frac{4a^2}{c^2}{y}_0 \\ \left(-4-\frac{4b^2}{c^2}-\frac{4a^2}{c^2}\right)y=-\frac{4bd}{c^2}+\frac{4ab}{c^2}{x}_0+\frac{4b}{c}{z}_0-4y_0-\frac{4a^2}{c^2}{y}_0 \\ \left(\frac{-4c^2-4b^2-4a^2}{c^2}\right)y=\frac{-4bd+4ab{x}_0+4bc{z}_0-4c^2{y}_0-4a^2{y}_0}{c^2} \\ -\frac{4}{c^2}\left(a^2+b^2+c^2\right)y=-\frac{4}{c^2}\left(bd-ab{x}_0-bc{z}_0+c^2{y}_0+a^2{y}_0\right) \\ \left(a^2+b^2+c^2\right)y=bd-ab{x}_0-bc{z}_0+c^2{y}_0+a^2{y}_0 \end{gathered}$$

$y=\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}$

Substitute $y=\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}$ in $\left(4\right)$

$$\begin{gathered} \frac{2ab}{c^2}x+\left(2+\frac{2b^2}{c^2}\right)\left(\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\right)=2y_0+\frac{2bd}{c^2}-\frac{2b}{c}{z}_0 \\ \frac{2}{c^2}\left\{abx+\left(c^2+b^2\right)\left(\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\right)\right\}=\frac{2}{c^2}\left(c^2{y}_0+bd-bc{z}_0\right) \\ abx+\left(c^2+b^2\right)\left(\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\right)=c^2{y}_0+bd-bc{z}_0 \\ abx=c^2{y}_0+bd-bc{z}_0-\left(c^2+b^2\right)\left(\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\right) \end{gathered}$$

$$\begin{gathered} =\frac{c^2{a}^2{y}_0+c^2\left(c^2+b^2\right)y_0+a^{2}bd+\left(c^2+b^2\right)bd-a^{2}bc{z}_0-\left(c^2+b^2\right)bc{z}_0}{a^2+b^2+c^2} \\ =\frac{c^2{a}^2{y}_0+a^{2}bd-a^{2}bc{z}_0+\left(c^2+b^2\right)ab{x}_0-\left(c^2+b^2\right)a^2{y}_0}{a^2+b^2+c^2} \\ =\frac{c^2{a}^2{y}_0+a^{2}bd-a^{2}bc{z}_0+c^{2}ab{x}_0+b^{2}ab{x}_0-c^2{a}^2{y}_0-b^2{a}^2{y}_0}{a^2+b^2+c^2} \\ =\frac{a^{2}bd-a^{2}bc{z}_0+c^{2}ab{x}_0+b^{2}ab{x}_0-b^2{a}^2{y}_0}{a^2+b^2+c^2} \\ =ab\left(\frac{ad-ac{z}_0+c^2{x}_0+b^2{x}_0-ab{y}_0}{a^2+b^2+c^2}\right) \\ x=\frac{ad-ab{y}_0-ac{z}_0+b^2{x}_0+c^2{x}_0}{a^2+b^2+c^2} \end{gathered}$$

Hence, maximum or minimum exists at

$\left(\frac{ad-ab{y}_0-ac{z}_0+b^2{x}_0+c^2{x}_0}{a^2+b^2+c^2},\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\right)$

Step 4: Calculation

To determine the nature point calculate $D_{xx}(x,y),D_{yy}(x,y)$ and $D_{xy}(x,y)$

Differentiate $( 2 )$ partially with respect to $x$

$$\begin{gathered} \frac{\partial }{dx}{D}_x(x,y)=\frac{\partial }{dx}\left\{2\left(x-x_0\right)-2·\frac{a}{c}\left(\frac{d-ax-by}{c}-z_0\right)\right\} \\ {D}_{xx}(x,y)=2\frac{\partial }{dx}\left(x-x_0\right)-2·\frac{a}{c}\frac{\partial }{dx}\left(\frac{d-ax-by}{c}-z_0\right) \\ =2-2·\frac{a}{c}\left(-\frac{a}{c}\right) \\ =2+\frac{2a^2}{c^2} \end{gathered}$$

Differentiate $( 3 )$ partially with respect to $y$

$$\begin{gathered} \frac{\partial }{dy}{D}_y(x,y)=\frac{\partial }{dy}\left\{2\left(y-y_0\right)-2·\frac{b}{c}\left(\frac{d-ax-by}{c}-z_0\right)\right\} \\ {D}_{yy}(x,y)=2\frac{\partial }{dy}\left(y-y_0\right)-2·\frac{b}{c}\frac{\partial }{dy}\left(\frac{d-ax-by}{c}-z_0\right) \\ =2-2·\frac{b}{c}\left(-\frac{b}{c}\right) \\ =2+\frac{2b^2}{c^2} \end{gathered}$$

Differentiate ( 2 ) partially with respect to $y$

$$\begin{gathered} \frac{\partial }{dy}{D}_x(x,y)=\frac{\partial }{dy}\left\{2\left(x-x_0\right)-2·\frac{a}{c}\left(\frac{d-ax-by}{c}-z_0\right)\right\} \\ {D}_{xy}(x,y)=2\frac{\partial }{dy}\left(x-x_0\right)-2·\frac{a}{c}\frac{\partial }{dy}\left(\frac{d-ax-by}{c}-z_0\right) \\ =0-2·\frac{a}{c}\left(-\frac{b}{c}\right) \\ =\frac{2ab}{c^2} \end{gathered}$$

Now, find the discriminant of $D(x, y)$

$$\begin{gathered} det\left(H_f\right)=D_{xx}{D}_{yy}-{\left(D_{xy}\right)}^2 \\ =\left(2+\frac{2a^2}{c^2}\right)\left(2+\frac{2b^2}{c^2}\right)-{\left(\frac{2ab}{c^2}\right)}^2 \\ =4+\frac{4b^2}{c^2}+\frac{4a^2}{c^2}+{\left(\frac{2ab}{c^2}\right)}^2-{\left(\frac{2ab}{c^2}\right)}^2 \\ =4+\frac{4b^2}{c^2}+\frac{4a^2}{c^2} \\ =\frac{4c^2+4b^2+4a^2}{c^2} \end{gathered}$$

Since square of a umber is always positive, so $det\left(H_f\right)>0$ and $D_{xx}(x,y)>0$ and $D_{yy}(x,y)>0$

As a result, the supplied function $D(x, y)$ has a minimum in my second-order partial derivative test.

$\left(\frac{ad-ab{y}_0-ac{z}_0+b^2{x}_0+c^2{x}_0}{a^2+b^2+c^2},\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\right)$