Chapter 4: Partial Differentiation

Section 4.6: Surface Normal and Tangent Plane

Example 4.6.3

$$ Derive the form of N for the surface given explicitly by $z\=f\left(x\,y\right)$. (See Table 4.6.1.) $$ Solution $$ •  The coordinate curves ${\mathbf{R}}_1\=\left[\begin{array}{c}x\\ b\\ f\(x\,b\)\end{array}\right]$ and ${\mathbf{R}}_2\=\left[\begin{array}{c}a\\ y\\ f\(a\,y\)\end{array}\right]$ project onto the grid lines $y\=b$ and $x\=a$, respectively.   •  Tangents to these curves, namely, ${\mathbf{T}}_1\=\left[\begin{array}{c}1\\ 0\\ f_x\end{array}\right]$ and ${\mathbf{T}}_2\=\left[\begin{array}{c}0\\ 1\\ f_y\end{array}\right]$, are distinct vectors tangent to the surface.   •  Their cross product   ${\mathbf{T}}_1\times {\mathbf{T}}_2\=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& 0& f_x\\ 0& 1& f_y\end{array}\right|$ = $\left[\begin{array}{c}-f_x\\ -f_y\\ 1\end{array}\right]$    is then orthogonal to the surface at the point $\left(a\,b\,f\left(a\,b\right)\right)$, that is, it is the normal N listed in Table 4.6.1. $$

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