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Alja [10]
3 years ago
10

Find the rate of change of f(x,y,z)=xyzf(x,y,z)=xyz in the direction normal to the surface yx2+xy2+yz2=120yx2+xy2+yz2=120 at (3,

4,3)(3,4,3). (Use symbolic notation and fractions where needed.) Rate of change =
Mathematics
1 answer:
Anon25 [30]3 years ago
6 0

Answer:

Rate of change of function in the direction of normal to the given surface at ( 3 , 4 , 3 )  is \frac{573}{\sqrt{985}}

Step-by-step explanation:

Given:

Function, f( x , y , z ) = xyz

Equation of surface, yx² + xy² + yz² = 120

To find: Rate of change of function in the direction of normal to the given surface at ( 3 , 4 , 3 )  

The Gradient of the normal to the surface

(\bigtriangledown_x\:,\:\bigtriangledown_y\:,\:\bigtriangledown_z)

\implies\:(2xy+y^2+0\:,\:x^2+2xy+z^2\:,\:0+0+2zy)

\implies\:(2xy+y^2\:,\:x^2+2xy+z^2\:,\:2zy)

Gradient at ( 3 , 4 , 3 ) =\:(2(3)(4)+(4)^2\:,\:(3)^2+2(3)(4)+(3)^2\:,\:2(4)(3))

\implies\:(40\:,\:42\:,\:24)

The Change in the directional derivative of f in given direction is,

\bigtriangledown f_{(3,4,3)}.\frac{(40,42,24)}{\sqrt{40^2+42^2+24^2}}=(yz,xz,xy)_{(3,4,3)}.\frac{(40,42,24)}{\sqrt{1600+1764+576}}=((4)(3),(3)(3),(3)(4)).\frac{(40,42,24)}{\sqrt{3940}}

=\frac{(12,9,12).(40,42,24)}{\sqrt{3940}}=\frac{480+378+288}{\sqrt{3940}}=\frac{1146}{2\sqrt{985}}=\frac{573}{\sqrt{985}}

Therefore, Rate of change of function in the direction of normal to the given surface at ( 3 , 4 , 3 )  is \frac{573}{\sqrt{985}}

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