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Let's say we have an equation of a curve as $y = f(x)$.
I want to find the curve $y = g(x)$ where $(x_1, f(x_1))$ has a perpendicular distance of $d$ from that curve.

Doing this with straight lines is pretty easy. For example:
If $f(x) = mx + b$, then $g(x) = mx + b \pm \frac{d}{sin(atan(1 / m))}$

But how would I be able to do this with curves?

Sorry if it's a bit hard to understand since I'm not too sure how to word this question.

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You have to use the normal vector, that is, perpendicular to the tangent vector. The tangent vector is given by $v(t)=(1,f'(t))$, since that $u(t)=(t,f(t))$ is the position vector. Then, you obtain the normal vector $n(t)$ (unitary) and use it to get the point at distance $d$ from the point $u(t)$ on the direction of $n(t)$. – Sigur Jul 6 '12 at 16:44
You know what parallel curves are? – J. M. Jul 6 '12 at 16:48
Something quite relevant... – J. M. Jul 6 '12 at 16:49
up vote 3 down vote accepted

If you insist that the second curve is given in the form $\gamma_d:\ y=g(x)$ it can be done only in very special cases, e.g., when $\gamma$ is a line or a circle. But it is easy to give a parametric representation for $\gamma_d\ $:

Assume the curve $\gamma$ is given by a parametric representation of the form $$\gamma:\quad t\mapsto z(t)=\bigl(x(t),y(t)\bigr)\qquad(a\leq t\leq b)\ .$$ (The form $y=f(x)$ can easily be recoded as $t\mapsto\bigl(t,f(t)\bigr)$.) Then the tangent vector at $z(t)$ is given by $\dot z(t)=\bigl(\dot x(t),\dot y(t)\bigr)$, and the unit normal vector pointing to the left by $$n(t)=\Bigl({-\dot y(t)\over|\dot z(t)|},{\dot x(t)\over|\dot z(t)|}\Bigr)\ ,$$ where $|\dot z(t)|=\sqrt{\dot x^2(t)+\dot y^2(t)}$. Therefore a parametric representation of the curve $\gamma_d$ at distance $d$ to the left of $\gamma$ reads as follows: $$\gamma_d:\quad t\mapsto z(t)+d\ n(t)=\Bigl(x(t)-d{\dot y(t)\over|\dot z(t)|}, \ y(t)+ d {\dot x(t)\over|\dot z(t)|}\Bigr)\ .$$

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