area of triangle = 1/2 (a)(b)sin(dtheta), then assume a = b = f(theta), so (1/2) f(theta)^2 sin(dtheta), then use small angle approximation for sin(dtheta)=dtheta to get the final expression. Problem is that small angle approximation is (in the geometrical proof, although there are probably alternatives) derived from that triangle=sector limit that you don't like, but most people find small angle approximation acceptable
Wouldn't be easier just to split the shape from the origin to the curve above in many triangles and sum the area of each one of them? Using the formula to find the area of a circle's sector sounds inappropriate.
Ricardo Mazeto the problem is if you try to evaluate a bunch of triangles, you have two variables: base and height. Height is given by f(θ) but the base is not given by θ. This means you don't have a variable that you can increase linearly to infinity, which is what θ allows one to do. Of course you could write the base in terms of theta and height, but that seems like a long workaround.
This formula assumes that the curved line is mostly convex. What if you got a more concave line? A straight line is the most unbiased way. If you got the location of the vertex and it's angle, you can find where it crosses the curved line. Therefore you can split it in how many segments you want and then sum the area of all the triangles.
No, the formula does not assume it is convex, because it is a limit as the size of the angle of each section becomes infinitely small. Just like a Riemann sum does not assume a curve is mostly flat, because the size of the base of each section becomes infinitely small. Here are the calculations to infinity of the triangle method: Using an isosceles triangle of height r and the unique angle theta height=h=r=f(theta) adjacent=A=r base=b=2*Atan(theta/2) base=2r*tan(theta/2)=2f(theta)*tan(theta/2) SA=bh/2=f(theta)^2*tan(theta/2) To calculate the area under the curve lim n->infinity of sum from i=1 to n of (f(a+(b-a)i/n)^2*tan((b-a)/(2n)) Which I would not find a nice solution for because of the n hidden in the tangent, converging to 0 as n goes to infinity and not cancelling nicely with another n due to it being in the argument for tan. And the circle method: r=f(theta) SA=f(theta)^2*(theta)/2 to calculate the area under the curve lim n->infinity of sum from i=1 to n of (f(a+(b-a)i/n)^2*(b-a)/n)/2 Notice how (b-a)/n varies as inversely linear as n increases whereas tan(b-a)/2n does not. That is what I was describing as the reason for it being very challenging to do triangles, and why it is a "very long workaround." The trick to doing Riemann sums is to get a polynomial and the triangle method is not doing that in any obvious way.
Ok, but what if you decide to split the shape in, let's say, only 10 segments to speed up the calculation. That convex curve will make a small difference. Plus, Infinitely small segments would take infinitely forever to sum.
You know your a beast when you do math in sharpie.
PatrickJMT you're the man
thank you sir, your videos are soo helpful 👍
Great, I really appreciate your videos, thanks!
explanation using simple graphics gets the job done....Nice job!
do cal 3 vidz professor
tysm for showing me this I was soo confused about this derivation
i hope it made some sense :)
yup ,tysm once again
Isn't it true that the derivative of the area with respect to theta is always r/2?
you da real mvp
Thank you sir!
Mind blowing🗿👌
Can you also approximate the area with triangles instead of sectors of circles?
i don't see why you couldn't! try it and report back.
Massimiliano Tron a triangle an a sector of a circle with an infinitesimal angle are the same
area of triangle = 1/2 (a)(b)sin(dtheta), then assume a = b = f(theta), so (1/2) f(theta)^2 sin(dtheta), then use small angle approximation for sin(dtheta)=dtheta to get the final expression. Problem is that small angle approximation is (in the geometrical proof, although there are probably alternatives) derived from that triangle=sector limit that you don't like, but most people find small angle approximation acceptable
what is the maths book of USA board?
Do you ever take requests for topics? :o How do you decide what to explain next?
i take requests from people who are supporting me on patreon. otherwise, i do whatever i want and feel like doing.
Ahhhh the painful calc 2 days. Never want to relive that nightmare again.
cuando podran traducir su canal al español, al menos que lo subtitularan? , gracias. Espero una respuesta. Gracias
What's JMT?
I am left handed too
Wouldn't be easier just to split the shape from the origin to the curve above in many triangles and sum the area of each one of them? Using the formula to find the area of a circle's sector sounds inappropriate.
Ricardo Mazeto but a sector of a circle with an infinitesimal angle IS THE SAME as a triangle
Ricardo Mazeto the problem is if you try to evaluate a bunch of triangles, you have two variables: base and height. Height is given by f(θ) but the base is not given by θ. This means you don't have a variable that you can increase linearly to infinity, which is what θ allows one to do. Of course you could write the base in terms of theta and height, but that seems like a long workaround.
This formula assumes that the curved line is mostly convex. What if you got a more concave line? A straight line is the most unbiased way.
If you got the location of the vertex and it's angle, you can find where it crosses the curved line. Therefore you can split it in how many segments you want and then sum the area of all the triangles.
No, the formula does not assume it is convex, because it is a limit as the size of the angle of each section becomes infinitely small. Just like a Riemann sum does not assume a curve is mostly flat, because the size of the base of each section becomes infinitely small.
Here are the calculations to infinity of the triangle method:
Using an isosceles triangle of height r and the unique angle theta
height=h=r=f(theta)
adjacent=A=r
base=b=2*Atan(theta/2)
base=2r*tan(theta/2)=2f(theta)*tan(theta/2)
SA=bh/2=f(theta)^2*tan(theta/2)
To calculate the area under the curve
lim n->infinity of sum from i=1 to n of (f(a+(b-a)i/n)^2*tan((b-a)/(2n))
Which I would not find a nice solution for because of the n hidden in the tangent, converging to 0 as n goes to infinity and not cancelling nicely with another n due to it being in the argument for tan.
And the circle method:
r=f(theta)
SA=f(theta)^2*(theta)/2
to calculate the area under the curve
lim n->infinity of sum from i=1 to n of (f(a+(b-a)i/n)^2*(b-a)/n)/2
Notice how (b-a)/n varies as inversely linear as n increases whereas tan(b-a)/2n does not. That is what I was describing as the reason for it being very challenging to do triangles, and why it is a "very long workaround." The trick to doing Riemann sums is to get a polynomial and the triangle method is not doing that in any obvious way.
Ok, but what if you decide to split the shape in, let's say, only 10 segments to speed up the calculation. That convex curve will make a small difference. Plus, Infinitely small segments would take infinitely forever to sum.