How using a mirror to find the tangent at a point on the curve works

lokifenrir96

Hi, I recently learned that to find the tangent at a point on any curve, you can simply place a mirror on that point and reflect the part of the curve on one side of that point such that the reflection flows smoothly into the other part of the curve on the other side. Once this is done, draw a line along the mirror, and this line would be perpendicular to the tangent.

However, I do not understand the principle behind this method. How exactly does ensuring that the reflection of one side of the curve flows smoothly into the other side result in the mirror being perpendicular to the tangent?

My best guess as of now is that the point must be the point about which the curve is exactly symmetrical, in which case it would make sense that if you reflect one part of the curve into the other part, the perpendicular line which you draw along the mirror would be the line of symmetry. But doesn't this mean that this method would not work if your point is not the point of symmetry?

Thank you!

Simon Bridge

You should actually do it. Draw a random curve and get a mirror.

If the mirror is not on the normal to the curve, then the reflection + the curve line will show a sharp bend at the mirror. Any curve that starts at that point, to be continuous, must join at the tangent angle so it does not matter that the actual curve is not a mirror reflection. Just try drawing a line where this does not happen.

lokifenrir96

Oh, so let me see if I understand you correctly.

If we have a curve, and the point we are taking the tangent about is X, and to the right of X is A, and to the left of X is B.

If we put a mirror at X which reflects A, the reflection must flow into A, but it does not need to flow into B. (Though my notes mentioned that the reflection must flow into both A and B?)

And as long as the A and the reflection are continuous, the mirror will definitely be at a normal to X.

So it does not matter whether the curve is symmetrical about X.

Is that right?

Okay, I can understand that if you perform this on a hundred different curves, the result will always be the same. But what I wish to know is not the rule derived from experimentation, but rather, the principle behind this rule. Or is this one of those scenarios where there is no explanation, but that it just works?

Mentallic

If you're interested, here's a video that explains it, and also gives some examples of its real life applications.

http://www.youtube.com/watch?v=dsRsap2_RAc

If you'd like to do some Maths on it yourself to see why it works, draw two non parallel lines that intersect, and label the acute angle between them as [itex]\alpha[/itex]. Now, draw a ray hitting the first line (imagining it's a mirror), and then at the point of contact between these two lines, draw a dotted perpendicular line. The angle between the perpendicular and the light ray can be labelled [itex]\theta[/itex].

Now as you should know due to the law of reflection, the angle of incidence equals the angle of reflection, so you'll need to label the angle between the perpendicular and the ray bouncing away from the line as [itex]\theta[/itex] as well.

Now do the same with the ray bouncing from the first line into the second, but this time with another angle, say, [itex]\phi[/itex].

What algebraic formulae can you up with to describe this situation? Think about the triangle with the angle [itex]\alpha[/itex], and the sum of the angles in a triangle. And then look for a second relationship.

Simon Bridge

Ah - thank you mentallic.
I was about to explain that this will only work for continuous curves.
I was suggesting the experiment because in the act of doing it realisation of the underlying principle will come.

Mentallic

Well of course if we found a tangent at a point where the curve isn't continuous, we're doing it wrong

Or if you can't get two mirrors that are free to rotate, some accurate diagrams will do just as well.

Simon Bridge

I think you can see the effect with just one plane mirror.