Tangent Planes

Tangent planes to graphs of functions

Assume that \( f(x,y) \) is a function of two variables. Its graph is a surface \( z=f(x,y) \). Our goal is to find an equation of the tangent plane passing trough a point \( A(x_0,y_0,z_0) \) on the surface.

picture

In the picture above, you can see a graph of a surface \( z=f(x,y) \) and a point \( A \) on that graph. Our goal is to find the equation of the shaded plane. In order to do this, we just need to find the coordinates of the normal vector \( \overrightarrow{n} \).

Recall that in the single variable calculus, the quantity \( f^{\prime}(a) \) represented the slope of the tangent line to the graph of \( f \) at the point \( a \). The tangent vector is then equal to \( \overrightarrow T=\langle 1, f^{\prime}(a)\rangle \).

In multivariable calculus, the tangent to the surface is a plane, rather than a vector. The quantity \( f_x(x_0,y_0,z_0) \) also represents a slope of a certain vector. This vector belongs to the tangent plane, and is parallel to the \( xz \) plane. Let us call this vector \( \overrightarrow T_x \). Let us find its coordinates, i.e. let us find real numbers \( p \), \( q \), and \( r \) such that \[ \overrightarrow T_x=\langle p,q,r\rangle.\] First of all \( q \) is easy to find. It is equal to \( 0 \). The coordinates \( p \) and \( r \) are obtained in the same way as in single-variable calculus. We may take \( p=1 \) and \( r=f_x(x_0,y_0) \). Therefore we have \[ \overrightarrow T_x(x_0,y_0)=\langle 1,0,f_x(x_0,y_0)\rangle.\] Similarly, we obtain another vector that belongs to the tangent plane: \[ \overrightarrow T_y(x_0,y_0)=\langle 0,1,f_y(x_0,y_0)\rangle.\] And now we use a magnificent trick: Normal vector can be obtained as a cross product of \( \overrightarrow T_x \) and \( \overrightarrow T_y \). Thus:

\[ \overrightarrow n(x_0,y_0)= \overrightarrow T_x\times\overrightarrow T_y=\left|\begin{array}{ccc} \overrightarrow i& \overrightarrow j & \overrightarrow k \\ 1 & 0 & f_x \\ 0 & 1 & f_y \end{array}\right|=\langle -f_x,-f_y,1\rangle. \] We have obtained the following theorem:

The Equation of the Tangent Plane
 
If a surface is defined as a graph of \( z=f(x,y) \), and if \( A(x_0,y_0,z_0) \) is a point on the surface, then a normal vector to the tangent plane at the point \( A \) is given by \[ \overrightarrow n(x_0,y_0)=\langle -f_x(x_0,y_0), -f_y(x_0,y_0),1\rangle.\] The equation of the plane tangent to the surface at the point \( A \) is given by: \[ -f_x(x_0,y_0)(x-x_0)-f_y(x_0,y_0)(y-y_0)+(z-z_0)=0.\]

Example 1
 
Consider the function: \[ f(x,y)=3-x^2-y^2-3x^3y+2xy.\] Find the equation of the tangent plane to the surface at the point \( M(0,1,2) \).

Example 2
 
Find the equation of the tangent plane to the ellipsoid \( x^2+4y^2+9z^2=1 \) at the point \( \left(\frac12,\frac14,\frac{\sqrt 2}3\right) \).

Tangent planes to surfaces given by parametric equations

Assume that the surface \( S \) is given by its parametric equations: \begin{eqnarray*} x&=& X(s,t)\\ y&=& Y(s,t)\\ z&=&Z(s,t), \end{eqnarray*} where \( X \), \( Y \), and \( Z \) are differentiable functions of two variables. Assume that we want to find the tangent plane to the surface \( S \) at the point \( (x_0,y_0,z_0) \). Since each point of \( S \) is obtained when a particular pair \( (s,t) \) is plugged into the equations above, we may assume that there exists a pair \( (x_0,y_0) \) such that \( (x_0,y_0,z_0)=(X(s_0,t_0),Y(s_0,t_0),Z(s_0,t_0)) \). Consider the function \( \overrightarrow R(s,t)=\langle X(s,t), Y(s,t), Z(s,t)\rangle \).

Theorem
 
The normal vector to the tangent plane of \( S \) at the point \( (x_0,y_0,z_0) \) is parallel to the vector \[ \overrightarrow N_0= \frac{\partial }{\partial s}\overrightarrow R(s_0,t_0)\times \frac{\partial }{\partial t}\overrightarrow R(s_0,t_0).\]

We will now solve the same problem as in Example 2 by parametrizing the ellipsoid.

Example 3
 
Find the equation of the plane tangent to the ellipsoid \( x^2+4y^2+9z^2=1 \) at the point \( \left(\frac12,\frac14,\frac{\sqrt 2}3\right) \).


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