Problem on the normal vector and tangent line to a curve
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Find a normal vector to the curve $\sqrt{x} + \sqrt{y} = 2$ at $(x,y) = (1,2)$. Use it to find the tangent line at $(1,1)$ expressed in the form $\bfn \cdot \bfx = b$.
Solution
Recall that
Normal vector to an implicitly defined surface
Gradient is perpendicular to level curves
Level curves and surfaces
To find a normal vector to a curve, view that curve as a level set of some function $g(x,y)$.
A normal vector to the implicitly defined curve $g(x,y) = c$ is $\nabla g(x,y)$.
We identify that the curve is a level set of the function $g(x,y) = \sqrt{x} + \sqrt{y}.$
Definition of the gradient
Hence, a normal vector is given by \begin{align}\nabla g(x,y) &= \left \langle \frac{1}{2} x^{-1/2}, \frac{1}{2} y^{-1/2} \right \rangle \\ \nabla g(1, 1) &= \left \langle -\frac{1}{2}, -\frac{1}{2} \right \rangle. \end{align}
Recall that
Lines in 2d can be specified by a point and a normal vector
The 2d line going through $\bfx_0$ with normal vector $\bfn$ is given by $\bfn \cdot \bfx = \bfn \cdot \bfx_0$.
Identifying $\bfn = \langle -1/2, -1/2 \rangle$, and $\bfx_0 = \langle 1, 1 \rangle$, we compute that $\bfn \cdot \bfx_0 = 1$. Hence, the 2d line is $$\langle -1/2, -1/2 \rangle \cdot \bfx = 1.$$
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