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Mathematics LibreTexts

1.7: Tangent Planes and Normal Lines

  • Page ID
    602
  • [ "article:topic", "Tangent Line", "authorname:green", "Tangent Planes", "Normal Lines", "Angle of Inclination" ]

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    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

    \[F_x(x, y, z) x' + F_y(x, y, z) y' + F_z(x, y, z) z'  =  0\]

    Notice that this is the dot product of the gradient function and the vector \(\langle x',y',z'\rangle \),

    \[\nabla F  \cdot \langle x', y', z'\rangle   =  0.\]

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

    Definition: Tangent Plane

    Let \(F(x,y,z)\) define a surface that is differentiable at a point \((x_0,y_0,z_0)\), then the tangent plane to \(F ( x, y, z )\) at \(( x_0, y_0, z_0)\) is the plane with normal vector

    \[ \nabla \, F(x_0,y_0,z_0) \]

    that passes through the point \((x_0,y_0,z_0)\) .  In particular, the equation of the tangent plane is

    \[ \nabla \, F(x_0,y_0,z_0) \cdot \langle x - x_0 , y - y_0 , z - z_0 \rangle   =  0. \]

    Example \(\PageIndex{1}\)

    Find the equation of the tangent plane to

    \[   z  =  3x^2 - xy \]

    at the point \((1,2,1)\).

    Solution

    We let

    \[F(x,y,z)  =  3x^2 - xy - z\]

    then

    \[\nabla F  =  \langle 6x - y, -x, -1\rangle . \]

    At the point \((1,2,1)\), the normal vector is

    \[\nabla F(1,2,1) = \langle 4, -1, -1\rangle . \]

    Now use the point normal formula for a plan

    \[\langle 4, -1, -1\rangle \cdot \langle x - 1, y - 2, z - 1\rangle   =  0\]

    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: Normal Line

    Let \(F(x,y,z)\) define a surface that is differentiable at a point \((x_0,y_0,z_0)\), then the normal line to \(F(x,y,z)\) at \((x_0,y_0,z_0)\) is the line with normal vector

    \[ \nabla \, F(x_0,y_0,z_0) .\]

    that passes through the point \((x_0,y_0,z_0)\) .  In Particular the equation of the normal line is

    \[          x(t) = x_0 + F_x(x_0,y_0,z_0) t,  \]

    \[          y(t) = y_0 + F_y(x_0,y_0,z_0) t,  \]

    \[          z(t) = z_0 + F_z(x_0,y_0,z_0) t.  \]

    Example \(\PageIndex{2}\)

    Find the parametric equations for the normal line to

    \[  x^2yz - y + z - 7  = 0 \]

    at the point \((1,2,3)\).

    Solution

    We compute the  gradient:

    \[\nabla F  =  \langle 2xyz, x^2z - 1, x^2y + 1\rangle   =  \langle 12, 2, 3\rangle .\]

    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

    \[\cos q = \dfrac{|\textbf{n} \cdot k|}{ ||\textbf{n} ||}. \]

    More generally, if  \( F(x,y,z)  =  0 \) is a surface, then the angle of inclination at the point \((x_0,y_0,z_0)\) is defined by the angle of inclination of the tangent plane at the point with

    \[ \cos\,q = \dfrac{  | \nabla F(x_0, y_0, z_0) \cdot \textbf{k}| }{|| \nabla F(x_0, y_0, z_0)||}. \]

    Example \(\PageIndex{3}\)

    Find the angle of inclination of

    \[ \dfrac{x^2}{4} + \dfrac{y^2}{4} + \dfrac{z^2}{8} = 1\]

    at the point \((1,1,2)\).

    Solution

    First compute

    \[  \nabla F  =  \langle \dfrac{x}{2}, \dfrac{y}{2}, \dfrac{z}{4}\rangle .\]

    Now plug in to get

    \[\nabla F(1,1,2)  =  \langle \dfrac{1}{2}, \dfrac{1}{2}, \dfrac{1}{2} \rangle .\]

    We have

    \[|\langle \dfrac{1}{2} , \dfrac{1}{2} , \dfrac{1}{2} \rangle \cdot \hat{\textbf{k}} |  =  \dfrac{1}{2} .\]

    Also,

    \[||\langle \dfrac{1}{2} , \dfrac{1}{2} , \dfrac{1}{2} \rangle   || = \dfrac{\sqrt{3}}{2}  .\]

    Hence

    \[ \cos q = \dfrac{\frac{1}{2}}{( \frac{\sqrt{3}}{2}  )} = \dfrac{1}{\sqrt{3}} . \]

    So the angle of inclination is

    \[q  =  \cos^{-1}(\dfrac{1}{\sqrt{3}}) = 0.955 \text{ radians} .\]

    The Tangent Line to a Curve

    Example \(\PageIndex{4}\)

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

    \[x^2 + y^2 + z^2  =  30\]

    and the paraboloid

    \[z  =  x^2 + y^2\]

    at the point \((1,2,5)\).

    Solution

    We find the gradient of the two surfaces at the point

    \[  \nabla(x^2 + y^2 + z^2) = \langle 2x, 2y, 2z\rangle = \langle 2, 4,10\rangle \]

    and

    \[\nabla (x^2 + y^2 - z)  =  \langle 2x, 2y, -1\rangle = \langle 2, 4, -1\rangle .\]

    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

    \[ \begin{vmatrix}  \hat{\textbf{i}} & \hat{\textbf{j}} &  \hat{\textbf{k}} \\ 2 & 4 & 10 \\ 2 & 4 & -1 \end{vmatrix} = -44 \hat{\textbf{i}}  + 22  \hat{\textbf{j}}. \]

    Hence the equation of the tangent line is

    \[x(t) = 1 - 44t       y(t) = 2 + 22t       z(t) = 5.\]

    Larry Green (Lake Tahoe Community College)

    • Integrated by Justin Marshall.