Find the normal vector of the tangent plane to x^2yz+3y^2-2xz^2+8z=0 x2yz+3y2−2xz2+8z=0 at (1,2,-1)(1,2,−1) ?
1 Answer
-6hat(i) +12hat(j)+14hat(k) −6ˆi+12ˆj+14ˆk
Explanation:
First we rearrange the equation of the surface into the form
x^2yz+3y^2-2xz^2+8z=0 x2yz+3y2−2xz2+8z=0
And so we define our surface function,
f(x,y,z) = x^2yz+3y^2-2xz^2+8z f(x,y,z)=x2yz+3y2−2xz2+8z
In order to find the normal at any particular point in vector space we use the Del, or gradient operator:
grad f(x,y,z) = (partial f)/(partial x) hat(i) + (partial f)/(partial y) hat(j) + (partial f)/(partial z) hat(k) ∇f(x,y,z)=∂f∂xˆi+∂f∂yˆj+∂f∂zˆk
remember when partially differentiating that we differentiate wrt the variable in question whilst treating the other variables as constant. And so:
grad f = ((partial)/(partial x) (x^2yz+3y^2-2xz^2+8z))hat(i) + ∇f=(∂∂x(x2yz+3y2−2xz2+8z))ˆi+
" " ((partial)/(partial y) (x^2yz+3y^2-2xz^2+8z))hat(j) + (∂∂y(x2yz+3y2−2xz2+8z))ˆj+
" " ((partial)/(partial z) (x^2yz+3y^2-2xz^2+8z))hat(k) (∂∂z(x2yz+3y2−2xz2+8z))ˆk
" "= (2xyz-2z^2)hat(i) + (x^2z+6y)hat(j) + (x^2y-4x+8)hat(k) =(2xyz−2z2)ˆi+(x2z+6y)ˆj+(x2y−4x+8)ˆk
So for the particular point
grad f(1,2,-11) = (-4-2)hat(i) +(-1+12)hat(j) +(2-4+8)hat(k) ∇f(1,2,−11)=(−4−2)ˆi+(−1+12)ˆj+(2−4+8)ˆk
" " = -6hat(i) +12hat(j)+14hat(k) =−6ˆi+12ˆj+14ˆk
We can confirm this graphically: Here is the surface with the normal vector:
