An easier method of doing that trig is to know that not only is cot = 1/tan, but tan(x) = cot(π/2 - x). It’s where the “co” comes from, same as the “co” in cosine and cosecant. “Co” stands for complementary and complementary angles add up to π/2.
As an electrical engineer, I always assume alpha=alpha' for simple reflections, and it’s cool to see that the lines meet at (0, 1/2a), which is the focal point. It’s fascinating how math ties into real world physics like this. I remember when I first learned about such concepts, and a SolutionInn tutor helped me understand the theory behind these reflections much better. Great video, this is definitely one of those ‘aha’ moments.
MythBusters took on Archimedes "Death Ray" three times, first in 2004 (S2E16), again in 2006 (S4E7), and again in 2010 in a special episode featuring President Obama. They busted the myth all three times.
As an electrical engeneer, I would assume that alpha=alpha' (simple reflection) and then prove that all lines will meet in (0, 1/2a) which is the focal point. Great video.
Why does the whole video sound like: Given a ray that get reflected to the focus of a parabola, then the ray get reflected by the parabola (skeleton head)
WHATT you’re kidding me. I had to prove this exact reflective property for one of the questions on a Math Olympiad I did recently! My solution is really similar to yours
@@PapaFlammy69 one of the other questions was "if you pick integers from the set {1, 2, 3, ..., 144}, what is the max. number of integers you could select before three of the integers correspond to sides of a triangle?". the solution is really beautiful and simple
This is showing that if the reflected beam passes through the focus point then the reflection law is satisfied, it's quite easy to visualise that if the beam passes above or below the focus point then the angle of reflection will be different to the angle of incidence meaning that the only way this law can be held true is if the reflected beam passes through the focus point which is what we wanted to show
@@harley_2305 but this just shows that given the reflected beam of light hits the focus point then the angles are the same. You would need a different proof to show the reflected beam of the vertical beam always does go to the focus in the first place since we used those coordinates to derive the angle. A priori I have no idea where the second reflected beam goes. We can take it as a fact, but we don’t it from this derivation. That’s what I’m trying to say but maybe I’m wrong and misunderstanding, so thank you for the help
@Happy_Abe I do understand what youre saying and yeah that is one possible way, but the reflection law is something that is ALWAYS true. Hence the name "reflection LAW". We showed that when the beam of light passes through the focus point, the angle of incidence and reflection are the same. This is something that ALWAYS needs to be true. Since the angle of reflection is exactly the same as the angle of incidence when it passes through the focus point, changing the direction of the reflected beam will change the angle it makes with the surface which means the reflection law is not true anymore meaning those aren't possible solutions.
Is it located in a haunted Streichelzoo? Is it the creation of Italian reggae singer Rasta Primavera? Or built on a dare by 90-year-old club promoter Fuji Howser, MD?
Math teachers finally have an answer for that timeless question: "When am I ever gonna use this stuff?" 😏
"We know what beta is."
Points to camera
An easier method of doing that trig is to know that not only is cot = 1/tan, but tan(x) = cot(π/2 - x). It’s where the “co” comes from, same as the “co” in cosine and cosecant. “Co” stands for complementary and complementary angles add up to π/2.
I was using exactly that
@ I meant directly 😅 without having to navigate through odd and even properties of cos and sin.
@@PapaFlammy69 I think shauna meant that those are known trig identities and you don't need the sin cos shifting.
As an electrical engineer, I always assume alpha=alpha' for simple reflections, and it’s cool to see that the lines meet at (0, 1/2a), which is the focal point. It’s fascinating how math ties into real world physics like this. I remember when I first learned about such concepts, and a SolutionInn tutor helped me understand the theory behind these reflections much better. Great video, this is definitely one of those ‘aha’ moments.
MythBusters took on Archimedes "Death Ray" three times, first in 2004 (S2E16), again in 2006 (S4E7), and again in 2010 in a special episode featuring President Obama. They busted the myth all three times.
As an electrical engeneer, I would assume that alpha=alpha' (simple reflection) and then prove that all lines will meet in (0, 1/2a) which is the focal point. Great video.
Just beautiful... and the demonstration too
thx Andres :)
Why does the whole video sound like:
Given a ray that get reflected to the focus of a parabola, then the ray get reflected by the parabola (skeleton head)
WHATT you’re kidding me. I had to prove this exact reflective property for one of the questions on a Math Olympiad I did recently! My solution is really similar to yours
ohhhhhh, very nice task for a competition!!!!
@@PapaFlammy69 yeah, it was an awesome competition and I secured a silver medal🙏
@@PapaFlammy69 one of the other questions was "if you pick integers from the set {1, 2, 3, ..., 144}, what is the max. number of integers you could select before three of the integers correspond to sides of a triangle?". the solution is really beautiful and simple
This is assuming that the ray passes through the focus right?
I wasn’t sure if that’s what we were also trying to show here
This is showing that if the reflected beam passes through the focus point then the reflection law is satisfied, it's quite easy to visualise that if the beam passes above or below the focus point then the angle of reflection will be different to the angle of incidence meaning that the only way this law can be held true is if the reflected beam passes through the focus point which is what we wanted to show
@@harley_2305 but this just shows that given the reflected beam of light hits the focus point then the angles are the same. You would need a different proof to show the reflected beam of the vertical beam always does go to the focus in the first place since we used those coordinates to derive the angle. A priori I have no idea where the second reflected beam goes. We can take it as a fact, but we don’t it from this derivation. That’s what I’m trying to say but maybe I’m wrong and misunderstanding, so thank you for the help
@Happy_Abe I do understand what youre saying and yeah that is one possible way, but the reflection law is something that is ALWAYS true. Hence the name "reflection LAW". We showed that when the beam of light passes through the focus point, the angle of incidence and reflection are the same. This is something that ALWAYS needs to be true. Since the angle of reflection is exactly the same as the angle of incidence when it passes through the focus point, changing the direction of the reflected beam will change the angle it makes with the surface which means the reflection law is not true anymore meaning those aren't possible solutions.
This is actually true for all of the curves obtained from a degree 2 equation(ellipse hyperbola parabola circle) and its really amazing 😮
indeed!!! :)
wow
though it was hentai 😕
Does Khan Academy have a course on Trig-Fuckery? 🤔
nah
@@PapaFlammy69 Does Papa Flammy teach trig fuckery tho? Absolutely! Dad 1, Sal Khan 0
🤪🤪
Hey, Flammable Maths, what's New York's hottest club?
Is it located in a haunted Streichelzoo? Is it the creation of Italian reggae singer Rasta Primavera? Or built on a dare by 90-year-old club promoter Fuji Howser, MD?