Is the Gradient Vector Only Applicable for Multivariable Functions?

[Telemachus]

Hi there. I have a doubt that I never cleared before, so I wanted your opinions on this. The thing is that when in vector calculus the gradient vector is presented, one of the "geometric" interpretation that is given is that it's a vector always perpendicular to the curve. So at first I've always tried to think in simpler cases when I have to face something new. I've tried going down one dimension. I always tried to think of this with a parabola of equation
$$y=x^2$$
The thing is that if we think of the gradient for this parabola what we get its only the derivative, which is the slope of the curve. Of course, I would need another variable for y to get a vector pointing on the normal direction to the curve. So, how does this must be reasoned? how one gets the gradient vector for one variable functions? I thought that maybe involving the implicit function theorem I could get on something, but didn't get too far. The other idea requires to "extend" the function on two variables, thinking of it as the intersection of a surface with a plane.

Is it that the gradient only exists for functions of more of one variable? now that I wrote all this I'm thinking that the interpretation of the gradient as a vector normal to the curve (or the surface) perhaps only holds for two dimensional curves, because if we go one dimesion over then we can't think in something like the normal vector, right? and with one dimension less we only get the slope for the curve.

So what you say?

Bye there, thanks for posting.

 

[SteamKing]

The first derivative of a function of a single variable gives an expression for the slope of the tangent to the curve of that function. From the text of your question, it appears that you are looking to find a normal to the curve instead of a tangent.

If y = f(x), then dy/dx| x = x0 gives the slope m of the tangent at x = x0

To find the slope n of the normal, n = -1/m = -1/dy/dx = -dx/dy provided that dy/dx is not equal to zero at x = x0.

 

[Char. Limit]

The gradient of a single-variable function would indeed be the derivative (or rather, a vector with the same magnitude as the derivative).

 

[Telemachus]

Thanks.