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  "body": "\n## Statement\n\nLet $U \\subseteq \\mathbb{R}^n$ be open, let $f \\colon U \\to \\mathbb{R}$, and let $c \\in \\mathbb{R}$.\nLet $I \\subseteq \\mathbb{R}$ be an open interval and let $\\gamma \\colon I \\to U$ be a curve that lies in the [[level set|0Y5Q8Q9WJ7]] $L_c(f)$, so that $f(\\gamma(t)) = c$ for each $t \\in I$.\nLet $t_0 \\in I$, and suppose that $\\gamma$ is differentiable at $t_0$ and $f$ is differentiable at $\\gamma(t_0)$.\nThen $\\nabla f(\\gamma(t_0)) \\cdot \\gamma'(t_0) = 0$.\n\n## Remarks\n\nThe proof is one step: $f \\circ \\gamma$ is constant, so its derivative is $0$, and by the [[chain rule for a curve|0A8TJ0VHC7]] that derivative is $\\nabla f(\\gamma(t_0)) \\cdot \\gamma'(t_0)$.\nSo the [[gradient|0NS2QYVC4Y]] is orthogonal to each tangent vector of the level set at the point. Let $\\nabla f \\ne 0$ and let $f$ be of class $C^1$. Then the level set near the point is a smooth hypersurface, and $\\nabla f$ is a normal vector to it.\nFor $f(x, y, z) = x^2 + y^2 + z^2$, the level set at $c = 1$ is the unit sphere, and $\\nabla f(p) = 2p$ is normal to the sphere at $p$.\n\n## Explanations\n\n### L1 human\n\nAlong a contour line the height does not change, so the slope along the line is $0$.\nThe gradient gives the slope in each direction by a dot product, so it must be at right angles to the contour line.\n",
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  "name": "Gradient is orthogonal to level sets",
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