Why do unit vectors appear in the scalar gradient?

[amaresh92]

greetings

in a scalar gradient why does the unit vector has appeared?scalar gradient only represent the change in that scalar quantity along x,y and z axis.then why unit vector along x, y and z comes in picture?
advanced thanks.

 

[Hootenanny]

greetings

in a scalar gradient why does the unit vector has appeared?scalar gradient only represent the change in that scalar quantity along x,y and z axis.then why unit vector along x, y and z comes in picture?
advanced thanks.

How would you know in which direction the x,y and z axes point, without using vectors?

 

[HallsofIvy]

I'm afraid it is not clear to me what you are asking. If you have a function, f(x,y,z), of three variables, then the "gradient" of f is the vector $\nabla f= \left<\partial f/\partial x, \partial f/\partial y, \partial f/\partial z\right>$. I don't recognize "scalar gradient" as a standard term but I would interpret it to mean the magnitude of the gradient vector:
$$\sqrt{\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2+ \left(\frac{\partial f}{\partial z}\right)^2}$$

There is no "unit vector" involved in that but, of course, you can also write any vector as its magnitude times a unit vector in its direction.

Or, since you mention unit vectors "along x, y, and z", you may mean separating the gradient vector into its components
$$\left(\frac{\partial f}{\partial x}\right)\vec{i}+ \left(\frac{\partial f}{\partial y}\right)\vec{j}+ \left(\frac{\partial f}{\partial z}\right)\vec{k}$$
where $\vec{i}$, $\vec{j}$, and $\vec{k}$ are the unit vectors in the direction of the coordinate axes. Again, any vector can be written in that way: "< a, b, c>" is just a different notation for $a\vec{i}+ b\vec{j}+ c\vec{k}$.