Hello!!!
Show that $n(t)=-g'(t) i+ f'(t) j$ and $-n(t)=g'(t) i- f'(t) j$ are both normal to the curve $r(t)=f(t) i+g(t) j$ at the point $(f(t), g(t))$.
$n(t) \cdot r(t)=-g'(t) f(t)+ f'(t) g(t)$
But is this equal to $0$ ? If so why?
Hello!!!
Show that $n(t)=-g'(t) i+ f'(t) j$ and $-n(t)=g'(t) i- f'(t) j$ are both normal to the curve $r(t)=f(t) i+g(t) j$ at the point $(f(t), g(t))$.
$n(t) \cdot r(t)=-g'(t) f(t)+ f'(t) g(t)$
But is this equal to $0$ ? If so why?
No.
$r$ is merely a vector from the origin to a point on the curve, and $r$ is generally not parallel or perpendicular to the curve.
See:
The fact that $n$ is normal to $r'$ means that $n$ is normal to the curve given by $r(t)$.