If Ax+By+Gz=D the equation of the tangent plane of the surface xy+yz+zx=3 at the point (1,1,1) then |A|+|B|+|G|=?
Any ideas how this prolem can be solved???
Hey rikelda91.
Hint: How do you calculate the derivatives with respect to a specific direction vector (think x, y, and z axes)? Given this what is the relationship to the tangent plane (i.e. how do you go from finding the partial derivatives for x,y,z at point (1,1,1) to tangent plane)?
$\displaystyle \nabla f$ is perpedicular to the surface f(x, y, z)= constant and so perpendicular to the tangent plane at any point. If <A, B, C> is tangent to a plane at point $\displaystyle (x_0, y_0, z_0)$ then the plane is given by $\displaystyle A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$.
Consider the surface given by z=f(xy). Let (x0y0z0) be any point on this surface. If f(xy) is differentiable at (x0y0), then the surface has a tangent plane at (x0y0z0). The equation of the tangent plane at (x0y0z0) is given by
fx(x0y0)(x−x0)+fy(x0y0)(y−y0)−(z−z0)=0