Suppose I have a continuously differentiable function $z=k(x,y)$ where $z : \Bbb R^3 \rightarrow \Bbb R$, then is it true that if $v = (x-x_0, y-y_0, z-z_0)$ is tangent to the graph at $(x_0,y_0,z_0)$ that $v$ will satisfy $z = z_0 + \frac {\partial k} {\partial x}(x_0,y_0,)(x-x_0) + \frac {\partial k} {\partial y} (x_0,y_0)(y-y_0)$ (i.e the equation of the tangent plane) and why? I can't seem to connect these two concepts. Hints and insights much appreciated.


Hint: The vector $v$ as you defined it is approximately tangent to the surface if $(x,y,z)$ is sufficiently close to $(x_0, y_0, z_0)$. See:

\v represents your vector, \nabla U represents the vector $\vec n$ that I define below

Gif from Google. Plus:

$$ -(z-z_0) + \frac{\partial k}{\partial x}(x-x_0) + \frac{\partial k}{\partial y}(y-y_0) = 0 $$

may be interpreted as a scalar product between $v$ as you defined it and the vector $\nabla U = \vec n = \frac{\partial k}{\partial x}\hat{x}+ \frac{\partial k}{\partial y}\hat{y}-\hat{z}$ equal to zero. $\vec n$ is normal to the plane.

  • $\begingroup$ Lol how did you insert that animation $\endgroup$ – IntegrateThis Jun 8 '17 at 2:44
  • $\begingroup$ Straight-forward, I'm surprised as well... $\endgroup$ – rrogerr Jun 8 '17 at 2:45

Your definition of v might not be exactly correct, since you seem to consider it a plane, and yet the set $(x - x_0, y - y_0, z - z_0)$ is just $\mathbb{R}^3$ translated.

This might be helpful: the equation of a plane through the origin is $(x, y, z) \cdot N = 0$, i.e. the set of all points with position vectors perpendicular to a given non-zero vector $N$ (the normal). For a plane through a given point $(x_0, y_0, z_0)$, the equation becomes $((x, y, z) - (x_0, y_0, z_0)) \cdot N = 0$. In the case of the tangent plane to a point $(x_0, y_0, z_0)$ on the graph of a function $k(x, y)$, the normal vector $N$ is nothing but the cross product of the two tangent vectors at that point.

Specifically, the points of the graph have the form $(x, y, k(x, y))$, and the tangent vectors are the derivatives of this w.r.t. $x$ and $y$, which are $T_1 = (1, 0, \partial k/\partial x(x, y))$ and $T_2 = (0, 1, \partial k/\partial y(x, y))$. Taking their cross product gets you $(-\partial k/\partial x(x, y), -\partial k/\partial y(x, y), 1)$, which is the same as the expressions above up to sign, which doesn't matter.

You can also probably get the same answer by considering the span of $T_1$ and $T_2$, which also defines the tangent plane.

  • $\begingroup$ $v$ is a vector in $\Bbb R^3$ $\endgroup$ – IntegrateThis Jun 8 '17 at 2:40
  • $\begingroup$ Ah, yes, that's clearer. In that case, thinking about it with relation to the normal may help you connect the two concepts. $\endgroup$ – Fred Akalin Jun 8 '17 at 2:48
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    $\begingroup$ also do you mean the gradient vector's z component to be -1? $\endgroup$ – IntegrateThis Jun 8 '17 at 2:48
  • $\begingroup$ Ah, yes. I had to think about it, and so wrote up a derivation. $\endgroup$ – Fred Akalin Jun 8 '17 at 2:58

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