Surface Normals and Tangent Planes to Level Surfaces
Because the equation of a plane requires a point and a normal vector to the
plane, finding the equation of a tangent plane to a surface at a given point
requires the calculation of a surface normal vector. In this section, we explore
the concept of a normal vector to a surface and its use in finding equations of
tangent planes.
To begin with, a level surface U( x,y,z) = k is said to be smooth if
the gradient ÑU =
áU_{x},U_{y},U_{z}
ñ is continuous and nonzero at each point on the surface.
Equivalently, we often write
the gradient as
ÑU = U_{x}e_{x}+U_{y}e_{y}+U_{z}e_{z} 

where e_{x} =
á 1,0,0
ñ , e_{y} =
á 0,1,0
ñ , and e_{z} =
á0,0,1
ñ .
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
Since k is a constant, the chain rule implies that
where v =
á x' (t), y' (t), z' (t)
ñ. However, v
is tangent to the surface because it is tangent to a curve on the surface,
which implies that ÑU is orthogonal to each tangent vector v at a given point on the surface.

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That is, ÑU(p,q,r) at a given point (p,q,r) is normal to the tangent plane
to the surface U( x,y,z) = k at the point (p,q,r).
We thus say that the gradient ÑU is normal to the surface U(x,y,z) = k
at each point on
the surface.
EXAMPLE 1 Find the equation of the tangent plane to the
hyperboloid in 2 sheets
x^{2 } y^{2 } z^{2} = 4 

at the point ( 3,2,1) .
Solution: To begin with, we identify U( x,y,z) = x^{2}y^{2}z^{2}, so that its gradient is
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( x3) 4( y2) 2( z1) = 0 

which simplifies to the equation z = 3x2y4.
EXAMPLE 2 Find the equation of the tangent plane to the
right circular cone
at the point (0.6, 0.8, 1) .
Solution: Since the equation of the surface can be written x^{2}+y^{2}z^{2} = 0, we let U( x,y,z) = x^{2}+y^{2}z^{2}. As a
result, the gradient of U is
At the point ( 0.6, 0.8, 1) , a normal vector is ÑU =
á 1.2, 1.6, 2
ñ , so that the equation of the tangent
plane is
1.2( x0.6) +1.6( y0.8) 2( z1) = 0 

Solving for z then yields z = 0.6x + 0.8y, 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?