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 $$f_x(p)(x-x_0)+f_y(p)(y-y_0)+f_z(p)(z-z_0)$$

  • $\begingroup$ Hint: gradient is normal to the level surface. $\endgroup$ – Jacky Chong Feb 11 '19 at 5:20
  • $\begingroup$ @JackyChong I got that, but how do you get that the gradient is perpendicular to $\textbf{$x-x_0$}$ $\endgroup$ – JB071098 Feb 11 '19 at 5:27
  • $\begingroup$ 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. $\endgroup$ – Jacky Chong Feb 11 '19 at 5:32
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    $\begingroup$ To have an equation (here of a plane) you need an equals sign and something on the other side of it. :) $\endgroup$ – 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 $p_0=(x_0,y_0,z_0)\in S$.

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