Question

) If the gradient of ff is ∇f=xi⃗ +yj⃗ +3zyk⃗ ∇f=xi→+yj→+3zyk→ and the point P=(−6,−6,−2)P=(−6,−6,−2) lies on the level surface f(x,y,z)=0f(x,y,z)=0, find an equation for the tangent plane to the surface at the point PP.

Answer 1

The equation for the tangent plane to the surface at the point P is

x + 3y + z + 38 = 0

Given the function f expressed as $g=x_i + y_j + 3zy_k$ passing through the point P(6, -6, -2). The equation for the tangent plane to the surface at the point P is expressed according to the expression:

$gx_{(x_0, y_0, z_0)}(x-x_0) + gy_{(x_0, y_0, z_0)}(y-y_0) + gz_{(x_0, y_0, z_0)} (z-z_0)= 0$

Substituting the differential function gx, gy, and gz and the point P (-6, -6, -2) into the formula, we will have;

$-6(x-(-6))+(-6)(y-(-6)) + -2(x-(-2))=0\\-6(x+6)-6(y+6)-2(x+2)=0$

Expand the result to have:

$-6x-36-6y-36-2z-4=0\\-6x-6y-2z-76=0$

Divide through by -2 to have:

$3x + 3y+z+38=0$

Hence the equation for the tangent plane to the surface at the point P is

3x + 3y + z + 38 = 0

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