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Rufina [12.5K]
2 years ago
10

Help plss this is due in an hour i’ll give brainliest

Mathematics
1 answer:
scZoUnD [109]2 years ago
4 0
Yss the answer is c
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Rewrite the expression in the form 9^n
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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)

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3 years ago
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