consider the paraboloid z = x^2+y^2. The plane3x-2y+z-5=0 cuts the paraboloid, i

Question

consider the paraboloid z = x^2+y^2. The plane3x-2y+z-5=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))

Explanation / Answer

x^2 + y^2 = z, and 3x-2y+z-5=0 means 2y - 3x + 5 = z

Setting the 2 values of z equal, we have x^2 + y^2 = 2y - 3x + 5, or x^2 + 3x + y^2 - 2y = 5, or, completing the square,

x^2 + 3x + 9/4 + y^2 - 2y + 1 = 5 + 9/4 + 1 = 33/4

(x+3/2)^2 + (y-1)^2 = 33/4

This is a circle of radius sqrt(33)/2 centered at -3/2, 1

This gives us the parametrization of x and y as (-3/2 + sqrt(33)/2 cos t, 1 + sqrt(33)/2 sin t)

Finally, we consider z

Let's use the second expression here, 2y - 3x + 5 = z.

Thus, as x = -3/2 + sqrt(33)/2 cos t, and y = 1 + sqrt(33)/2 sin t,

2y - 3x + 5 = 2(1 + sqrt(33)/2 sin t) - 3( -3/2 + sqrt(33)/2 cos t) + 5 =

2 + sqrt(33) sin t + 9/2 - 3 sqrt(33)/2 cos t + 5 = 23/2 + sqrt(33) sin t - 3 sqrt(33)/2 cos t

Thus, the complete parametrization is

(-3/2 + sqrt(33)/2 cos t, 1 + sqrt(33)/2 sin t, 23/2 + sqrt(33) sin t - 3 sqrt(33)/2 cos t) t = 0 to 2pi

Note: I actually created values of t from 0 to 2 pi, calculated x, y, and z from the parametrization, and verified that

z = x^2+y^2 and 3x-2y+z-5=0. I invite you to do likewise.

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