Let $α(s)=(x(s)),y(s))$ be a regular plane curve that is parameterized by arc length, and let $n(s)$ be its normal vector. Consider the family of curves:
$β(s,r)=α(s)+rn(s),−ϵ≤r≤ϵ$
I need to prove that if $s_0$ and $r_0$ are constants, then $β(s,r_0)$ and $β(s_0,r)$ are orthogonal.
So far, I've tried to show the dot product of $β(s,r_0)$ and $β(s_0,r)$ is equal to $0$, but I don't see where to go from here. Any hints?
Best Answer
For constant $r=r_0$, the tangent vector to $\beta(s,r_0)$ is $$ T_1(s)=\frac{\partial\beta}{\partial s}=\alpha'(s)+r_0n'(s). $$ For fixed $s=s_0$ the tangent vector to $\beta(s_0,r)$ is$$ T_2(r)=\frac{\partial\beta}{\partial r}=n(s). $$ Their dot product when $r=r_0$ and $s=s_0$ is $$ T_1\cdot T_2=\alpha'(s_0)\cdot n(s_0)+r_0n(s_0)\cdot n'(s_0) $$ The first product above is zero because $\alpha'(s)$ is tangent to the curve, and the second is zero because $n(s)$ is a unit vector.