The semi circle is y^2+x^2=2, using the first two quadrants. R=sqrt(2). If x goes from 0 to 1, the angle from the x-axis 45 degrees to 90 degrees. Thats a difference of 45 degrees or pi/4 radians. The arc length is sqrt(2)pi/4.
I have a much simpler and neater solution: Rewrite the equation to get y^2 + x^2 = 2. The radius of such circle is root(2), when x = 1, y will also = 1. Therefore, the point(1,1) is on the circle and we can draw a triangle with base 1, height 1 and hypotenuse root(2) (from the radius and Pythagoras). We can then use simple trigonometry to determine that the line form from the base of the triangle and the radius is pi/4 radians, thus the angle for the arc is also pi/4 radians. Then, just use the r theta formula to get root(2)pi/4. Much simpler
I did the exact same thing but I made the mistake of considering the length of the arc in the 4th quadrant. So I got pi/root2😅. The ans is root2pi/4 tho
Solution: y = √(2 - x²) → y² = 2 - x² → y² + x² = 2 The y²-equation is a circle centered at the origin (0|0), that goes through (1|1), (1|-1), (-1|-1) and (-1|1), with the radius √2 (easily calculated by setting x=0). But since we only have a y-equation, we only consider the top half of the circle The important points for this task are (1|1) as well as the point, where the semi-circle crosses the y axis, so where x = 0: (0|√2) Given this information, we can deduce, that the arc between x = 0 and x = 1 is exactly 45° or 1/8 of the full circle. Therefore the length is 2πr/8 = πr/4 = π/4 * √2 or π/√8
Love the videos; keep it up. As others have said, the angle 𝜃 introduced for the substitution actually corresponds to the angle from the vertical axis (x = 0) growing clockwise rather than the angle drawn, which is from the horizontal axis growing anticlockwise. Because in both cases the angle ranges over an eighth of the circle, the arc length isn't affected
I was with you up until the Trig sub... I had always been taught to use x = asin(theta) as the sub... doesn't work here though...maybe I'm not seeing it. Is 1-2sintheta^2 a trig identity? Maybe that would be how I get to what you had doing it that way. But if I'm understanding what you're saying.. DONT take asin theta without splitting the fraction to make a = 1, correct?
Does that mean when I substitute I am then NOT putting the sub as trig sub/a but rather trig sub = x/a? then I solve for dx/du by THEN moving it to atrigsub and taking the derivitive. Sorry, finals week epiphany. I think I just discovered a way to take tackle it doing it that way.
Wait I just realized something. If I'm treating 2 as something squared, could I make 2 s.r.2 and set that equal to my a? I feel like this is close to what needs to happen I'm just really confused as to what happens when a =/= 1
Hmmm... x^2 + y^2 = 2 - that's a circle around the coordinate origin with radius sqrt(2). If we speak of an "arc" over a limited range of positive x - how exactly do we have to interpret this wording? Does it refer to the arc in the upper right quadrant? Or that in the lower right? Or does it refer to both? In the title text and picture, there is no explanation to that... So, for the moment, i tend to interpret that as a reference to BOTH arcs, since i don't see any wording that limits the scope of the coordinates. Then, as soon as we have the angle at which the arcs end, the length calculation is trivial as a part of the circumference of the circle. And the angle is obviously 45°. Thus, the length of both arcs is 2*45°/360° * 2*pi*r = pi / sqrt(2). ========== In the end, the "original idea of the inventor" gets explained during the run of the video: He only wanted the upper right quadrant being considered. That leads to half the length, OK. (It's always interesting to see what evolves from the problem presentation, since i always try to solve those problems before watching the video. More than rarely, it turns out that there are greater discrepancies between what the author of the problems meant and what the presentation of the problems allows the recipient to interpret.) And as always: All roads lead to Rome :D
Sorry, wait, I think I have it. The a^2 is -already- the 2. So if I take a (square root of a)^2 and factor that out, I get 1 - x^2/([s.r. a]^2) and then I can make -that- into 1 - (x/s.r.a)^2? Sorry hope that makes sense.
I made the common mistake of moving from x and y to the r and theta limits of integration. I can't believe it! I went from the common 0 but looking at the 1 for x integration upper limit I was thinking of π/2 rather than π/4 because I forgot it was Arcsin(0) to Arcsin(1) of radius √2 thoughts vs angle of 0 to where Arcsin(1) is π/2 for unit circle thoughts. Be careful and slow down on your math in the future because errors like mine are easy to do! 😬👎 As a physics in electricity and magnetism studies (not just in mechanical motion in aviation of length of particle paths) we have similar problems in a current flowing in a bent wire with a magnetic field vector B derivative of a current carrying wire or transmission line length L or integral of B dL to either figure out the current or magnetic flux strength B. Physics in electricity and magnetism definitely need Calculus coursework math to solve such problems in 2D and 3D space.
Your drawing at 5:07 doesn't seem right. The originalproblem is the from where x=0 to x=1. But you're talking about x going from sqrt(2) to 1? Shouldn't it be the arc starting where the circle crosses the y axis (x=0) to the right until x=1?? The answer comes out the same, since it happens to be symetrical ( 1/8 of a circle), but for more complex curves, I think the concept of integrating from 0 to 1 would be the drawing of the top-right eighth of the curve?
Just so long as you don't compute negative length in your limits of integration "same errors as limits of integrand in areas!". He is correct, a theta of a circle radius √2 going from theta equal 0 to theta equal π/4 as he figured out. Don't make the common answer errors in physics such as those in electricity and magnetism finding direction of magnetic flux vector B in the current flow vector I of current carrying bent wire of length dL. 😬😩
@@jumpman8282 ... I figured out what he did wrong ... He inverted the limits of integration between the x-y world and the r-theta world. Yes from x = 0 theta is π/2 and x = 1 theta is π/4 to figure out completely what you were saying. Yes it is true from x=0 is not x=radius=√2 and then x=1 to have in the x-y world to match with his r,theta polar coordinates equivalently. Because he corrected a wrong xy limits of integration setup with the correct r,theta limits of integration I also wasn't seeing the "from x=√2 to x=1 to theta = 0 to theta equal π/4" limits of the r,theta integrands. The rsin(theta) substitution integration part was correct ... Just the limits of integration had to be figured out to avoid a negative length answer like we do in areas under curves in the xy world.
He forgot about dy/dx = -x/(✓(2 - x^2)) about the minus sign. For y=0 to y=1 is what the integrands should be. There is a minus sign mistake in his xy domain L = -√2/(✓(2-x))dx part of the integral that would affect the proper setup of the way the limits of integrand should be to get a negative or like a limit of integration from y=1 to y=0!? ... This is where proper graphical interpretation is required because inverting limits of integration and minus sign mistakes are problematic. Slow down ALL proceed with caution! 😬👎 😂
If you did not memorize the formula dl ^2= (dx)^2 +(dy)^2) dl =sqrt((dx)^2 +(dy)^2)) Pul dx outside the square root becase we want to integerate in term of x dl = sqrt(1+(dy / dx )^2) dx And when integerate we get the formula
Nice, I really like this new version of the solving. Much clear and simplest than the previous ambiguous one. :) I really like to live and never stop learning. :)
13:20 The square root of cosine squared is not cosine but the absolute value of cosine. Now with the limits that absolute value is not needed because cosine is positive on the interval, so you lucked out...But there are arclength and surface area integrals where you may pass a point of "non-differentiability" where this becomes an issue
Not really. If you figure out, that it is a semi-circle, centered at (0|0) with the radius of √2, that goes through (1|1), you can immediately see, that the arc between x = 0 and x = 1 is 45° of the full circle, hence it is 2πr/8 = πr/4. With r = √2 you get √2 * π/4. Done.
