# Find the equation for the plane tangent to each surface

Find the equation for the plane tangent to each surface $z=f\left(x,y\right)$ at the indicated point.
$\left(a\right)z={x}^{3}+{y}^{3}-2xy,\text{at the point}\left(1,2,-5\right)$
$\left(b\right)z=\left(\mathrm{cos}\left(x\right)\right)\left(\mathrm{cos}\left(y\right)\right),\text{at the point}\left(0,\frac{\pi }{2},0\right)$
$\left(c\right)z=\left(\mathrm{cos}\left(x\right)\right)\left(\mathrm{sin}\left(y\right)\right),\text{at the point}\left(0,\frac{\pi }{2},1\right)$

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Jeremy Merritt

Part (a): We have:
Surface $z={x}^{3}+{y}^{3}-2xy,\text{at the point}\left(1,2,-5\right)$
$\left({x}_{0},{y}_{0},{z}_{0}\right)=\left(1,2,-5\right)$
Used formula in equation:
An equation of the tangent plane to the surface $z=f\left(x,y\right)$ at the point $P\left({x}_{0},{y}_{0},{z}_{0}\right)$ is:
$z-{z}_{0}={f}_{x}\left({x}_{0},{y}_{0}\right)\left(x-{x}_{0}\right)+{f}_{y}\left({x}_{0},{y}_{0}\right)\left(y-{y}_{0}\right)$
Differentiating partially equation of surface with respect to x and y:

${f}_{x}=\frac{\partial z}{\partial x}=3{x}^{2}-2y$

$\frac{\partial z}{\partial x}=3{\left(1\right)}^{2}-2\left(2\right)=-1$

${f}_{y}=\frac{\partial z}{\partial x}=3{y}^{2}-2x$

$\frac{\partial z}{\partial x}=3{\left(2\right)}^{2}-2\left(1\right)=10$

Now, put the valuesin formula:
$z-{z}_{0}={f}_{x}\left({x}_{0},{y}_{0}\right)\left(x-{x}_{0}\right)+{f}_{y}\left({x}_{0},{y}_{0}\right)\left(y-{y}_{0}\right)$
$z-\left(-5\right)=\left(-1\right)\left(x-1\right)+\left(10\right)\left(y-2\right)$
$x-10y+z+24=0$

temnimam2

Part (b):Given:
Surface $z=\mathrm{cos}x×\mathrm{cos}y,\text{at the point}\left(\begin{array}{c}0,\frac{\pi }{2},0\end{array}\right)\left({x}_{0},{y}_{0},{z}_{0}\right)=\left(\begin{array}{c}0,\frac{\pi }{2},0\end{array}\right)$
Differentiating partially equation of surface with respect to x and y:
${f}_{x}=\frac{\partial z}{\partial x}=\left(-\mathrm{sin}x\right)\left(\mathrm{cos}y\right)$
$\frac{\partial z}{\partial x}=\left(-\mathrm{sin}0\right)\left(\mathrm{cos}\frac{\pi }{2}\right)=0$
${f}_{y}=\frac{\partial z}{\partial x}=\left(\mathrm{cos}x\right)\left(-\mathrm{sin}y\right)$
$\frac{\partial z}{\partial x}=\left(\mathrm{cos}0\right)\left(-\mathrm{sin}\mid \frac{\pi }{2}\right)=-1$
Now, put the values in formula:
$z-{z}_{0}={f}_{x}\left({x}_{0},{y}_{0}\right)\left(x-{x}_{0}\right)+{f}_{y}\left({x}_{0},{y}_{0}\right)$
$z-\left(0\right)=\left(0\right)\left(x-0\right)+\left(-1\right)\left(y-\frac{\pi }{2}\right)$
$y+z\frac{\pi }{2}=0$

nick1337

Part (c):Given:
Surface
$\left({x}_{0},{y}_{0},{z}_{0}\right)=\left(0,\frac{\pi }{2},1\right)$
Differentiating partially equation of surface with respect to x and y:
${f}_{x}=\frac{\partial z}{\partial x}=\left(-\mathrm{sin}x\right)\left(\mathrm{sin}y\right)$
$\frac{\partial z}{\partial x}=\left(-\mathrm{sin}0\right)\left(\mathrm{sin}\frac{\pi }{2}\right)=0$
${f}_{y}\frac{\partial z}{\partial x}=\left(\mathrm{cos}x\right)\left(\mathrm{cos}y\right)$
$\frac{\partial z}{\partial x}=\left(\mathrm{cos}0\right)\left(\mathrm{cos}\frac{\pi }{2}\right)=0$
Now, put the values in formula:
$z-{z}_{0}={f}_{x}\left({x}_{0},{y}_{0}\right)\left({x}_{0},{y}_{0}\right)+{f}_{y}\left({x}_{0},{y}_{0}\right)\left(y-{y}_{0}\right)$
$z-\left(1\right)=\left(0\right)\left(x-0\right)+\left(0\right)\left(y-\frac{\pi }{2}\right)$
$z-1=0$