Partial Derivatives of Parametric Survaces Polar Coordinates Rectangular Coordinates  Contents

If the coordinate functions of a parametric surface (x(u,v),y(u,v),z(u,v)) have partial derivatives with respect to u and v, then the partial derivative of the surface with respect to u is (xu(u,v),yu(u,v),zu((u,v)) and the partial derivative of the surface with respect to v is (xv(u,v),yv(u,v),zv((u,v)).

Exercises:  Find the u- and v- partial derivatives of the following parametric surfaces:

1.  The sphere
2.  The cone
3.  The torus

Critical Points of Parametric Surfaces 1D  Polar Coordinates  Rectangular Coordinates  Top of Page  Contents

A critical point of a parametric surface (x(u,v),y(u,v),z(u,v)) is a point
(x(u0,v0),y(u0,v0),z(u0,v0)) such that one of the partial derivatives is a
multiple of the other.

figure3     figure4     figure5


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Exercise

  • What are the critical points of the following parametric surfaces:
  • 1. The Sphere
  • 2. The Cone
  • 3. The Torus
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    Tangent Planes and Normal Vectors for Parametrized Surfaces 1D  Polar Coordinates  Rectangular Coordinates  Top of Page  Contents

    If both partial derivatives for each coordinate function of the parametrized surface (x(u,v), y(u,v), z(u,v)) exist at a point (x(u0,v0), y(u0,v0), z(u0,v0)), then the tangent plane of the surface at (u0,v0) is the plane determined by the vectors

    (xu(u0,v0), yu(u0,v0), zu(u0,v0)) and (xv(u0,v0), yv(u0,v0), zv(u0,v0)).

    In term of parameters s and t, the tangent plane will be given by

    (x(u0,v0), y(u0,v0), z(u0,v0)) + s(xu(u0,v0), yu(u0,v0), zu(u0,v0) + t(xv(u0,v0), yv(u0,v0), zv(u0,v0)).

    To find the normal line to the parametric surface at a point, we can use the cross-product of the partial derivative vectors to get a vector perpendicular to both of them.  We then get the normal line by adding all multiples of this cross product to the point.


    figure10


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    Differentiability in Parametric Equations 1D  Polar Coordinates  Rectangular Coordinates  Top of Page  Contents

    Figure20

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