The derivative of distance function with respect to time give velocity function in single variable calculus.
But how does gradient of a multivariable function point steepest ascent?
I have been reading about the concept and trying to make an analogy with single variable calculus examples to understand the logic behind that but I still don't understand.
The derivative of distance function with respect to time give velocity but the derivative of a scalar field with respect to position(x,y,z,) gives steepest ascent but how?
The rate of increase of the function $f(\vec{x})$ in the direction of $\vec{u}$, a unit vector, is $$D :=\lim_{t \to 0} \frac{f(\vec{x} + t\vec{u}) - f(\vec{x})}{t}$$ Now let $F(t) = f(\vec{a} + t\vec{u})$, then $$D = \frac{dF}{dt}(0) = \frac{\partial f}{\partial x_1}(\vec{a}) \, \frac{dx_1}{dt} + ... + \frac{\partial f}{\partial x_n}(\vec{a}) \, \frac{dx_n}{dt} = \frac{\partial f}{\partial x_1}(\vec{a}) \, u_1 + ... + \frac{\partial f}{\partial x_n}(\vec{a}) \, u_n$$ by the chain rule for partial derivatives. But this is just $$D = \nabla{f}(\vec{a}) \cdot \vec{u}$$ which is clearly maximised by $\vec{u} = \nabla{f}(\vec{a})$ (or in the direction of $\nabla f(\vec{a})$ as it is a unit vector) so the greatest increase is in the direction of $\nabla{f}$.