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  "body": "\n## Proof\n\nWe differentiate the constant function $f \\circ \\gamma$ in two ways.\n\n1. Let $U$, $f$, $c$, $I$, $\\gamma$ and $t_0$ be as in the statement. Put $\\varphi = f \\circ \\gamma \\colon I \\to \\mathbb{R}$.\n2. Since $\\gamma$ lies in the [[level set|0Y5Q8Q9WJ7]] $L_c(f)$, $\\varphi(t) = c$ for each $t \\in I$.\n3. So for $s \\ne 0$ with $t_0 + s \\in I$, the difference quotient $(\\varphi(t_0 + s) - \\varphi(t_0)) / s$ is $0$. Hence $\\varphi'(t_0) = 0$.\n4. By the [[chain rule for a composition with a curve|0A8TJ0VHC7]] with $m = 1$, $\\varphi'(t_0) = Df(\\gamma(t_0))\\big(\\gamma'(t_0)\\big)$.\n5. Since $f$ is differentiable at $\\gamma(t_0)$, the [[gradient|0NS2QYVC4Y]] $\\nabla f(\\gamma(t_0))$ is defined, and $Df(\\gamma(t_0))(v) = \\nabla f(\\gamma(t_0)) \\cdot v$ for each $v \\in \\mathbb{R}^n$.\n6. By steps 3, 4 and 5, $\\nabla f(\\gamma(t_0)) \\cdot \\gamma'(t_0) = \\varphi'(t_0) = 0$.\n\n## Idea\n\nAlong a curve inside a level set, the value of $f$ does not change, so its rate of change is $0$. The chain rule computes that rate as the dot product of the gradient with the velocity of the curve. So this dot product is $0$.\n",
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  "name": "Proof that the gradient is orthogonal to level sets",
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