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Two other approaches (both already exploiting the vertical symmetry so the whole area is just the right half of 2/3): First: compute the bottom half of the area by decomposing into a rectangle of sides c and sqrt(1-c) added to the area under the curve (to the x-axis) between x = c and x = 1. So that's c*sqrt(1-c) + ∫(1-x^2)dx between c and 1. Equate that to 1/3, solve for c. Second: work in terms of strips from the curve to y-axis. So ∫xdy between y = c to y = 1, change variable, ∫xd(1-x^2) = ∫x(-2x)dx = ∫-2x^2dx between x =sqrt(1-c) to x = 0. Again, equate to 1/3, solve for c.

Can you plz explain the topic of banking of road? my physics textbook has this topic that gives you a safe turning speed at a sprcific bank angle ,{related to application of centripetal force } its really confusing and there is no video on yt that could help me

What i do intuitively is that when you have integrate f(x) w.r.t x , we have the formula , but we can't integrate f(2x) w.r.t x , so we scale the dx by a factor of 2 and divide by 2 . Now integral of f(2x) w.r.t to 2x is possible.

The second part is a new trick for me 👍

Very good example! Thank you ❤❤❤

What would you do here if you were given the maximum velocity (at angle 0) and needed to find the tension in general as a function of the angle? Also, if you tried to do this without involving conservation of energy, what would that look like?

hey is the link up for that uncertainty item you said you'd link to? This helps a ton!!

Babe wake up zaks lab new upload

Great videos! I learn so much

This is a great way to show the methods of calc II, but it is not actually a proof, since the values of sine (which are used in the end when you plug the 1 and 0 to the inverse sine function) depend on the fact that π is the ratio between the circumference and the diameter in the first place, so it is kind off circular reasoning. Great content to someone who is studying calculus nonetheless.

this is brilliant, this should have way more views,,, beautifully explained

This is pretty damn interesting.......

What justifies the implicit assumption that the speed is the same in each of the two exit pipes?

Love form india

so it's the same concept even if it's double angle

lm already here ,and that's a 👌🏽 from me

u made it years ago , buy it helped a soul today appreciate u man

Sir, I was wondering if you could recommend the software you use for writing, recording, and video editing? Your insight would be greatly helpful.

Great video and explanations. Thank you.

yo fire video

Thx sir, love from india.... I came here after your 3 years old video of spring mass problem thx for that too it helped a lot and you also got many informational videos of my interest:) I would come here again after 2 day to get info about some problems on magnetic field from your channel :)

Brainrot I

You can also take u^2 = x^2 + 1, then udu = xdx, so you get 3udu/sqrt(u^2) and if you change boundaries appropriately 0 -> 1, 2 -> sqrt 5, the integrand simplifies to 3du, so the solution is 3(sqrt 5 - 1).

Thankss, its really helpful :)

those who struggle you can just simplify the equation with trigonometry. cos(2x) = 2cos^2(x) -1 sin(2x) = 2sin(x)cos(x) so you get: 2sin(x)cos(x) 2sin(x)cos(x) sin(x) ------------------------- = --------------------------- = ------------ = tan(x) 1+ 2cos^2(x) -1 2*cos(x)*cos(x) cos(x) the integral of tan(x) is: -ln(abs(cos(x)) -ln(cos(pi/4)) + ln(cos(0)) = -ln(sqrt(2)/2) + ln(1) = ln((sqrt(2)/2))^(-1)) =ln(2/sqrt(2)) = ln(sqrt(2)) = (ln(2))/2

What about the diameter of the valve opening?

To add, inverse cotangent could have been used, which would not have necessitated the flipping of the bounds of integration but as I've realized, just made intuiting the answer more difficult😢

To add, inverse cotangent could have been used, which would not have necessitated the flipping of the bounds of integration but as I've realized, just made intuiting the answer more difficult😢

Greetings, nice short. I use Maxima a lot.

Cool man, Nice explanation

Sooo its A(d/dx 1+cos2x) +B= sin2x .....am i right

Good video! good explanation, that cleared up some inconsistencies for me.

Nice work.

Why doesn't the first one work?,please explain

thank you so much!

❤

Super awesome!👍

Investigation Jags Ch Unfinished. Copenhagen. V? (32 49.36 40.40 411.44 42.48,😍)

Neat stuff!

Given I have a number of measurements x1,x2,…,xn, I get that to get the uncertainty of the mean related to these uncertainties I’ll have to calculate the uncertainty of the sum, and then divide by n, i.e. sqrt(deltax1^2, deltax2^2,…deltaxn^2)/n. However I am interested in the entire uncertainty of the mean, including the statistical uncertainty (i.e. standard deviation/sqrt(n)) and not just the measurement uncertainty. My understanding is that here I need to again apply the sum formula to add them both together, i.e. sqrt((standard error of mean)^2+(propagated measurement error of mean)^2), am I getting this correctly?

how would this work as a definite integral? I can only find this kind of problem as indefinite

Walker Kevin Rodriguez Dorothy Williams Jeffrey

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That was explained so beautifully, great visualization! Good job!

He just showed how to derived those formulas. Good if you're getting your masters but its too complicated for students. Just memorize those formulas and you're good to go.

because gravity is aiding to the compression of the spring via PE, wouldnt it be a positive number? isnt d just a scalar?

Can we solve it if time is unknown and Velocity is unknown as well but height at end is defined?

Thank you so much I would have died without this 😭

i want this code

Thank you so very ❤