Tangent Planes and Normal Lines

Tangent Planes

Let z = f(x,y) be a function of two variables.  We can define a new function F(x,y,z) of three variables by subtracting z.  This has the condition

F(x,y,z)  =  0

Now consider any curve defined parametrically by

x  =  x(t),      y  =  y(t)      z  =  z(t)

We can write,

F(x(t), y(t), z(t))  =  0

Differentiating both sides with respect to t, and using the chain rule gives

Fx(x, y, z) x' + Fy(x, y, z) y' + Fz(x, y, z) z'  =  0

Notice that this is the dot product of the gradient function and the vector <x',y',z'>,

GradF . <x', y', z'>  =  0

In particular the gradient vector is orthogonal to the tangent line of any curve on the surface.  This leads to

 Definition Let F(x,y,z) define a surface that is differentiable at a point (x0,y0,z0), then the tangent plane to F ( x, y, z ) at ( x0 , y0 , z0 ) is the plane with normal vector           Grad F(x0,y0,z0) that passes through the point (x0,y0,z0).  In Particular the equation of the tangent plane is        Grad F(x0,y0,z0) . < x - x0 , y - y0 , z - z0 >  =  0

Example

Find the equation of the tangent plane to

z  =  3x2 - xy

at the point (1,2,1)

Solution

We let

F(x,y,z)  =  3x2 - xy - z

then

Grad F  =  <6x - y, -x, -1> At the point (1,2,1), the normal vector is

Grad F(1,2,1) = <4, -1, -1>

Now use the point normal formula for a plan

<4, -1, -1> . <x - 1, y - 2, z - 1>  =  0

or

4(x - 1) - (y - 2) - (z - 1)  =  0

Finally we get

4x - y - z  =  1

Normal Lines

Given a vector and a point, there is a unique line parallel to that vector that passes through the point.  In the context of surfaces, we have the gradient vector of the surface at a given point.  This leads to the following definition.

 Definition Let F(x,y,z) define a surface that is differentiable at a point (x0,y0,z0), then the normal line to F(x,y,z) at ( x0 , y0 , z0 ) is the line with normal vector           GradF(x0,y0,z0) that passes through the point (x0,y0,z0).  In Particular the equation of the normal line is           x(t) = x0 + Fx(x0,y0,z0) t             y(t) = y0 + Fy(x0,y0,z0) t             z(t) = z0 + Fz(x0,y0,z0) t

Example

Find the parametric equations for the normal line to

x2yz - y + z - 7  = 0

at the point (1,2,3).

Solution

Grad F  =  <2xyz, x2z - 1, x2y + 1>  =  <12, 2, 3>

Now use the formula to find

x(t) = 1 + 12t      y(t) = 2 + 2t      z(t) = 3 + 3t

The diagram below displays the surface and the normal line. Angle of Inclination

Given a plane with normal vector n the angle of inclination, q is defined by

 |n . k|     cosq =                                   ||n||

More generally, if

F(x,y,z)  =  0

is a surface, than the angle of inclination at the point  (x0, y0, z0) is defined by the angle of inclination of the tangent plane at the point.

 |Grad F(x0, y0, z0) . k|      cosq =                                                                   ||Grad F(x0, y0, z0)||

Example

Find the angle of inclination of

x2             y2          z2
+            +             =  1
4            4            8

at the point (1,1,2).

Solution

First compute

Grad F  =  <x / 2, y / 2, z / 4>

Now plug in to get

Grad F(1,1,2)  =  <1/2, 1/2, 1/2>

We have

|<1/2, 1/2, 1/2> . k|  =  1/2

Also,

||<1/2, 1/2, 1/2>||  = / 2

Hence

cosq = (1/2)/[( )/2] = 1/ So the angle of inclination is

q  =  arccos(1/ @   0.955 radians

The Tangent Line to a Curve

Example

Find the tangent line to the curve of intersection of the sphere

x2 + y 2 + z2  =  30

and the paraboloid

z  =  x2 + y2

at the point (1,2,5).

Solution

We find the Grad of the two surfaces at the point

Grad (x2 + y 2 + z2) = <2x, 2y, 2z> = <2, 4,10>

and

Grad (x2 + y 2 - z)  =  <2x, 2y, -1> = <2, 4, -1>

These two vectors will both be perpendicular to the tangent line to the curve at the point, hence their cross product will be parallel to this tangent line.  We compute Hence the equation of the tangent line is

x(t) = 1 - 44t       y(t) = 2 + 22t       z(t) = 5

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