Let \(f(x, y)\) be a differentiable function from \(\mathbb{R}^2\) to \(\mathbb{R}\) and let \(g(t) = (x(t), y(t))\) be a differentiable curve from \(\mathbb{R}\) to \(\mathbb{R}^2\). State the multivariable chain rule for the derivative of the composite function \(F(t) = (f \circ g)(t) = f(x(t), y(t))\).
Let \(f(x, y)\) be a continuously differentiable function on \(\mathbb{R}^2\), and let \(P = (x_0, y_0) \in \mathbb{R}^2\) be a point. Assume that in a neighborhood of \(P\), the level curve equation \(f(x, y) = f(P)\) implicitly defines \(y\) as a differentiable function of \(x\), say \(y = h(x)\) with \(y_0 = h(x_0)\). Show that the tangent vector to the graph of \(y = h(x)\) at \(P\) is perpendicular (orthogonal) to the gradient vector \(\nabla f(P)\).
Solution
Goal: State the multivariable chain rule and prove that level curve tangent lines are orthogonal to the gradient vector \(\nabla f(P)\).
1. Part (a): Multivariable Chain Rule:Proof:
1.1. Let \(f: \mathbb{R}^2 \to \mathbb{R}\) and \(g: \mathbb{R} \to \mathbb{R}^2\) given by \(g(t) = (x(t), y(t))\) be differentiable functions.
1.2. The derivative of the composite function \(F(t) = (f \circ g)(t) = f(x(t), y(t))\) is given by:\[\frac{dF}{dt}(t) = \frac{\partial f}{\partial x}(x(t), y(t)) \frac{dx}{dt}(t) + \frac{\partial f}{\partial y}(x(t), y(t)) \frac{dy}{dt}(t).\]
1.3. In vector notation using the gradient vector \(\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\) and velocity vector \(g'(t) = (x'(t), y'(t))\):\[\frac{d}{dt}(f(g(t))) = \nabla f(g(t)) \cdot g'(t) = \langle \nabla f(g(t)), g'(t) \rangle.\]
2. Part (b): Perpendicularity of \(\nabla f(P)\) to the Tangent Line:Proof:
2.1. Parameterize the graph of the implicit function \(y = h(x)\) as a smooth curve \(\gamma: I \to \mathbb{R}^2\):\[\gamma(x) = (x, h(x)).\]
2.2. The point \(P = (x_0, y_0) = \gamma(x_0)\).
2.3. The tangent vector to the graph at \(P\) is given by the derivative:\[\mathbf{T} = \gamma'(x_0) = \left( 1, h'(x_0) \right).\]
2.4. Since the graph lies entirely on the level set \(f(x, y) = f(P) = c\) for all \(x\) in a neighborhood of \(x_0\):\[f(\gamma(x)) = f(x, h(x)) = c \quad \text{for all } x \in I.\]
2.5. Differentiating both sides of the identity \(f(\gamma(x)) = c\) with respect to \(x\) using the Chain Rule from Part (a):\[\frac{d}{dx} [f(\gamma(x))] = \nabla f(\gamma(x)) \cdot \gamma'(x) = 0.\]
2.6. Evaluating at \(x = x_0\) where \(\gamma(x_0) = P\):\[\nabla f(P) \cdot \gamma'(x_0) = \nabla f(P) \cdot \mathbf{T} = 0.\]
2.7. Alternatively, by implicit differentiation:\[\frac{\partial f}{\partial x}(P) \cdot 1 + \frac{\partial f}{\partial y}(P) \cdot h'(x_0) = 0 \implies h'(x_0) = -\frac{f_x(P)}{f_y(P)} \quad (\text{if } f_y(P) \ne 0).\]Then:\[\nabla f(P) \cdot \mathbf{T} = (f_x(P), f_y(P)) \cdot \left( 1, -\frac{f_x(P)}{f_y(P)} \right) = f_x(P) - f_x(P) = 0.\]
2.8. Since the dot product \(\nabla f(P) \cdot \mathbf{T} = 0\), the gradient \(\nabla f(P)\) is perpendicular (orthogonal) to the tangent vector \(\mathbf{T}\) of the level curve at \(P\).
3. Conclusion:The chain rule gives \(\frac{d}{dt} f(g(t)) = \nabla f(g(t)) \cdot g'(t)\); differentiating the level curve identity \(f(x, h(x)) = c\) yields \(\nabla f(P) \cdot (1, h'(x_0)) = 0\). Q.E.D.