Gradient Vector, Tangent Planes and Normal Lines — Question 6

Question 6 – Solution

Strategy. Contrast a regular point, where the gradient is nonzero, with the cone vertex, where many generator directions meet and the gradient vanishes.

Step 1: The regular point $$\nabla F=\left\langle 2x,2y,-2z\right\rangle, \qquad \nabla F(1,0,1)=\left\langle 2,0,-2\right\rangle.$$ Thus $$2(x-1)-2(z-1)=0, \qquad \boxed{x-z=0},$$ and a normal line is $$\boxed{(x,y,z)=(1,0,1)+t(1,0,-1)}.$$

Step 2: The vertex At $O=(0,0,0)$, $\nabla F(O)=\left\langle 0,0,0\right\rangle$, so the usual plane equation reduces to $0=0$ and supplies no normal.

Step 3: Generator argument For every $\theta$, the curve $$\mathbf r_\theta(s)=\left\langle s\cos\theta,s\sin\theta,s\right\rangle$$ lies on the cone and has tangent $\left\langle\cos\theta,\sin\theta,1\right\rangle$ at $O$. The directions for $\theta=0,\pi,$ and $\pi/2$ span all of $\mathbb R^3$: the first two yield the $x$- and $z$-directions by differences and sums, and the third then yields the $y$-direction. No two-dimensional plane can contain them all. Hence .