Distance between point and sine wave

Question

I have a project where I need to know the exact minimal distance between a point $(e, f)$ and a sine wave $y = a + b\cdot\sin(cx+d)$

Is there any way of calculating this? If not, is there a way to approximate this?

Thanks in advance!

Edit

Following Ross' and Ben's answers, The distance between a point $(e,f)$ and a sine wave $y = a + b\cdot\sin(cx+d)$ can be calculated by defining it as a distance between two points: $$D=\sqrt{(x-e)^2+(a+b\cdot\sin(cx+d)-f)^2}$$ The $x$ where the distance is minimal can then be calculated by $E=D^2$ and then solving $E'=0$ $$E'= 2(bc\cdot\cos(cx+d)(a+b\cdot\sin(cx+d)-f)-e+x)=0$$ $$x=e-bc\cdot\cos(cx+d)(a+b\cdot\sin(cx+d)-f)$$ This last formula can only be answered exactly in certain cases.

Answer 1

Following Ross' answer, I will use $(e,f)$ to denote your point.

First, some simplifications:

• The minimal distance between $(e,f)$ and $y = a + b\sin(cx + d)$ is the same as the minimal distance between $(e,f - a)$ and $y = b\sin(cx + d)$. So if I can solve the problem with $a = 0$, I can easily solve it for other cases as well.
• Similarly, the minimal distance between $(e,f)$ and $y = b\sin(cx + d)$ is the same as the minimal distance between $(e + d/c,f)$ and $y = b\sin cx$, as long as $c \not= 0$. If $c = 0$ then the problem is not very difficult.
• Again, assuming $c \not= 0$, the minimal distance between $(e,f)$ and $y = b\sin cx$ is the same as $1/c$ times the minimal distance between $(ec,fc)$ and $y = bc \sin x$.

So it's enough to solve the problem in the form "find the minimal distance between $(e,f)$ and $y = a\sin x$".

The points on the sine curve are of the form $(x,a\sin x)$. The distance to such a point is \[D = \sqrt{(x - e)^2 + (a\sin x - f)^2}.\] When this reaches a minimum, its derivative will be zero. As Jack M points out in the comments, you can throw away the square root, if you like: $D$ will reach its minimum exactly when $D^2$ does.

I've no patience for algebra so I leave the rest to you (unless you really get stuck), but I suspect you'll not end up with a neat expression, because there'll be too many trig functions in there. There are various ways to proceed from there, but I don't currently have time to go into them.

Answer 2

Let your point be $(e,f)$ so we don't reuse $x,y$. An approximate approach is as follows: First, find the limits of the half wave of interest. Let's say we are above the curve. You want the local minimum nearest $e$ and the local maximum on the other side of $e$. The perpendicular at a point has slope that is the inverse reciprocal of the derivative. The perpendicular from the maximum will be vertical and on one side of $(e,f)$, the perpendicular from the minimum will be on the other side of $(e,f)$ Call up your favorite one-dimensional root finder to find the $x$ value where the perpendicular goes through $(e,f)$. We have bracketed the root, so it should be easy to find. Now find the distance from $(x,y)$ to $(e,f)$