# sqrt( 1)

• ban ryu: alright then han sung. since you say you're that smart, solve this without using any calculator. find the arc length of three x times the square root of seven minus two from x equals zero to one half.
• han sung: *stares at equation while deeply thinking* well, there are two 'twos' in the equation- minus two and the denominator from the one half. the number two sounds like the letter 'u', and u is between the letters 't' and 'v' in the alphabet.
• han sung: but tvs aren't really relevant anymore because everybody has computers now, so it's kind like... if you have two tvs, what is it even-
• han sung: *out loud* four.
• ban ryu:
• han sung:
• ban ryu: ... that's right...

This is my trick for remembering trigonometric values. Forget memorizing those little charts or triangles, this is so much easier.

I learnt this in 10th grade from my favorite teacher (who wasn’t actually my maths teacher at the time) and I still used it for the rest of High School and will probably continue to use it in University.

So what do you do? You draw the chart in the picture. That one.

OK, but how do you read it? Like a table. (If you’re unfamiliar with radians, do not worry. It also works in degrees, just write: 0, 30, 45, 60, 90 instead.)

For example, if you want to find the sin(pi/2), you simply start at sin(x) and then move over until you’re at pi/2 (or 90 degrees). It says sqrt(4). Now you put that over 2, and you have it. Sin(pi/2) = (sqrt(4))/2, also known as 2/2, or 1.

If you want to find the cos(pi/4), you do the same thing: start at cos(x), move until (pi/4). It says sqrt(2), now put that over 2, and you have cos(pi/4) = sqrt(2)/2.

For tan(x), just take the sin(x) value and divide it by the cos(x) value!

So tan(pi/6) = sqrt(1) / sqrt(3) = 1/sqrt(3).

I find this so easy to remember because the first line starts at zero and increases, and the second line starts at 4 and decreases. It works for any values (yes, even ones above pi/2!) and it’s so simple. I encourage you to start by writing it out on every paper, when you’re studying, and on your exams.

The sines as numbers
• sin(pi/6): 1/2
• sin(pi/4): 1/sqrt(2)
• sin(pi/3): sqrt(3)/2
• sin(pi/2): 1
• sin(pi): 0
• sin(3pi/2): -1
• sin(2pi): 0

I sometimes forget how common the “simplify your answers” mantra is in math classes.

My students are always so confused when I tell them I don’t want them to “simplify” their derivatives after taking them.

I think requiring simplification is often defensible in a few situations. The big advantage is that it forces practice on the steps involved in the simplification—that is, requiring all answers to be “simplified” forces people to practice doing basic algebraic manipulations.

And for that reason, when I teach calculus I require trigonometric and logarithmic expressions to be simplified, so that people get used to knowing what ln(e^3) and sin(pi/3) mean.

Second, sometimes it’s useful to require people to jump through formatting hoops just to learn how to: the ability to deliver exactly what your boss/client/whatever asked for is actually pretty useful in life. But this also doesn’t need to be every assignment.

Third, simplifying answers can in some cases make problems easier to grade, because it requires everyone to put the answers into the same format.

But there are a few reasons I don’t like requiring simplification in most cases. One is that it often feels like busywork—if I’m testing your ability to take a derivative, and you can take the derivative, I don’t want to find myself taking off points because you didn’t describe it the way I want. (I’d really like my students to internalize the idea that (x-3)(x-2) and x^2-5x+6 are the same thing).

Second and related, “simplification” is often fairly arbitrary. The exemplar here is the high-school rule that you can’t have any radicals in the denominator. Is (sqrt(5)+1)/4 really simpler than 1/(sqrt(5)-1)? So if you’re docking points for lack of simplification

And third is the counterpoint of the third argument above: sometimes simplification makes things harder to grade. If you leave your derivative unsimplified it’s easy for me to check what you did and where you might have screwed up. I’d way prefer to grade quotient rules left in the form (x dy - y dx)/y^2 to grading ones with those things multiplied out and factored and canceled, because that reflects your actual thought processes and work better.

And god forbid I ask anyone to simplify answers to the massive seven-layer-deep chain rule problems I like to give.

Send this 2 to ur crush and tell her to graph it on google

5+(-sqrt(1-x^2-(y-abs(x))^2))cos(30(1-x^2-(y-abs(x))^2))

Related to the black hole question. What would an observer see as they looked away from the black hole as they approached the event horizon? And is there an equation that can tell you how your time changes dependant on the strength of the gravitational field?

If you were to falling into a black hole facing out, you would see the night sky bend and distort as if you were looking at a fun-house mirror, eventually collecting into a small circle. Here’s a simulation for what it would look like: https://gfycat.com/gifs/detail/UntidySpottedAdouri

There is a relatively simple equation to model gravitational time dilation, which is derived from the Schwarzschild metric. The equation is t0/tf = sqrt(1-(2GM/rc^2)), where t0 is the proper time, and tf is the coordinate time for an observer far away. Of course, if you wanted to then include how fast the object was traveling, the equation quickly becomes much more complicated.

I'm a 4th year Undergraduate student and today in class we broke the transitive property, since sqrt(1) is simultaneously equal to -1 and 1. So 1 = sqrt(1) = -1 but 1 does not = -1. This makes me SO MAD. I LEARNED THAT PROPERTY A DECADE AND A HALF AGO. WHY ISNT THE PEPPERONI ON THE EARTH

Simple, because any equation that utilizes a square root (that isn’t geometric and requires it to be positive) is labelled as +/- (like the quadratic equation). In technicality, the answer is that it isn’t equal to both of them seperately, but both of them together. Technically the same square root function is used in geometry (Pythagorean Theorem), but since you can’t have a hypotenuse that’s -5 feet long, the negative root is omitted from the answer

So tempting to put my favorite trolly limit problem on the major’s comprehensive exam. But I really shouldn’t.

(The problem in question is the limit as x goes to positive infinity of $x/ \sqrt{x^2+1}$. If you’re not sure why it’s so trolly, try using L'Hospital’s rule on it and see what happens—that’s what everyone tries first, and the result is great.)

But what if that was just the third duck trying to trick you?

there is the old trick about releasing three pigs in a school but numbering them one, two, four, so people keep looking for the “missing” pig

go big or go home! number them, 0, 1, e, sqrt(-1), imply that you’ve covered the complex plane with pigs,

What does y = .75x^(2/3) + sqrt(1-x^2) and y = .75x^(2/3) - sqrt(1 - x^2) equal?

Why do you have the same equation twice?

Building off my previous ask about what it would be like to be in a ship going at light speed to Alpha Centauri, which you pointed out was impossible, if a ship was traveling at exactly 99% the speed of light how long would a trip to Alpha Centauri take from the perspective of those in the ship taking into account the distance to the star, time dilation and length contraction?

That’s a fairly simple calculation. The equations we need to solve are:

(Time passed on Earth) = (Time passed on ship)*1/sqrt(1-.99^2)

(Distance to star) = (Time passed on Earth)*(.99*c)

Plugging in the numbers, the time passed on Earth would be 4.4 years, while the time passed on the ship would only be about 7.5 months. Of course, this isn’t taking into account that the ship would have to speed up and slow down, which is probably the biggest restriction in space flight, since humans can only survive so much acceleration. Thanks for asking!

what's the natural domain of h(x) = 1 / sqrt (3x+1) ???

7 and a half

sqrt((sqrt(1+(10^(-7)/(8.854187817*10^(-12)))^2)-1)/(sqrt(1+(10^(-7)/(8.854187817*10^(-12)))^2)+1))

another homework problem and the answer was 1 lol…. well that’s enough for tonight

Fact: According to legend, Saint Valentine refused to renounce his faith and on the 14th day of February was beaten to death with clubs and beheaded. Over his bones, Hallmark has made a fortune in stale chocolates and constant reminders that if you are single on this day, you’ll be single forever.

Fact: The word holiday stems from the term “holy day,” meaning a day of religious reverence. It turns out you can build an entire empire on sacred grounds, structures of cards and dying flowers. It turns out you can sell love in a package. I know this because one of my exes once opened her mouth and let the wasps fall out, a whole hoard of grievances about how a small handmade card wasn’t proof enough of how I felt - I wanted to explain that my love is not a roller coaster that climbs to the top on the fourteenth of the month, that I love steady and hard - but instead I sat with my head in my hands and a bottle of vodka by my side and apologized for ruining a day that was made up by a company.

It’s just that I find love in better places than in balloons and teddy bears. I find love in better places than romance movies. I find love, and maybe that’s what’s wrong with me: I don’t need a wedding ring to know that I’m happy.

I find love in the girl with hair like the sunset, always changing with the weather, a girl that picks up her soul and shares it with me no matter how many times I hurt her, I find love in her calm voice and her constant assurances that one day I will be better, I find love in her loyalty and friendship and honesty, I find love in her writing, I find love in our four-hour phone conversations that are only cut short because my battery is always dying, I find love.

I find love in a boy who is no longer with us but still brings a smile to my lips when I think of him, I find love in how his family still talks to me, how his sister is basically related to me, how once in a while she and I get drunk and cry a lot and feel better for it. I find love in healing, in getting over it. I find love in learning you can be hurt but still one day get better - that even if you’re not the same, you learn how to deal with being different.

I find love in a lady who can speak fluent spanish better than I can as a native speaker, I find love in how she holds herself together, I find love in how we have both stared into the darkness and neither of us will let the other one flinch, I find love in her hands, soft and stained with ink, I find love in her dreams that inspire me to find freedom, I find love in how she skins her knees but always stands back up again, I find love in how she opens her ribs for anyone who wants in because she is naturally trusting, I find love in how it doesn’t matter how many times other people will scar her, she always still opens her heart.

I find love in a boy who is all that I love and if he doesn’t know how I feel, I haven’t done my job well enough.

I find love in one of my oldest friends, a girl with blue eyes and brown hair and who carries the sun in the palm of her hands. I find love in how she knows I’m bad at connecting and still finds ways to bring me back home again. I find love in her endless cheer and incredible mind, I find love in how dance is really more of her language than it is of mine but at parties the two of us start doing combinations, I find love in her laughter that follows her like a train, how she finds light in the smallest things, I find love in her acceptance of working to make this society change. I find love.

Fact: Love does not come in a card. Love does not come in a basket. The harder you look, the more you will glance over it. Love is just a chemical imbalance, love is formulated as 5 + (-sqrt(1-x^2-(y-abs(x))^2))*cos(30*((1-x^2-(y-abs(x))^2))), x is from -1 to 1, y is from -1 to 1.5, z is from 1 to 6.

Fact: Somebody loves you. You are enough.

—  Fact: Valentine’s day is about love, not couples. You can still find love even if you’re single. /// r.i.d

Sorry, this a super hypothetical question but would tachyons need to have a negative mass in order to travel faster than light & if so how would they interact with the Higgs Field (if at all)? Thanks! :)

Good question! We can easily find the rest mass of a particle given its energy and velocity using the equation:

E = mc^2/sqrt(1-(v/c)^2)

If we state that v/c is greater than 1, then sqrt(1-(v/c)^2) becomes an imaginary number. If we want to still have a real energy, the rest mass must also be an imaginary number. This also implies that energy is negative, which creates a major problem.

In particle physics, particles take on a mass that lowers their potential energy in the Higgs field, as if they were a ball stuck in a valley. Since the energy of a tachyon is negative, the energy will continue going down to negative infinity like a ball rolling down an infinite hill. This would mean that a tachyon would continue making more and more tachyons until it fills up all of space in a process called “tachyon condensation”. Since the entire universe hasn’t been consumed by tachyons yet, imaginary-mass particles are probably impossible. Thanks for asking!

2

The Official Guide to the MCAT Exam says to know the graphical representations of these dependences …

@ = Proportional

1) y @ x
2) y @ -x
3) y @ sqrt x
4) y @ 1/x
5) y @ 1 x^2
6) y @ sin x
7) y @ cos x

All of theses ^ are in the beautiful dance moves picture of above so practice these dance moves and MCAT graphs won’t stand a chance ;)

proportional (Linear)
inversely proportional (slow curve)
exponential (fast Curve)
Ect.

If the imaginary number i=sqrt(-1) why don’t we have other imaginary numbers derived from functions other that radicals/roots?

I want c=arcsin(2) and g=ln(0) even though they’d almost certainly be useless

Copy and paste this into google: 5 + (- sqrt(1- x^2- (y- abs(x))^2))cos(30((1-x^2-(y-abs(x))^2))), x is from -1 to 1, y is from -1 to 1.5, z is from 1 to 5
4

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