Tangent Plane

Intuitively, it seems clear that, in a plane, only one line can be tangent to a curve at a point. However, in three-dimensional space, many lines can be tangent to a given point. If these lines lie in the same plane, they determine the tangent plane at that point. A tangent plane at a regular point contains all of the lines tangent to that point. A more intuitive way to think of a tangent plane is to assume the surface is smooth at that point (no corners). Then, a tangent line to the surface at that point in any direction does not have any abrupt changes in slope because the direction changes smoothly.

Definition of Tangent Plane

Let

P_{0}=(x_{0},y_{0},z_{0})

be a point on a surface

S

, and let

C

be any curve passing through

P_{0}

and lying entirely in

S

. If the tangent lines to all such curves

C

at

P_{0}

lie in the same plane, then this plane is called the tangent plane to

S

at

P_{0}

.

For a tangent plane to a surface to exist at a point on that surface, it is sufficient for the function that defines the surface to be differentiable at that point, defined later in this section. We define the term tangent plane here and then explore the idea intuitively.

Equation of Tangent Plane

Let

S

be a surface defined by a differentiable function

z=f(x,y)

, and let

P_{0}=(x_{0},y_{0})

be a point in the domain of

f

. Then, the equation of the tangent plane to

S

at

P_{0}

is given by

z=f(x_{0},y_{0})+f_{x}(x_{0},y_{0})(x-x_{0})+f_{y}(x_{0},y_{0})(y-y_{0})

To see why this formula is correct, let’s first find two tangent lines to the surface S. The equation of the tangent line to the curve that is represented by the intersection of S with the vertical trace given by $x=x_{0}$ is $z=f(x_{0},y_{0})+f_{y}(x_{0},y_{0})(y-y_{0})$ . Similarly, the equation of the tangent line to the curve that is represented by the intersection of $S$ with the vertical trace given by $y=y_{0}$ is $z=f(x_{0},y_{0})+f_{x}(x_{0},y_{0})(x-x_{0})$ . A parallel vector to the first tangent line is $a=\mathbf {j} +f_{y}(x_{0},y_{0})\mathbf {k}$ ; a parallel vector to the second tangent line is $b=\mathbf {i} +f_{x}(x_{0},y_{0})\mathbf {k}$ . We can take the cross product of these two vectors:

a\times b=(\mathbf {j} +f_{y}(x_{0},y_{0})\mathbf {k} )\times (\mathbf {i} +f_{x}(x_{0},y_{0})\mathbf {k} )

=f_{x}(x_{0},y_{0})\mathbf {i} +f_{y}(x_{0},y_{0})\mathbf {j} -\mathbf {k}

This vector is perpendicular to both lines and is therefore perpendicular to the tangent plane. We can use this vector as a normal vector to the tangent plane, along with the point $P_{0}=(x_{0},y_{0},f(x_{0},y_{0}))$ in the equation for a plane:

$n\cdot ((x-x_{0})\mathbf {i} +(y-y_{0})\mathbf {j} +(z-f(x_{0},y_{0}))\mathbf {k} )=0$

$(f_{x}(x_{0},y_{0})\mathbf {i} +f_{y}(x_{0},y_{0})\mathbf {j} -\mathbf {k} )\cdot ((x-x_{0})i+(y-y_{0})\mathbf {j} +(z-f(x_{0},y_{0}))\mathbf {k} )=0$