Two Neat Properties of Cardioids
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- Опубліковано 29 тра 2024
- We prove two properties of cardioids of the form r = a(1 + cosθ). Firstly, we show that all chords passing through the origin have the same length. Secondly, we show that the tangents to the cardioid at either end of such a chord are perpendicular.
00:00 Property 1
02:36 Property 2
02:51 Differentiating
06:29 Perpendicularity
10:12 Simplifying
These properties confirm I can't sketch cardiods.
Some nice trig identitis appear in the final step of the second property:
cos(t)^2 - sin(t)^2 = cos(2t)
2sin(t)cos(t) = sin(2t)
And then cos(2t) +/- cos(t) and sin(2t) +/- sint(t) can be transformed to products. It doesn't simplify the calculations but is interesting to use them. Great video
Entertaining as always
This begs the question... Is every curve satisfying these two properties a cardioid?
Exactly my question as well
The double angle identities were screaming at me the entire time. Don't know if they would have been useful and had made the calculations easier, but it was such a nice pattern:
cos(2θ)+cos(θ)
---------------------------
-(sin(2θ)+sin(θ))
This is a good point! I didn't manage to find a way to use these when simplifying, but it would be interesting to see if there is another way of doing the calculations with the double angle identities.
You can simplify to - cot(3 theta /2)
These will be fun properties to add to my son’s list of derivations. Thanks for posting!
At 6m15 imho an error with the minus sign as you pulled that minus out of the brackets so a plus is needed before the sin theta!
I think proving the perpendicularity using vectors is a bit more elegant.
First note that the tangent lines are perpendicular iff the tangent vectors d(x,y)/dθ = (dx/dθ, dy/dθ) are.
I'll use the double angle identities to simplify your expressions for dx/dθ and dy/dθ (as others have also pointed out):
dx/dθ = a(-sin(2θ)-sin(θ))
dy/dθ = a(cos(2θ)+cos(θ))
Then, evaluate them at θ+π:
dx/dθ[θ+π] = a(-sin(2θ)+sin(θ))
dy/dθ[θ+π] = a(cos(2θ)-cos(θ))
Then, the dot product of the two tangent vectors is
dx/dθ · dx/dθ[θ+π] + dy/dθ · dy/dθ[θ+π]
= a²((-sin(2θ)-sin(θ))(-sin(2θ)+sin(θ)) + (cos(2θ)+cos(θ))(cos(2θ)-cos(θ)))
= a²(sin²(2θ) - sin²(θ) + cos²(2θ) - cos²(θ))
= a²(sin²(2θ) + cos²(2θ) - (sin²(θ) + cos²(θ)))
= a²(1-1) = 0.
Thus, the tangent lines are always perpendicular.
This approach has the added benefit of also working when dx/dθ = 0.
Thank you very much for all of your efforts Dr🎉🎉❤❤
dy/dx = - (cos(2t)+cos(t))/(sin(2t)+sin(t))
dy/dx = -(2cos(3t/2)cos(t/2))/(2sin(3t/2)cos(t/2))
dy/dx = -cot(3t/2)
I used your calculations but simplified the result of this derivative
After simplification we will have
(-cot(3t/2))*(-cot(3(t+π)/2)) = -cot(3t/2)*(-cot(3t/2+π+π/2))
(-cot(3t/2))*(-cot(3(t+π)/2)) = -cot(3t/2)*(-cot(3t/2+π/2))
(-cot(3t/2))*(-cot(3(t+π)/2)) = -cot(3t/2)*(tan(3t/2))
(-cot(3t/2))*(-cot(3(t+π)/2)) = -1
Conclusion: Lines are perpendicular
Very nice, a neat use of the sum to product identities!
Does 1/(df/dx) always equal dx/df? My maths teacher insisted _ad infinitum_ derivatives are not fractions.
This does work, so long as df/dx ≠ 0, but it doesn't mean the derivates should actually be thought of as a fraction. If you're interested, I have an old video which explores why this property doesn't work for second derivatives: ua-cam.com/video/DCocnAxCFgk/v-deo.html
using angle sum for cos(theta+pi) is probably an overkill
Could've just use the period of cos function to get cos(theta+pi) is equal to cos(theta)
... -cos(theta) you mean?
Stellar explanations, really clear and approachable. Thank you!
What's the orthoptic of a cardioid?
I prefere r=r0(1-sin(deta)),
which I figured out somewhere at 2008, and the first time I found on the internet, was like 2018.
Or 2014
ua-cam.com/video/1VNlWXSJjN4/v-deo.html
Too much brain hurt. I know this is important in the 3d waves I'm looking at, but it's too much. The cycle doesn't repeat, but there are cardioids. Providing, I pick and choose the correct numbers.
The way you pronounce the trig functions gets on my nerves. Why not just say ‘cosine’ instead of saying ‘coz’? If you’re going to insist on ‘coz’ then you should say ‘sin’ too, not ‘sine’ as you do say.
Please just say ‘sine’, ‘cosine’, & ‘tangent’. You’ll make at least one channel fan happy.
he is British that is how it is done, don't ask what shine and cosh mean...
OK Karen