Calculus III
Contents
3 Dimensional space
Partial derivatives
Multiple integrals
Vector Functions
Line integrals
Surface integrals
Vector operators
Applications
© The scientific sentence. 2010
|
|
Calculus III:
Parametric surfaces
1. Definitions and method
We have seen that the parameterization of a curve consists in
taking values of t from an interval [a, b] and substitute them into
the vector function :
 (t) =
x(t)  +
y(t)  + z(t) 
The resulting set of vectors is the position vectors for the points on the curve.
The parameterization of surfaces is similar.
We take two points (u, v) out of a two-dimensional
space D and substitute them into
the vector function
(u,v) =
x(u,v) +
y(u,v) + z(u,v)
The resulting set of vectors is the position vectors for the points on the surface S to parameterize.
The vector (u,v) is called the parametric representation of the parametric surface S.
The parametric equations for a surface are the components of the parametric representation:
x = x(u,v) , y = y(u,v), z = z(u,v)
Example 1
Determine the surface given by the following parametric representation:
(u,v) =
u cos v +
u sin v + u
The parametric equations are:
x = u cos v , y = u sin v, z = u
Let's square x, y and z ; we obtain:
z2 = x2 + y2
We have then eliminated the parameters to obtain an equation
in x, y, and z. This quadric surface is a cone
open along the z-axis.
Example 2 : Sphere
Give a parametric representation of the surface:
x2 + y2 + z2 = 16
Converting Cartesian coordinates into spherical coordinates gives
the parametric representation:
(θ,φ) =
4 sin θ cos φ +
4 sin θ sin φ + 4 cos θ
With the two restrictions:
0 ≤ θ ≤ π and 0 ≤ φ ≤ 2π
Example 3 : Cylinder
Give a parametric representation of the surface:
x2 + y2 = 16
Converting Cartesian coordinates into cylindrical coordinates gives
the parametric representation:
(z,θ) =
4 sin θ +
4 cos θ + z
With the restriction:
0 ≤ θ 2π
Example 4 : Elliptic paraboloid
Give a parametric representation of the surface:
y = 2x2 + 3z2 - 16
The surface is in the form y = f(x, z). So x and z becomes parameters:.
So the parametric equations are:
x = x , z = z , y = 2x2 + 3z2 - 16
The parametric representation is then:
(x, z) =
x +
(x2 + 3z2 - 16 ) + z
Note that, in the same manner, we can have: (x, y),
or (y, z)
Application: tangent plane to the parametric
surface S
Find the tangent plane to the parametric surface S given by:
(u,v) =
x(u,v) +
y(u,v) + z(u,v)
Let's determine the partial derivatives of this
parametric representation :
 (u,v) =
(∂x(u,v)/∂u) +
(∂y(u,v)/∂u) + (∂z(u,v)/∂u)
 (u,v) =
(∂x(u,v)/∂v) +
(∂y(u,v)/∂v) + (∂z(u,v)/∂v)
The non-zero vector cross product (u,v) x (u,v) is orthogonal to the
surface S, then normal to the tangent plane to this surface.
Example 5
Find the tangent plane to the parametric surface S given by:
(u,v) =
u +
v2 + (u2 + v)
at the point P (- 1, 1, 0)
The partial derivatives are:
(u,v) =
+ 2u
(u,v) =
2v +
The cross product gives the normal vector  :
= (u,v) x (u,v) = 〈 - 4uv, - 1, 2v 〉
Now, we determine the value of u and v at the point P :
To do this, we set the components of the parametric representation equal to the coordinates of the point (- 1, 1, 0). That is:
u = - 1
v2 = 1 → v = - 1 or v = + 1
u2 + v = 0 → v = - 1
Hence the correct values are :
u = - 1 , and
v = - 1
Substituting these two values in, the normal vector becomes:
= (u,v) x (u,v) = 〈 - 4 , - 1, - 2 〉
Therefore, the equation of the tangent plane is:
M(x, y, z), P (- 1, 1, 0) , so  = 〈 x + 1, y - 1, 0 - z 〉
. = 0
〈 x + 1, y - 1, 0 - z 〉 . 〈 - 4 , - 1, - 2 〉 = 0
- 4(x + 1) - 1(y - 1) - 2(0 - z) = 0
- 4 x - y + 2 z - 3 = 0
|
|