Functions of 3 variables. Linear approximation Suppose we want to find the tangent plane to the surface whose equation is x2+y2−z2=4 at the point (2, 1, 1). 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑝𝑙𝑎𝑛𝑒 𝑡𝑜 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 2 2 2 𝑥 +𝑦 −𝑧 = 4 𝑎𝑡 (2, 1, 1) ? A way to do this is to solve for z in the equation z = f(x, y); once that is done, we can write z as a function of x and y, then find a tangent plane approximation of the graph of that function. But the best way to do it, now that we have the gradient vector, is actually to directly say, oh, we know the normal vector of this plane. The solution to this problem is quite simple: So here you have the surface x2 + y2 – z2 = 4. This is called a hyperboloid, because it looks like what you get when you spin a hyperbola around an axis. And here's its tangent plane at the given point. So it doesn't look very tangent, because it crosses the surface. But it's really, you'll see that's really the plane that's approximating the surface in the best way that you can at this given point. It is really the tangent plane. So how do we find this plane? Well, we can plot it on a computer, but that's not exactly how we would look for it in the first place. So the way to do it is that we compute the gradient of this function. Thus, the equation of this level set is w equals 4x2 + 2y2 – 2z2 and its gradient is 2x, 2y and –2z. 𝑤 = 4, 2 2 2 𝑤 =𝑥 +𝑦 −𝑧 ∇𝑤 = 〈2𝑥, 2𝑦, − 2𝑧〉 At this point, the system has reached a stable equilibrium where x equals 2 and y and z are both 1. ∇𝑤 = 〈4, 2, − 2〉 One way to define a tangent plane for a surface is to use its normal vector. Or, you could say that a tangent plane has the same normal vector as the surface. Or, you could say that if a vector is perpendicular to a surface, then it's also perpendicular to all tangent planes to that surface. The equation is , well, 4x plus 2y minus 2z equals something, we should just plug in that point. We'll get 8 plus 2 minus 2. It looks like we'll get 8. 4𝑥 + 2𝑦 − 2𝑧 = 8 And of course, we could simplify by dividing everything by 2, but such a simplification is not appropriate here. Now, if you have a surface given by an equation of the form , and the point lies on this surface, then you know how to find a potential plane at that point.