3 years ago

Help plz:))) I’ll mark u brainliest 10 ez POINTS

Mathematics

1 answer:

Komok [63]3 years ago

7 0

I think is 24, 27, 2.
I hope is correct.

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Find equations of the tangent plane and the normal line to the given surface at the specified point. x + y + z = 8exyz, (0, 0, 8

Dima020 [189]

Let $f(x,y,z)=x+y+z-8e^{xyz}$ . The tangent plane to the surface at (0, 0, 8) is

$\nabla f(0,0,8)\cdot(x,y,z-8)=0$

The gradient is

$\nabla f(x,y,z)=\left(1-8yze^{xyz},1-8xze^{xyz},1-8xye^{xyz}\right)$

so the tangent plane's equation is

$(1,1,1)\cdot(x,y,z-8)=0\implies x+y+(z-8)=0\implies x+y+z=8$

The normal vector to the plane at (0, 0, 8) is the same as the gradient of the surface at this point, (1, 1, 1). We can get all points along the line containing this vector by scaling the vector by $t$ , then ensure it passes through (0, 0, 8) by translating the line so that it does. Then the line has parametric equation

$(1,1,1)t+(0,0,8)=(t,t,t+8)$

or $x(t)=t$ , $y(t)=t$ , and $z(t)=t+8$ .

(See the attached plot; the given surface is orange, (0, 0, 8) is the black point, the tangent plane is blue, and the red line is the normal at this point)