I want to find the vector normal to the curve $x=3/4-y^2$ at the point (0,$\sqrt{3}/2)$ I know that the tangent line at point (0,$\sqrt{3}/2)$ is (1)y-$\sqrt{3}/2 = -1/ \sqrt{3} (x-0). $ I'm guessing that the normal vector $n_1$ = -1/$\sqrt{3}$ $\hat{i}$ + $\hat{j}$. Am I correct? I really have no idea how to find the vector normal to a curve. I got the signs for the normal vector from the signs of the slope and the value of y of the tangent line equation
[Math] Find the vector normal to a curve at some point
vectors
Best Answer
$1=-2yy'$, which gives $y'=-\frac{1}{2y}$.
Thus, a slop of the normal it's $2\cdot\frac{\sqrt3}{2}=\sqrt3$.
Thus, the equation of the normal it's $y-\frac{\sqrt3}{2}=\sqrt3(x-0)$ or $y=\sqrt3x+\frac{\sqrt3}{2}$.
An equation of the tangent it's indeed, $$y-\frac{\sqrt3}{2}=-\frac{1}{\sqrt3}x$$ or $$\frac{1}{\sqrt3}x+y-\frac{\sqrt3}{2}=0,$$ which gives a vector normal $\frac{1}{\sqrt3}\vec{i}+\vec{j}$.