00:01
In this question, we're asked to find the equation of the tangent plane to the graph of this function at this point.
00:08
So the first thing that we typically do is to write the graph of the function as a level curve.
00:16
So f of x, y, z is something like z minus x over y squared.
00:23
So the graph that we are worried about is the set of points x, y, z, such that f of x, y, z is equal to zero.
00:32
And now, because we wrote this as a level curve, you know how the equation for the normal plane is.
00:41
You need a normal vector.
00:44
So this will be something of the form ax plus by plus cz equals d, where a, b, and c are coordinates of the normal vector, and d we just calculate.
00:56
And because we wrote our graph as the level curve of a function, we know that this normal vector can, for example, be taken to be the gradient of f at the point minus four, two, and minus one, or the symmetric of this, or a multiple of it.
01:12
So let's start by calculating the gradient of the vector of f at x, y, z.
01:20
So the first element is df dx.
01:23
So with respect to x, no derivative in here, and from here we get minus one over y squared...
