Consider the paraboloid z=x2+y2. the plane 4x−8y+z−10=0 cuts the paraboloid, its intersection being a curve. find "the natural" parametrization of this curve.

The curve parameter is indeed a map r(t) = <x(t), y(t)> of the interval R= [a, b] parameter to a plane. Co - ordinate functions are considered the functions x(t), y(t). It defines a curve, a surface, etc. with one or even more variables that could be used in a specified range, and the calculation can be defined as follows:

Consider the paraboloid [tex]z=x^2+y^2[/tex], and the plane [tex]4x-8y+z-10=0[/tex]

[tex]\to \bold{z= -4x+8y+10}[/tex]

Calculating the intersection curve of these curves:

[tex]\to \bold{x^2+y^2= -4x+8y+10}\\\\\to \bold{x^2+y^2+4x-8y-10=0}\\\\\to \bold{x^2+4x+4 +y^2-8y+16-10 -16-4=0}\\\\\to \bold{(x+2)^2+(y-4)^2-30=0}\\\\\to \bold{(x+2)^2+(y-4)^2=30}\\\\\to \bold{\frac{(x+2)^2}{30}+\frac{(y-4)^2}{30}=1}\\\\\to \bold{(\frac{(x+2)}{\sqrt{30}})^2+(\frac{(y-4)}{\sqrt{30}})^2=1}\\\\\therefore \\\\\bold{\cos^2 t+ \sin^2 t=1}\\\\[/tex]

Comapring the above equation:

[tex]\to \bold{\frac{(x+2)}{\sqrt{30}}= \cos t \ \ \ \ \ , \ \ \ \ \ \frac{(y-4)}{\sqrt{30}} =\sin t}\\\\\to \bold{(x+2)= \sqrt{30}\cos t \ \ \ \ \ , \ \ \ \ \ \ (y-4) = \sqrt{30}\sin t}\\\\\to \bold{x=-2+\sqrt{30}\cos t \ \ \ \ \ , \ \ \ \ \ \ y = 4+\sqrt{30}\sin t}\\\\ \therefore \\\\\bold{z= -4x+8y+10}[/tex]

Putting the x and y value in the above equation then:

[tex]\to \bold{z= -4(-2+\sqrt{30}\cos t)+8(4+\sqrt{30}\sin t)+10}\\\\\to \bold{z= 8-4\sqrt{30}\cos t+32+8\sqrt{30}\sin t+10}\\\\\to \bold{z= -4\sqrt{30}\cos t+8\sqrt{30}\sin t+50}\\\\[/tex]

Hence the pararnetric equations of the intersecting curve are:

[tex]x(t)=-2+\sqrt{30}\cos t \\\\y(t)= 4+\sqrt{30}\sin t \\\\z(t)= -4\sqrt{30}\cos t+8\sqrt{30}\sin t+50\\\\[/tex]

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