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Find the differential $dz$ of the function $f\left(x,y\right)=4{y}^{2}+{x}^{2}y-2xy$ and use it to approximate $\text{\Delta}z$ at point $\left(1,\mathrm{-1}\right).$ Use $\text{\Delta}x=0.03$ and $\text{\Delta}y=\mathrm{-0.02}.$ What is the exact value of $\text{\Delta}z?$
$\begin{array}{ccc}\hfill dz& =\hfill & 0.18\hfill \\ \hfill \text{\Delta}z& =\hfill & f\left(1.03,\mathrm{-1.02}\right)-f\left(1,\mathrm{-1}\right)=0.180682\hfill \end{array}$
All of the preceding results for differentiability of functions of two variables can be generalized to functions of three variables. First, the definition:
A function $f\left(x,y,z\right)$ is differentiable at a point $P\left({x}_{0},{y}_{0},{z}_{0}\right)$ if for all points $\left(x,y,z\right)$ in a $\delta $ disk around $P$ we can write
where the error term E satisfies
If a function of three variables is differentiable at a point $\left({x}_{0},{y}_{0},{z}_{0}\right),$ then it is continuous there. Furthermore, continuity of first partial derivatives at that point guarantees differentiability.
For the following exercises, find a unit normal vector to the surface at the indicated point.
$f(x,y)={x}^{3},(2,\mathrm{-1},8)$
$\left(\frac{\sqrt{145}}{145}\right)\left(12i-k\right)$
$\text{ln}\left(\frac{x}{y-z}\right)=0$ when $x=y=1$
For the following exercises, as a useful review for techniques used in this section, find a normal vector and a tangent vector at point $P.$
${x}^{2}+xy+{y}^{2}=3,P(\mathrm{-1},\mathrm{-1})$
Normal vector: $i+j,$ tangent vector: $i-j$
${\left({x}^{2}+{y}^{2}\right)}^{2}=9\left({x}^{2}-{y}^{2}\right),P(\sqrt{2},1)$
$x{y}^{2}-2{x}^{2}+y+5x=6,P(4,2)$
Normal vector: $7i-17j,$ tangent vector: $17i+7j$
$2{x}^{3}-{x}^{2}{y}^{2}=3x-y-7,P(1,\mathrm{-2})$
$z{e}^{{x}^{2}-{y}^{2}}-3=0,$ $P(2,2,3)$
$\mathrm{-1.094}i-0.18238j$
For the following exercises, find the equation for the tangent plane to the surface at the indicated point. ( Hint: Solve for $z$ in terms of $x$ and $y.)$
$\mathrm{-8}x-3y-7z=\mathrm{-19},P(1,\mathrm{-1},2)$
$z=\mathrm{-9}{x}^{2}-3{y}^{2},P(2,1,\mathrm{-39})$
$\mathrm{-36}x-6y-z=\mathrm{-39}$
${x}^{2}+10xyz+{y}^{2}+8{z}^{2}=0,P(\mathrm{-1},\mathrm{-1},\mathrm{-1})$
$z={e}^{7{x}^{2}+4{y}^{2}},$ $P(0,0,1)$
${x}^{2}+4{y}^{2}={z}^{2},P(3,2,5)$
${x}^{3}+{y}^{3}=3xyz,P\left(1,2,\frac{3}{2}\right)$
$4x-5y+4z=0$
$z=axy,P\left(1,\frac{1}{a},1\right)$
$z=\text{sin}\phantom{\rule{0.2em}{0ex}}x+\text{sin}\phantom{\rule{0.2em}{0ex}}y+\text{sin}(x+y),P(0,0,0)$
$2x+2y-z=0$
$h(x,y)=\text{ln}\sqrt{{x}^{2}+{y}^{2}},P(3,4)$
$z={x}^{2}-2xy+{y}^{2},P(1,2,1)$
$\mathrm{-2}\left(x-1\right)+2\left(y-2\right)-\left(z-1\right)=0$
For the following exercises, find parametric equations for the normal line to the surface at the indicated point. (Recall that to find the equation of a line in space, you need a point on the line, ${P}_{0}\left({x}_{0,}{y}_{0},{z}_{0}\right),$ and a vector $n=\u27e8a,b,c\u27e9$ that is parallel to the line. Then the equation of the line is $x-{x}_{0}=at,y-{y}_{0}=bt,z-{z}_{0}=ct.)$
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