An equation of a tangent plane

Functions of several variables: Tangent vector and tangent plane

An equation of a tangent plane

Interlude: an equation of a plane We will determine an equation of a tangent plane. This requires that you know how to determine an equation of an arbitrary plane in a three-dimensional space when you know two direction vectors of the plane. In general, the equation of a plane is of the from

$ax+by+cz+d=0$ with $a,b,c$ numbers that are not all equal to $0$ , and a fourth number $d$ . Note that the numbers $a$ , $b$ , $c$ and $d$ are not unique: each multiple of the quartet describes the same plane in the three-dimensional space. So we should really speak of an equation of a plane. The plane with equation $ax+by+cz+d=0$ is parallel to the plane through the origin given by the equation

$ax+by+cz=0$ This equation can also be written as inner product of two vectors, namely,

$\left(\begin{array}{c} a \\ b \\ c\end{array}\right)\mathbf{\cdot} \left(\begin{array}{c} x\\ y\\ z\end{array}\right) = 0$ In other words, the vector starting at the origin and ending at $(x,y,z)$ points toward a point on the plane through the origin if the vector is perpendicular to the vector with components $a$ , $b$ and $c$ . In other words, the vector $\left(\begin{array}{c} a \\ b \\ c\end{array} \right)$ is orthogonal to any vector in the plane. This is a normal vector of the plane. Given two vectors $\mathbf {a} = \left( \begin{array}{c}a_1 \\ a_2 \\ a_3\end {array} \right)$ , and $\mathbf{b} = \left(\begin{array}{c}b_1 \\ b_2 \\ b_3\end {array} \right)$ that do not lie in line with each other in a plane through the origin, an admissible normal vector is the cross product $\mathbf{a}\times \mathbf{b}$ of $\mathbf{a}$ and $\mathbf{b}$ (in this order!), defined as

$\mathbf{a}\times \mathbf{b}=\left(\begin{array}{c} a_2b_3-a_3b_2 \\ a_3b_1-a_1b_3 \\ a_1b_2-a_2b_1 \end{array}\right)$

After the above interlude we return to the subject of functions of two variables: let $z=f(x,y)$ be a function of two variables $x$ and $y$ . We consider a point $P=(a,b,c)$ with $c=f(a,b)$ and assume that the graph of $f$ is a smooth surface near point $P$ . We know that the vectors

$\mathbf{r}=\left(\begin{array}{c} 1 \\ 0 \\ f_x(a,b)\end{array}\right)\quad \text{and}\quad \mathbf{s}=\left(\begin{array}{c} 0 \\ 1 \\ f_y(a,b)\end{array}\right)$ span the tangent plane at $P$ . To find an equation of the tangent plane it is sufficient to find a normal vector of this plane. This can be done via the cross product of $\mathbf{r}$ and $\mathbf{s}$ :

$\mathbf{n}=\mathbf{r}\times \mathbf{s}= \left(\begin{array}{c} -f_x(a,b) \\ -f_y(a,b) \\ 1\end{array}\right)$ In this way we find the following equation of the tangent plane.

An equation of the tangent plane at the point $P=(a,b,c)$ with $c=f(a,b)$ is

$-f_x(a,b)(x-a)-f_y(a,b)(y-b)+(z-c)=0$ and thus

$z=f(a,b) + f_x(a,b)(x-a)+f_y(a,b)(y-b)$

Find an equation of the tangent plane at the point $P=\bigl(0.3, 0.5, f(0.3, 0.5)\bigr)$ on the graph of the function

$f(x,y)=\tfrac{1}{2}\!(1-\sin(2x^2-y-1)$

The graph of the function $f(x,y)=\tfrac{1}{2}\!(1-\sin(2x^2-y-1)$ is a curved surface. In the figure below, a wire frame of coordinate curves is drawn for $x=0.0, 0.1, \ldots , 0.9, 1.0$ and for $y=0.0, 0.1, \ldots , 0.9, 1.0$ , and the tangent plane at the point $P=\bigl(0.3, 0.5, f(0.3, 0.5)\bigr)$ and a normal vector $\mathbf{n}$ are sketched.

normal vector

Rounded to five decimal places, we have:

$\begin{aligned}P &= (0.3, 0.5, 0.98436)\\ f_x(0.3, 0.5) &= -0.14891\\ f_y(0.3, 0.5) &= 0.12409\end{aligned}$ The normal vector $\mathbf{n}$ shown is computed as a cross product of two direction vectors and is

$\mathbf{n}=\left(\begin{array}{c} \phantom{-}0.14891 \\ -0.12409 \\ 1\end{array}\right)$ An equation of the tangent plane is

$z = 0.96699-0.14891x+0.12409y$

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