Consider the paraboloid z = x2 + y2. The plane 6x - 8y z - 4 = 0 cuts the parabo

Question

Consider the paraboloid z = x2 + y2. The plane 6x - 8y z - 4 = 0 cuts the paraboloid, its intersection being a curve. Find "the natural" parametrization of this curve. Hint: The curve which is cut lies above a circle in the xy-plane which you should parametrize as a function of the variable t so that the circle is traversed counterclockwise exactly once as t goes from 0 to 2*pi, and the paramterization starts at the point on the circle with largest x coordinate. Using that as your starting point, give the parametrization of the curve on the surface. c(t) = (x(t),y(t),z(t)), where x(t) = y(t) = z(t) =

Explanation / Answer

intersecting those curves,

z= x^2 + y^2 and 6x - 8y +z-4 =0,

6x-8y + x^2 + y^2 -4=0

(x+3)^2 + (y-4)^2 = 29,

so,

x(t) = -3 + sqrt (29) cos t

y(t) =4 + sqrt (29) sin t

z(t) = 8 y- 6x + 4 = 32 + 8 sqrt (29) sin t -18 + 6 sqrt(29) cos t + 4

z(t) = 18 + 8 sqrt (29) sin t + 6 sqrt(29) cos t

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