Consider a smooth surface given by the function $z = f(x,y)$, such that the partial derivatives of $f(x,y)$ exist. Suppose $Q$ is a point that does not lie on the surface, and $P$ is the nearest point on the surface to $Q$. Show that the line through $P$ and $Q$ is perpendicular to the surface at $P$.
Proving line is perpendicular using partial derivatives
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partial-derivative
vector-analysis
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1"...through $\;P,\,Q\;$ is perpendicular" ...perpendicular **to what**? – 2017-01-02
1 Answers
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At the point $P(x_p,y_p,z_p)$; the tangent plane to the surface $z=f(x,y)$ has following cartesian equation :
$$z-z_p=\frac{\partial f}{\partial x}(x_p,y_p)(x-x_p)+\frac{\partial f}{\partial y}(x_p,y_p)(y-y_p)$$
thus the normal vector is $$\vec{PQ}=\lambda (\frac{\partial f}{\partial x}(x_p,y_p),\frac{\partial f}{\partial y}(x_p,y_p),-1)$$
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0How do you justify that the normal to the tangent plane is $\;\vec{PQ}\;$ ? I mean, how does $\;Q\;$ enters there? – 2017-01-02