![]() ![]() The gradient vectors mapped to (x 1, y 1, z 1) and (x 2, y 2, z 2) show the direction of fastest increase. If we could fold the xy plane so each input (x, y) maps to their corresponding output value (x, y, z) on the paraboloid, then each corresponding gradient vector can also be mapped onto the paraboloid such that its initial point is at (x, y, z). Note that the gradient vectors are calculated in component form but are translated to their respective input points on the xy plane to better show the direction associated with those points. The gradient vectors ∇f(x 1, y 1) and ∇f(x 2, y 2) drawn in the xy-plane have their initial points placed at (x 1, y 1) and (x 2, y 2) respectively. The graph of z = f(x, y) is a paraboloid opening upward along the z-axis whose vertex is at the origin. ![]() Geometric interpretation of the gradient for a function of two variablesĬonsider the following graph with gradient vectors denoted in red. The minimum value for the directional derivative at any point in the xy plane is -||∇f(x, y)|| in the direction of -∇f(x, y). ![]() That is, if ∇f(x, y) ≠ 0, then the maximum of D u(x, y) is ||∇f(x, y)||. Its maximum value is the magnitude of its gradient. The directional derivative for any point in the xy plane has its maximum increase when it is in the direction of its gradient.That is, if ∇f(x, y) = 0, then D u(x, y) = 0 for any u. If the gradient for f is zero for any point in the xy plane, then the directional derivative of the point for all unit vectors is also zero.Let y = f(x, y) be a function for which the partial derivatives f x and f y exist. ![]()
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