What is the Tangent Plane at a Given Point on a Level Surface?

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SUMMARY

The discussion focuses on determining the tangent plane at a specific point on a level surface defined by the equation f(x, y, z) = 0. The normal vector to the surface is given by the gradient ∇f = f_x i + f_y j + f_z k. The equation for the tangent plane at the point (x_0, y_0, z_0) is expressed as f_x(x_0, y_0, z_0)(x - x_0) + f_y(x_0, y_0, z_0)(y - y_0) + f_z(x_0, y_0, z_0)(z - z_0) = 0. The key question raised is identifying the conditions under which the point (0, 0, 0) lies on this tangent plane.

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  • Knowledge of tangent planes in the context of differential geometry.
  • Ability to solve equations involving partial derivatives.
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  • Study the properties of level surfaces in multivariable calculus.
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f(x, y, z) = 0 f(x, y, z) = ...

Homework Statement



[PLAIN]http://www.netbookolik.com/wp-content/uploads/2010/07/q12.png

Homework Equations





The Attempt at a Solution



Sorry have no idea:(
 
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If a surface is given by f(x,y,z)= 0 (or any constant) we can think of it as a "level surface" so \nabla f= f_x\vec{i}+ f_y\vec{j}+ f_z\vec{k} is normal to the surface. The tangent plane at (x_0,y_0,z_0) is of the form f_x(x_0,y_0,z_0)(x- x_0)+f_y(x_0,y_0,z_0)(y-y_0)+ f_z(x_0,y_0, z_0)(z- z_0)= 0.

For what (x_0, y_0, z_0) does (0, 0, 0) lie on that tangent plane?
 

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