13.2 Tangent Planes

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Earlier (Chapter 4), we saw the role of the tangent line in locating local maxima and local minima of a differentiable function $y=f(x).$ For a surface, $z=f(x,y)$, the tangent plane plays a similar role.

Now suppose the surface $S$ is defined implicitly by the equation $F(x,y,z)=0.$ Then $S$ is equivalent to a level surface of the function $w=F(x,y,z) .$ Let $P_{0}=(x_{0},y_{0},z_{0})$ be a point on the surface $S$. Suppose $F$ is differentiable at $P_{0}$ and ${\boldsymbol \nabla} F(x_{0},y_{0},z_{0})\neq \mathbf{0}.$ Then the gradient ${\boldsymbol \nabla} F(x_{0},y_{0},z_{0})$ is normal to the level surface that contains the point $P_{0}.$ That is, for any smooth curve $C$ on the surface $S$ that contains the point $P_{0}$, the tangent lines to $C$ at $P_{0}$ are orthogonal to the vector ${\boldsymbol \nabla}F(x_{0},y_{0},z_{0}).$ As a result, the tangent lines to $C$ at the point $P_{0}$ all lie in the same plane, and the surface $S$ has a tangent plane at $P_{0}$.

See Figure 10. Suppose $(x,y,z)$ is any point on the tangent plane to $S$ at the point $P_{0}=(x_{0},y_{0},z_{0})$. Since ${\boldsymbol \nabla }F(x_{0},y_{0},z_{0})$ is normal to the tangent plane at $P_{0}$, an equation of the tangent plane is $${\boldsymbol \nabla }F(x_{0},y_{0},z_{0})\,{\bf\cdot}\, \lbrack (x-x_{0})\mathbf{i} +(y-y_{0})\mathbf{j}+(z-z_{0})\mathbf{k}]=0$$

where $(x-x_{0})\mathbf{i}+(y-y_{0})\mathbf{j}+(z-z_{0})\mathbf{k}$ is a vector in the direction of the tangent line to a smooth curve $C$ at the point $P_{0}.$ Since $${\boldsymbol \nabla}F(x_{0},y_{0},z_{0})=F_{x}(x_{0},y_{0},z_{0})\mathbf{i} +F_{y}(x_{0},y_{0},z_{0})\mathbf{j}+F_{z}(x_{0},y_{0},z_{0})\mathbf{k}$$

an equation of the tangent plane is $$F_{x}(x_{0},y_{0},z_{0})(x-x_{0})+F_{y}(x_{0},y_{0},z_{0})(y-y_{0})+F_{z}(x_{0},y_{0},z_{0})(z-z_{0})=0$$

1 Find a Tangent Plane to a Surface

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Figure 11 shows the hyperboloid and part of the tangent plane at the point $(3,-4,1)$.

2 Find a Normal Line to a Tangent Plane

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The line normal to the tangent plane to a surface $F(x,y,z)=0$ at the point $P_{0}=(x_{0},y_{0},z_{0})$ is called the normal line to the surface at $P_{0}.$ Since the gradient ${\boldsymbol \nabla }F(x_{0},y_{0},z_{0})$ is normal to the tangent plane at $P_{0}$, it follows that the normal line is in the direction of the gradient. The vector equation of the normal line is $${\bbox[#FAF8ED]{\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{r}(t)=\mathbf{r}_{0}+t{\boldsymbol \nabla }F(x_{0},y_{0},z_{0}) }}}$$

where $\mathbf{r}_{0}=x_{0}\mathbf{i}+y_{0}\mathbf{j}+z_{0}\mathbf{k}$ is the position vector of $P_{0}$ and $\mathbf{r}$ is the position vector of any point $P$ on the normal line.

The corresponding parametric equations of the normal line are $${\bbox[#FAF8ED]{\bbox[5px, border:1px solid black, #F9F7ED]{ x=x_{0}+at\qquad y=y_{0}+bt\qquad z=z_{0}+ct }}}$$

where $a=F_{x}(x_{0},y_{0},z_{0})$, $b=F_{y}(x_{0},y_{0},z_{0})$, and $c=F_{z}(x_{0},y_{0},z_{0})$. If $abc\neq 0$, symmetric equations of the normal line are $${\bbox[#FAF8ED]{\bbox[5px, border:1px solid black, #F9F7ED]{ \dfrac{x-x_{0}}{a}=\dfrac{y-y_{0}}{b}=\dfrac{z-z_{0}}{c} }}}$$

Surfaces Defined Explicitly by $z=f( x,y)$

To find an equation of the tangent plane to the surface $z=f(x,y)$ at $(x_{0},y_{0},z_{0})$, $z_{0}=f(x_{0},y_{0}),$ we write the equation of the surface implicitly as $$F(x,y,z)=z-f(x,y)=0$$

Then $$F_{x}=-f_{x}\qquad F_{y}=-f_{y}\qquad F_{z}=1$$

An equation of the tangent plane to $z=f(x,y)$ at $(x_{0},y_{0},z_{0})$ is $$\begin{eqnarray*} &&F_{x}( x_{0},y_{0},z_{0}) (x-x_{0}) +F_{y}(x_{0},y_{0},z_{0}) ( y-y_{0}) +F_{z}(x_{0},y_{0},z_{0}) ( z-z_{0}) =0 \\[4pt] &&\hspace{42pt}-f_{x}(x_{0},y_{0})(x-x_{0})-f_{y}(x_{0},y_{0})(y-y_{0})+(z-z_{0}) =0 \end{eqnarray*}$$ $${\bbox[#FAF8ED]{\bbox[5px, border:1px solid black, #F9F7ED]{ z-z_{0}=f_{x}(x_{0},y_{0})(x-x_{0})+f_{y}(x_{0},y_{0})(y-y_{0}) }}}$$

where $z_{0}=f( x_{0},y_{0}).$ The parametric equations of the normal line to $z=f(x,y)$ at the point $(x_{0},y_{0},z_{0})$ are $${\bbox[#FAF8ED]{\bbox[5px, border:1px solid black, #F9F7ED]{ x=x(t)=x_{0}+tf_{x}(x_{0},y_{0})\qquad y=y(t)=y_{0}+tf_{y}(x_{0},y_{0})\qquad z=z(t)=z_{0}-t }}}$$