Chapter 2.4 Exercise 1 in Do Carmo (Tangent Plane)

Show that the equation of the tangent plane at $$p=(x_0,y_0,z_0)$$ of a regular surface is given by $$f(x,y,z)=0$$ where $$0$$ is a regular value of $$f$$, is $$f_x(p)(x-x_0)+f_y(p)(y-y_0)+f_z(p)(z-z_0)$$

• Hint: gradient is normal to the level surface. – Jacky Chong Feb 11 '19 at 5:20
• @JackyChong I got that, but how do you get that the gradient is perpendicular to $\textbf{$x-x_0$}$ – JB071098 Feb 11 '19 at 5:27
• Note that $x-x_0$ is a tangent vector to the surface, then you can find a curve $\gamma(t)$ that lies on the surface such that $\gamma(0) = x_0$ and $\gamma'(0) = x-x_0$. Finally, consider $\frac{d}{dt}f(\gamma(t))$ which equals zero since $\gamma$ lies on the level surface. – Jacky Chong Feb 11 '19 at 5:32
• To have an equation (here of a plane) you need an equals sign and something on the other side of it. :) – Ted Shifrin Feb 11 '19 at 17:15

Let $$S=\{(x,y,z)\in\mathbb{R}^3; f(x,y,z)=0\}$$, and $$p_0\in S$$. Now, there exist a curve $$\alpha:(-\epsilon,\epsilon)\to S$$, such that $$\alpha(0)=p_0$$ and $$\alpha'(0)=w\in T_{p_0}S$$, with $$w=p-p_0$$. Note that $$(f\circ\alpha)(t)=f(x(t),y(t),z(t))=0\implies df_{p_0}(w)=0$$ So the inner product of an element of $$S$$ and any element of $$T_{p_0}S$$ is $$0$$, i.e., $$\langle(f_{x}(p_0),f_{y}(p_0),f_{z}(p_0)),w\rangle=\langle(f_{x}(p_0),f_{y}(p_0),f_{z}(p_0)),p-p_0\rangle=0$$
Where $$p=(x,y,z)\in T_{p_0}S$$ and $$p_0=(x_0,y_0,z_0)\in S$$.