Tangent Planes to Level Surfaces   

Suppose that r( t) = á x( t) ,y(t) ,z( t) ñ is a curve that lies on a smooth surface U(x,y,z) = k. Applying the derivative with respect to t to both sides of the equation of the level surface yields

 dU
dt
 =  d
dt
 k
Since k is a constant, the chain rule implies that
ÑU ·    dx
dt
 ,  dy
dt
 ,  dz
dt
  = 0
However, the vector v = á dx/dt, dy/dt, dz/dt ñ  is tangent to the surface, which implies that ÑU is orthogonal to each tangent vector v at a given point on the surface.  

Thus, ÑU(p,q,r) is normal to the tangent plane to the surface at a given point (p,q,r).   

We say that the gradient ÑU( x,y,z) is normal to the surface U( x,y,z) = k at each point ( x,y,z) on the surface.

     

EXAMPLE 4    Find the equation of the tangent plane to the hyperboloid in 2 sheets
x2 - y2 - z2 = 4
at the point ( 3,2,1) .      

Solution: To begin with, we identify U( x,y,z) = x2-y2-z2, so that its gradient is
ÑU = á 2x,-2y,-2z ñ
As a result, at the point ( 3,2,1 ) a normal to the tangent plane is given by
n = ÑU( 3,2,1) = á 6,-4,-2 ñ
It follows that the equation of the tangent plane is
6( x-3) -4( y-2) -2( z-1) = 0
which simplifies to the equation z = 3x-2y-4.

Maple/Javaview Figure

EXAMPLE 5    Find the equation of the tangent plane to the right circular cone
x2+y2 = z2
at the point ( 3,4,5) .       

Solution: Since the equation of the surface can be written x2+y2-z2 = 0, we let U( x,y,z) = x2+y2-z2. As a result, the gradient of U is
ÑU = á 2x,2y,-2z ñ
At the point ( 3,4,5) , a normal vector is ÑU = á 6,8,-10 ñ , so that the equation of the tangent plane is
6( x-3) +8( y-4) -10( z-5) = 0
Solving for z then yields z = ( 3x+4y) /5, which is shown in the figure below:

               

Check your Reading: What degenerate conic section is formed by the intersection of the cone with the tangent plane in example 3?