Question

Find an equation of the tangent plane to the given

Negative numbers and coordinate plane
ANSWERED
asked 2021-08-17
Find an equation of the tangent plane to the given surface at the specified point.
\(\displaystyle{z}={2}{\left({x}-{1}\right)}^{{2}}+{6}{\left({y}+{3}\right)}^{{2}}+{4},\ {\left({3},-{2},{18}\right)}\)

Expert Answers (1)

2021-08-18
Find an equation of the tangent plane to the given surface at the specified point.
\(\displaystyle{z}={2}{\left({x}-{1}\right)}^{{2}}+{6}{\left({y}+{3}\right)}^{{2}}+{4},{\left({3},-{2},{18}\right)}\)
\(\displaystyle\Rightarrow{z}_{{x}}={4}{\left({x}-{1}\right)}\)
\(\displaystyle={4}{x}-{4}\),
\(\displaystyle{z}_{{y}}={12}{\left({y}+{3}\right)}\)
\(\displaystyle={12}{y}+{36}\)
At the point (3,-2,18):
\(\displaystyle{z}_{{x}}={4}{x}-{4}\)
\(\displaystyle={4}{\left({3}\right)}-{4}\)
\(\displaystyle={12}-{4}\)
\(\displaystyle={8}\)
\(\displaystyle{z}_{{y}}={12}{y}+{36}\)
\(\displaystyle=-{12}{\left(-{2}\right)}+{36}\)
\(\displaystyle=-{24}+{36}\)
\(\displaystyle={12}\)
Equation of the tangent plane is,
\(\displaystyle{z}-{z}_{{1}}={z}_{{x}}{\left({x}-{x}_{{1}}\right)}+{z}_{{y}}{\left({y}-{y}_{{1}}\right)}\)
\(\displaystyle{z}-{18}={8}{\left({x}-{3}\right)}+{12}{\left({y}-{\left(-{2}\right)}\right)}\)
\(\displaystyle{z}-{18}={8}{x}+{12}{y}\)
\(\displaystyle{z}={8}{x}+{12}{y}+{18}\)
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