I need some help here.
Let $f(x, y, z)$ be a differentiable function and suppose that $c(t)$ is a path which lies on the surface $f(x, y, z) = 17.$ If $c(5) = <1, 4, 2>$ show that $c′(5)$ is orthogonal to $∇f(1,4,2).$
2026-04-17 18:06:07.1776449167
show that $c′(5)$ is orthogonal to $\nabla f(1,4,2).$
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$f$ constant imply
$0=\dfrac{f(c(t))}{dt}=\nabla f(c(t))\cdot{c'(t)}$