Math, asked by ashaalka3333, 4 months ago

Find the rate of change of o = xyz in the direction normal to the surface xy + y'x + yz = 3 at the point (1, 1, 1).

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Answered by aman52380
12

Answer:

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Answered by prachikalantri
1

Given- xy + y'x + yz = 3 at the point (1, 1, 1).

​Find the rate of change

Rate of change of f(x,y,z) in direction normal to \frac{yx^2+xy^2+yz^2}{g(x,y,z)}=3

at (1,1,1).

\hat n unit normal vector

\hat n=\frac{\bigtriangledown g }{|\bigtriangledown g|}

{\bigtriangledown g }=(2xy+y^2)\hat i+(2xy+x^2+z^2) \hat j +2y=\hat k

{\bigtriangledown g } =\frac{\delta g}{\delta x}\hat i, \frac{\delta g}{\delta y} \hat j+\frac{\delta g}{\delta 2} \hat k (for clarity)

\bigtriangledown g at (1, 1, 1)=3\hat i+4\hat j+2\hat k

\hat n=\frac{3\hat i+4\hat j+2\hat k}{\sqrt{29} }

f(x,y,z)=xyz

\frac{rate of change along \hat n }{\bigtriangledown f,\hat n}

=(yz\hat i+xy \hat k+xz \hat j).(\frac{3\hat i+4\hat j+2\hat k}{\sqrt{29} } )

=\frac{3yx+2xy+4xz}{\sqrt{29} } units

Since (x,y,z)=(1,1,1)

Rate of change =\frac{9}{\sqrt{29} } units

#SPJ2

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