Let \(z=f(x,y)\) be a smooth surface. Show that the gradient is perpendicular to the level curves.
Solution
- Definitions:
\(f: \RR^2 \to \RR, \quad (x,y) \mapsto z = f(x,y)\)
Level curves are given by \(f(x, y) = c\);
\(\nabla f: \RR^2 \to \RR^2, \quad \nabla f(\vector p) = \thevector{f_x(\vector p), f_y(\vector p)}\)
\(\gamma(t): \RR \to \RR^2, \quad \gamma(t) = \thevector{x(t), y(t)}, \quad \gamma_t(t) = \thevector{x_t(t), y_t(t)}\)
\(g(t) = f \circ \gamma:\RR \to \RR, \quad (f \circ \gamma)(t) = f(x(t), y(t)), \quad g_t(t) = f_t(\gamma(t))\cdot \gamma_t(t)\)
A vector \(v\) is perpendicular to a surface at a point \(p\) iff \(v\) is perpendicular to the tangent vector of every curve passing through \(p\).
Proof of actual statement
Let \(\vector p = \thevector{x_0, y_0}\) be a point on the level surface, so \(f(\vector p) = c\) for some constant.
Let \(\gamma(t) = \thevector{x(t), y(t)}\) be a curve on the level surface, so \(\gamma(t_0) = \vector p\) for some \(t_0\). Let \(\gamma'(t)\) be its tangent vector.
Let \(g(t) = f(x(t), y(t)) = (f \circ \gamma)(t)\), and note that for some \(\vector x\) on the level surface, and so we have \(g(t_0) = f(\vector x) = c\) and thus \(\dd{g}{t}(t) = 0\).
By the chain rule, compute \[\begin{align*} \dd{g}{t}(t) &= \left(\dd{f}{x}\dd{x}{t} + \dd{f}{y}\dd{y}{t}\right)(t) \\ &= \inner{ \thevector{\dd{f}{x}(x(t), y(t)), \dd{f}{y}(x(t), y(t))} } { \thevector{\dd{x}{t}(t), \dd{y}{t}(t)} } \\ &= \inner{\nabla f(\gamma(t))}{\dd{\gamma}{t} (t)}. \end{align*}\]
From above, we know that this is zero. Now note that we have \[ \dd{g}{t}(t_0) = \inner{\nabla f(\gamma(t_0))}{\dd{\gamma}{t}(t_0)} = \inner{\nabla f(\vector p)}{\gamma'(t_0)} \] but by the previous statement, \(\dd{g}{t}(t_0) = 0\), which exacty says that the gradient of \(f\) is orthogonal to \(\gamma\) at \(\vector p\). But \(\vector p\) was an arbitrary point on the level surface, and \(\gamma\) was an arbitrary curve through it. So \(\nabla f\) is orthogonal to every level curve through \(\vector p\), and this orthogonal to the tangent plane at \(\vector p\), and thus normal to the surface at \(\vector p\). Since \(\vector p\) was an arbitrary point on the level curve, this holds everywhere on the level curve, and for arbitrary level curves. Thus \(\nabla f\) is orthogonal to every level curve. \(\qed\)
Solution
Goal: Prove that for a continuously differentiable function \(f: \mathbb{R}^2 \to \mathbb{R}\), the gradient vector \(\nabla f(\mathbf{p})\) at any point \(\mathbf{p}\) on a level curve \(C = \{(x,y) \in \mathbb{R}^2 \mid f(x,y) = c\}\) is orthogonal to the tangent vector of every smooth curve in \(C\) passing through \(\mathbf{p}\).
1. Let \(\mathbf{p} = (x_0, y_0) \in C\) such that \(f(\mathbf{p}) = c\). Let \(\gamma: (-\varepsilon, \varepsilon) \to \mathbb{R}^2\) be a differentiable parametrized curve such that \(\gamma(0) = \mathbf{p}\) and \(\gamma(t) \in C\) for all \(t \in (-\varepsilon, \varepsilon)\).
By definition of a smooth curve on the level set \(C\).
2. The composite function \(g(t) = (f \circ \gamma)(t) = f(\gamma(t))\) is constant on \((-\varepsilon, \varepsilon)\) with value \(c\).
Since \(\gamma(t) \in C\) for all \(t\), \(f(\gamma(t)) = c\) identically.
3. \(g'(0) = 0\).
The derivative of a constant function is identically zero.
4. By the multivariable Chain Rule, \(g'(0) = \nabla f(\gamma(0)) \cdot \gamma'(0) = \nabla f(\mathbf{p}) \cdot \gamma'(0)\).
Since \(f\) is continuously differentiable and \(\gamma\) is differentiable at \(t=0\), the chain rule applies: \[g'(0) = \left.\frac{d}{dt} f(x(t), y(t))\right|_{t=0} = \frac{\partial f}{\partial x}(\mathbf{p}) x'(0) + \frac{\partial f}{\partial y}(\mathbf{p}) y'(0) = \nabla f(\mathbf{p}) \cdot \gamma'(0).\]
5. \(\nabla f(\mathbf{p}) \cdot \gamma'(0) = 0\).
Combining 3 and 4 gives \(\nabla f(\mathbf{p}) \cdot \gamma'(0) = g'(0) = 0\).
6. Conclusion: \(\nabla f(\mathbf{p})\) is orthogonal to the tangent vector \(\gamma'(0)\) of any curve lying on the level set through \(\mathbf{p}\).
Since two vectors with zero dot product are orthogonal by definition, and \(\gamma\) was arbitrary. Q.E.D.