# Homework Help: Gradients and Tangents to Level Curves

1. Jul 17, 2010

### Quincy

1. The problem statement, all variables and given/known data
Find the gradient of the function at the given point, then sketch the gradient together with the level curve that passes through the point.

f(x,y) = y - x (2,1)

2. Relevant equations

gradient of f = (df/dx)i + (df/dy)j

3. The attempt at a solution

df/dx = -1
df/dy = 1

gradient of f = (df/dx)i + (df/dy)j = -i + j

I don't really know where to go from here; what exactly is a level curve?

2. Jul 17, 2010

### Staff: Mentor

Keep in mind that these are really partial derivatives.
A level curve is a curve for which all points on the curve have the same function value (z-value in this case. Since the graph of your function is a plane in R3, level curves will be straight lines.

If you are familiar with topographical maps, each curve shown on the map is a level curve. Each point on a given curve is at the same altitude.

3. Jul 17, 2010

### Quincy

So, the value of a function with two variables is the value of z? In other words, f(x,y) = z? What about functions with 3 variables (f(x,y,z))?

4. Jul 17, 2010

### Staff: Mentor

Typically w is used for functions of three variables, with w = f(x, y, z). In fact, the particular letter doesn't matter; it just has to be different from the others.

5. Jul 17, 2010

### Quincy

But a graph can only have 3 dimensions, x, y, and z, so what is w supposed to represent?

6. Jul 17, 2010

### vela

Staff Emeritus
It's true that we can't readily visualize a graph in more than 3 dimensions, but there's no reason we can't generalize the notion to higher dimensions.

7. Jul 18, 2010

### HallsofIvy

Well, a graph can have more than 3 dimensions, it is just hard to draw in a 3 dimensional space!

One important property of the gradient, by the way, is that it always point in the direction of fastest increase and is always perpendicular to "level curves" (or level surfaces for functions of two dimensions). If you know a vector perpendicular to a line at a point, it is easy to give the equation of the line perpendicular to that vector and so tangent to the curve. And, of course, it is easy to use a vector perpendicular to a surface, at a point, to find the tangent plane to the surface at that point.