Hello Math StackExchange!
I had this particular question in my paper, and I can't, for the life of me, understand it.
Can someone help me with explaining why the answer is what it is? Also, on KhanAcademy, 3Blue1Brown/Grant Sanderson says that for vector fields, we use them when we have n input spaces and n output spaces. That is, when the number of input spaces and output spaces are the same. Why is it so that in here, the input space has x and y, and the output space has only z?
Thanks!
Chris
I think you may be mislead by the notations, the title of the plot should have been simply $\vec{grad(f)}$ (it is the gradient of the function f that is plotted and not f).
If you want to see things in terms of input and output spaces, then here the input space is the 2D scalar field f(x,y) and your output space is the 2D vector field grad(f)=grad(f(x,y))
It might be helpful to understand what is going on to calculate explicitly the gradient so that you see its relationship with x and y and z:
$\vec{grad(f)}=\frac{\partial f(x,y)}{\partial x}\vec{u_x} + \frac{\partial f(x,y)}{\partial y}\vec{u_y} $
$=\frac{\partial sin(x)*sin(y)}{\partial x}\vec{u_x} + \frac{\partial sin(x)*sin(y)}{\partial y}\vec{u_y}$
$= cos(x)*sin(y)\vec{u_x} + sin(x)*cos(y)\vec{u_y} $