Level Surfaces in Cylindrical Coordinates

In cylindrical coordinates, a level surface is of the form U( r,q,z) = k. To find the tangent plane to a level surface in cylindrical coordinates, we must first determine a formula for the gradient in cylindrical coordinates.

To begin with, if er = á cos(q), sin(q), 0 ñ, eq = á -sin(q) ,cos(q), 0ñ , and ez = k, then er, eq, and ez are 3 mutually orthogonal unit vectors that form a basis for cylindrical coordinates, much as er and eq are a basis in polar coordinates and i, j, and k are a basis for Cartesian coordinates.  However, notice that the basis vectors er, eq, and k change direction from point to point, unlike the i, j, and k vectors (drag the red point in each of the following applets).
LiveGraphics3d Applet              LiveGraphics3d Applet

Since cylindrical coordinates are simply polar coordinates in the xy-plane, the gradient in cylindrical coordinates reduces to the gradient in polar in the xy-plane.  Thus the gradient in cylindrical is simply
ÑU = Ur er+  1
r
 Uq eq+Uz ez
Moreover, ÑU is normal to U( r,q,z) = k.    

EXAMPLE 5    The equation r = 1 defines a right circular cylinder in cylindrical coordinates.  What is the tangent plane to the cylinder at
P1: ( r,q,z) = ( 1,p,2)
Solution: If we let U( r,q,z) = r, then
ÑU = 1er+0 eq+0 ez = er
Since er( 1,p,2) = á cos( p) ,sin( p) ,0 ñ = á-1,0,0 ñ , we have
ÑU( 1,p,2) = á -1,0,0 ñ
Since ( r,q,z) = ( 1,p,2) is equivalent to ( x,y,z) = ( 1cos( p) ,1sin( p) ,2) = ( -1,0,2) , the tangent plane is
-1( x-1) +0( y-0) +0( z-2) = 0
which reduces to x = -1.
LiveGraphics3d Applet

Place cursor over surface to animate.

EXAMPLE 6    The sphere of radius 3 centered at the origin is given by
r2+z2 = 9
in cylindrical coordinates. Find the equation of the tangent plane to the sphere at the point (r, q, z) = (4, p/4 , 3).        

Solution: Let us let U( r,q,z) = r2+z2. Then Ur = 2r, Uq = 0, and Uz = 2z. Thus,
ÑU
=
Ur er+  1
r		 Uq eq+Uz ez
=
2rer+0 eq+2z ez
=
2r á cos( q) ,sin( q),0 ñ +0+2z á 0,0,1 ñ
=
á 2rcos( q) ,2rsin( q),2z ñ
At (r, q, z) = (4, p/4 , 3), the gradient is
ÑU(4, p/4 , 3)  = 
2(4)cos(
p
4
), 2(4)sin( p
4		 ),2(3)   =  á 4Ö2, 4Ö2, 6 ñ

If we let P1 denote the point (4, p/4 , 3) in cylindrical, then in Cartesian coordinates we have
P1  : 
4 cos(
p
4
), 4 sin( p
4		 ), 3   =  á 2Ö2, 2Ö2, 3 ñ
Thus, the equation of the plane is
4Ö2(x - 2Ö2) + 4Ö2 (y - 2Ö2) + 6 (x - 3) = 0   
which upon solving for z leads to
z =     25
3		   -   2Ö2
3		  x    -   2Ö2
3		  y
LiveGraphics3d Applet
ÑU drawn to 1/2 scale

 

It is important to note that we can also obtain ÑU in cylindrical coordinates by calculating ÑU in Cartesian coordinates and then converting the resulting gradient vector into cylindrical coordinates.  To illustrate, consider that the sphere in example 3 is given by x2+y2+z2 = 25,  which has a gradient of   
ÑU = á 2x, 2y, 2z ñ
Converting to cylindrical leads to ÑU = á 2rcos( q),  2rsin( q),  2z ñ.

Check Your Reading: Does ez ever change directions?