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

Answer:

-3x-3y-z=38

Step-by-step explanation:

To find the equation of the tangent plane you can use

$\bigtriangledown f_x(x_0,y_0,z_0)(x-x_0)+\bigtriangledown f_y(x_0,y_0,z_0)(y-y_0)+\bigtriangledown f_z(x_0,y_0,z_0)(z-z_0)=0$

the point P is (-6,-6,-2). Hence you have

$(-6)(x-(-6))+(-6)(y-(-6))+(-2)(z-(-2))=0\\-6x-36-6y-36-2z-4=0\\-6x-6y-2z=76\\-3x-3y-z=38$

hope this helps!!