As illustrated in the picture, the thickness $d$ is known and the parabolic formula of the inner (red) curve is given, say $y = 0.1x^2$. Then, how to obtain the formula for the outer (green) curve?
I think the formula should be in this form: $y_2 = kx^2 - d$ but I have no idea how to derive the value of $k$.
Can anyone give me some hint?
Thank you!
For any point it $(t,at^2)$ on the parabola $y=ax^2$, the corresponding point on the outer curve is
$$x=t+d\sin \theta,\>\>\>\>\> y = at^2 - d\cos \theta$$ where $\theta $ is the tangent angle at $(t,at^2)$, given by $\tan \theta = y’ =2ax$. Thus, the outer curve is
$$x=t+\frac {2adt}{\sqrt{1+4a^2t^2}}, \>\>\>\>\> y= at^2- \frac d{\sqrt{1+4a^2t^2}} $$
which is in a parametrized form with $t$ as the parameter. Note that it is no longer a parabola.
