Gradient Vector, Tangent Planes and Normal Lines — Question 9

Question 9 – Solution

Strategy. Differentiate the constant identity $F(\mathbf r(t))=c$ and interpret the chain rule geometrically.

Step 1: Tangent vectors Suppose $\mathbf r(t_0)=P$ and the curve lies on $S$. Then $$F(\mathbf r(t))=c.$$ Differentiating at $t_0$ by the multivariable chain rule gives $$0=\frac d{dt}F(\mathbf r(t))\bigg|_{t=t_0} =\nabla F(P)\cdot\mathbf r'(t_0).$$ Thus every surface-tangent velocity is perpendicular to $\nabla F(P)$.

Step 2: Tangent plane Because $\nabla F(P)$ is nonzero, it supplies a valid normal. Therefore $$\boxed{F_x(P)(x-x_0)+F_y(P)(y-y_0)+F_z(P)(z-z_0)=0}.$$

Step 3: Normal line The line through $P$ in the gradient direction is $$\boxed{\mathbf r(t)=\left\langle x_0,y_0,z_0\right\rangle+t\,\nabla F(P)}.$$ The condition $\nabla F(P)\ne 0$ ensures a genuine normal direction and a two-dimensional tangent plane. At a singular point, the zero vector cannot determine either object; additional geometric analysis is required.