# Gradient and equation for curve in space

1. Sep 18, 2012

### dumbQuestion

Let's say we have a function f(x,y,z)=k which is a level surface for a function of 3 variables. Now say at some point P we want to find the derivative in the direction of some vector, u. (the change in z in the direction of u at point P). We can easily find this direction derivative using

∇f|p ° u/||u|| (where u/||u|| is the unit vector in the direction of u)

My thinking is this: so if we took a "slice" of this surface, namely, the intersection of a straight vertical plane containing P and facing in direction of u, this "slice" would be a curve and the direction derivative would be no more than the slope of the tangent line to this curve. So I was just curious, from our equation of directional derivative, can we somehow get the equation for this curve? I just didn't know if there was any relation between the two. Also, say for some reason we could not compute the gradient or were unable to know exactly u (the direction we want our derivative to be), but say we could somehow find the equation for this curve. Could we then just take the derivative of this curve to arrive at the directional derivative?

2. Sep 18, 2012

### dumbQuestion

An example of what I'm talking about

let's say we have

f(x,y)=2x+2y2 and we want the directional derivative at point (1,2) in the direction u=<4,-3>

OK, so ∇f=<2,4y> and u/||u|| = <4/5,-3/5>

So the general vector for the directional derivative in the direction of u would be

∇f ° u/||u|| = 8/5-12/5y

So its my understanding this would also be the general equation for the derivative of the curve defined as the intersection of a vertical plane in direction u. So for example if I plugged in the point (1,2), it would give the slope of the tangent line to the curve defined as the intersection of a vertical plane containing point (1,2) in the direction of u. So if I integrate something, will I somehow generate a general equation for the curve? What would I be integrating?