 One of the reasons we learned more math is that more powerful tools allow us to solve old problems more easily. So earlier we found the plane tangent to a surface at some point by taking x equal to x0, and this gave us a curve in the yz plane, so we could find dz dy. And we can interpret this as a vector. Then we let y equals y0, and that gave us a curve in the xz plane, and so we could find dz dx, and that would also give us a second vector. Now this actually gave us the direction vectors for two tangent lines, so we took the cross product to get a vector normal to the plane. And once we have that normal vector to the plane, we can write the equation. But now we know a little more calculus. Since holding one variable constant and differentiating with respect to the other is the same as finding the partial derivative, we can restate our process. First, find the vector 1, 0, the partial of f with respect to x. Also we'll find the vector 0, 1, the partial of f with respect to y. And again, since these are the direction vectors for two tangent lines, the cross product will give us a vector normal to the plane. So for example, let's find the equation of the plane tangent to z equals x cubed minus y cubed at the point 2, 1, 7. So we'll find our partial derivatives. And at the point 2, 1, 7, those values will be. And so two lines tangent to the plane have direction vectors. Well, this partial of z with respect to x, we can think about that as a rise over run. So z has increased by 12 when x has increased by 1. And since it's a partial of z with respect to x, y hasn't changed. And so that's the vector 1, 0, 12. Similarly, our partial of z with respect to y, again, rise over run. z has increased by, well, negative 3 while y has increased by 1. And again, since this is a partial of z with respect to y, x hasn't changed. And so that's our vector 0, 1, negative 3. To find a vector orthogonal to both of these, we can find their cross product. Now there's two different cross products we can take, and it doesn't really matter which one. And so we'll find the second. And so the normal vector will be 12, 3, negative 1. So remember the normal vector will be orthogonal to every vector in the plane. And so if x, y, z is a point in the plane, then a vector from 2, 1, 7, which we know is in the plane, to the point x, y, z will be x minus 2, y minus 1, z minus 7. And the requirement that this vector be orthogonal to this vector 12, 3, negative 1, well, that gives us the equation where the dot product is equal to 0. And if we expand that, we get the equation of the plane. Now one of the advantages to being able to speak the language of math is we can solve a problem generically once, and then come up with a formula that will solve the problem for all possibilities. So let's think about this. The plane tangent to z equals f of x, y, it'll include the vector 1, 0, and the partial of f with respect to x. Likewise, it will include the vector 0, 1, the partial of f with respect to y. And since these are two vectors in the plane, we can find the normal vector, and again we'll take the cross product this way, and we get, and so the normal vector will be the partial of f with respect to x, the partial of f with respect to y, negative 1. And here's an important idea I'd like all of you future engineers to think about, because one day I might be driving over a bridge you design. If somebody says, here's a new way of doing things, always test a new method on a problem where you already know the answer. This allows you to check to see whether the method actually works. So we've already solved this problem the old way, let's solve it the new way. A vector normal to the plane will be the partial of z with respect to x, partial of z with respect to y, negative 1. And so we find our partial derivatives, which gives us our normal vector, and so the equation of the plane will be, which is the same answer we got before, and so we have some faith that this new method will actually give us correct results. So let's try it on a case where we don't already know the answer. So again a vector normal to the plane will be the partial derivatives with respect to x and y, negative 1. So we find those partial derivatives evaluated at the point, that gives us our normal vector, and then we can write the equation of the plane.