Math, asked by prafull212000, 7 months ago

A unit normal to the surface z = - 2xy at the point (2,1, 4) is​

Answers

Answered by lipun707763
6

A unit normal to the surface ans 10

Answered by anjumanyasmin
0

Given:

z = - 2xy

at the point (2,1, 4)

\phi=z + 2xy

\frac{\partial \phi}{\partial x}=\frac{\partial\left(z+2xy\right)}{\partial x}=2y\\\\\frac{\partial \phi}{\partial y}=\frac{\partial\left(z+2xy\right)}{\partial x}=2x\\\\\frac{\partial \phi}{\partial z}=\frac{\partial\left(z+2xy\right)}{\partial x}=1

\nabla \phi=\frac{\partial \phi}{\partial \mathrm{x}} \hat{\mathrm{i}}+\frac{\partial \phi}{\partial \mathrm{y}} \hat{\mathrm{j}}+\frac{\partial \phi}{\partial z} \hat{\mathrm{k}}=(2y) \hat{\mathrm{i}}+\mathrm2x \hat{\mathrm{j}}+1 \mathrm{} \hat{\mathrm{k}}

\text { At point }(2,1,4)

\nabla \phi=2\times1 \hat{\mathrm{i}}+2\times 2 \hat{\mathrm{j}}+1 \times 4\hat{\mathrm{k}}\\\nabla \phi=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+4\hat{\mathrm{k}}

|\nabla \phi|=\sqrt{(2)^{2}+(4)^{2}+(4)^{2}}

=\sqrt{36}=6

\text { Unit normal vector }=\hat{\mathrm{n}}=\frac{\nabla \phi}{|\nabla \phi|}=\frac{2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}}{6}=\frac{1}{3}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})

Hence the answer is \frac{1}{3}(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})

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