# 4D tangent planes

From my understanding, tangent planes in 3d space have the form ax+by+cz+d = 0.

If we have a function

$$f(x,y,z) = w$$

how would we represent a tangent plane in 4D space for 4D points of the form below? $$(x,y,z,f(x,y,z))$$

otherwise written:

$$(x,y,z,w)$$

(the gradient exists in 3 axis for the value w).

Thanks!

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The formula for the tangent hyperplane in $\mathbb{R}^n$ always takes a simple form so long as the function $g(\vec x)$ is differentiable. At a point $\vec x_0$, the equation for the tangent hyperplane is

$$\nabla g(\vec x_0) \cdot (\vec x - \vec x_0) = 0.$$

To be explicit, this is the tangent to the set $g(\vec x ) = 0$ (or any other constant). The intuition is that the gradient should always be normal to the tangent plane. If you're interested in the tangent to some graph of a function $f$ (as you are), just take $g(\vec x) = x_n - f(x_1,\ldots,x_{n-1})).$ Writing this out in lower dimensions, and also in the case you're interested in $(n=4)$, we get the familiar formulas:

$$a(x-x_0) + (y-f(x_0)) = 0, \qquad a= {\partial f \over \partial x}(x_0)$$ (the above is the tangent line to the point $(x_0,f(x_0))$ on the curve $f(x) = y$.) $$a(x-x_0) + b(y-y_0) + (z-f(x_0,y_0)) = 0, \qquad a= {\partial f \over \partial x}(x_0,y_0), \qquad b= {\partial f \over \partial y}(x_0,y_0)$$ (the above is the tangent line to the point $(x_0,y_0,f(x_0,y_0))$ on the surface $f(x,y) = z$.) And finally, $$a(x-x_0) + b(y-y_0) + c(z-z_0) +(w-f(x_0,y_0,z_0)) = 0, \qquad a= {\partial f \over \partial x}(x_0,y_0,z_0), \qquad b= {\partial f \over \partial y}(x_0,y_0,z_0),\qquad c= {\partial f \over \partial z}(x_0,y_0,z_0)$$ gives the tangent hyperplane to the hypersurface in $\mathbb{R}^4$ defined by $(x,y,z,f(x,y,z))$.

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very cool answer. thanks for that! – JayDee Jan 6 '13 at 3:23

How about $ax+by+cz+dw = e$?

I guess the only problem comes when you have 13 dimensions, then you run out of alphabets ... Hopefully by then, you let the coordinates be $x_i$, and the coefficients be $\alpha_i$.

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could we thus rewrite it: ax+by+cz+dw+e = 0? – JayDee Jan 6 '13 at 3:36
@JayDee Of course, since $a, b, c, d, e$ are variables, you could write it as $ax+by+cz+dw+(-e)=0$, and do a substitution $(-e)=E$. – Calvin Lin Jan 6 '13 at 15:59
thanks Calvin, too bad i can't accept two answers. yours is equally right to the accepted one. I just give it to the others for the good explanatory build-up. thanks for your help though :) – JayDee Jan 6 '13 at 20:08

Generally, if $F:M^k\to\mathbb{R}^n$ is an embedding, if $v_i$ span $T_pM$, then we can represent the tangent space to $p$ by the affine space $F(p) + \mbox{span}\{dFv_1,dFv_2,\ldots,dFv_k\}$.

In this case, $F:\mathbb{R}^3\to\mathbb{R}^4$ maps $\mathbb{R}^3$ to its graph. Particularly, $(x,y,z)\mapsto F(x,y,z) = (x,y,z,f(x,y,z))$. The differential is $$dF = \begin{pmatrix} I \\ \nabla f\end{pmatrix}$$ and so the tangent space to $F(x,y,z)$ is given by $$F(x,y,z) + \mbox{span}\{(1,0,0,f_x),(0,1,0,f_y),(0,0,1,f_z)\}$$

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