Gradients and Level Curves
The gradient has a special relationship with level curves that reveals important geometric information. Let’s explore this connection.
Recall that if a curve is defined parametrically by [latex](x(t), y(t))[/latex], then the vector [latex]x'(t){\bf{i}}+y'(t){\bf{j}}[/latex] is tangent to the curve for every value of [latex]t[/latex] in the domain.
Now suppose [latex]z=f(x, y)[/latex] is a differentiable function, and [latex](x_0, y_0)[/latex] is in its domain. Assume that [latex]x_0=x(t_0)[/latex] and [latex]y_0=y(t_0)[/latex] for some value of [latex]t[/latex], and consider the level curve [latex]f(x, y)=k[/latex].
Define [latex]g(t)=f(x(t), y(t))[/latex] and calculate [latex]g'(t)[/latex] on the level curve. By the Chain Rule:
[latex]g'(t)=f_x(x(t),y(t))x'(t)+f_y(x(t),y(t))y'(t)[/latex]
But [latex]g'(t)=0[/latex] because [latex]g(t)=k[/latex] for all [latex]t[/latex] (the function value is constant on a level curve). Therefore:
[latex]f_x(x(t),y(t))x'(t)+f_y(x(t),y(t))y'(t)=0[/latex]
We can rewrite the left side as a dot product:
[latex]f_x(x(t),y(t))x'(t)+f_y(x(t),y(t))y'(t)=\nabla{f}(x,y)\cdot\langle{x}'(t),y'(t)\rangle[/latex]
Therefore:
[latex]\nabla{f}(x,y)\cdot\langle{x}'(t),y'(t)\rangle=0[/latex]
The dot product of these vectors equals zero, which means they’re orthogonal. Since the vector [latex]\langle{x}'(t),y'(t)\rangle[/latex] is tangent to the level curve, the gradient must be normal (perpendicular) to the level curve. This gives us an important theorem.
gradient is normal to the level curve
Suppose the function [latex]z=f(x, y)[/latex] has continuous first-order partial derivatives in an open disk centered at a point [latex](x_0, y_0)[/latex]. If [latex]\nabla{f}(x_0,y_0)\ne0[/latex], then [latex]\nabla{f}(x_0,y_0)[/latex] is normal to the level curve of [latex]f[/latex] at [latex](x_0, y_0)[/latex].
We can use this theorem to find tangent and normal vectors to level curves of a function.
