Posted on 24 February 2017 - 06:09 PM

Nessie

Posted on 24 February 2017 - 09:43 PM

what is calculus and how do u do this
http://aleph0.clarku.edu/~ma120/FTC.jpg
(taken from google)

no im not joking btw, havent done calculus yet in school and dont care to learn it online, and everyone seems to claim calculus is hard, so we'll see about that!!

most importantly, what the hell does that fancy symbol mean

Posted on 24 February 2017 - 09:51 PM

Am I smart

Posted on 25 February 2017 - 12:02 AM

Calculus is the study of rates of changes and infinitesimals. Isaac Newton (and Gottfried Leibniz) came up with calculus as a tool to solve physics problems. I think he called it the study of "fluxions" or something like that.

There are two main branches of elementary calculus: differential calculus and integral calculus.

If you've ever had any physics, you've seen rates of changes as delta(x)/delta(y) or something like that, and they represent "change" and rates. Differential calculus makes this more concrete and useful.

The classic example is finding the slope of a curve. I'm sure you've done problems where you had to calculate the slope of a line (m = delta(y)/delta(x)). But how does one find the slope of a curve?

Well, first of all, a curve most likely doesn't doesn't have the same slope at every point. If you use the slope formula between two points, you're getting an approximation for the slope at a point, but not the exact slope. Calculus allows one to actually find the slope at a point by making the delta(y) and delta(x) so small that it essentially zooms in on a point so much that it resembles a line. That's the derivative: dy/dx (or f'(x) for a function f(x)).

The fundamental laws of physics are often given in terms of derivatives, such as Newton's Second Law: F = ma = m (d^2y/dt^2), where you're taking TWO derivatives of your position (y) with respect to t. So if you have a function for your position, you can take the derivative twice and find what the acceleration and the force is.

An example is the derivative of y = x^2: dy/dx = 2x. I challenge you to go on any point of that curve and plug in the value of x, and check that the slope of the parabola at that point is 2x.

Another application of calculus is the integral, which is that symbol that looks like a skinny S. Integral calculus allows one to find areas under curves. One method of doing this is create a whole bunch of little rectangles under the curve, making them so skinny that it actually gets to the area of the curve (since it's easy to calculate the area of a rectangle). As you make them skinny, you take their width to dx while their height remains the value of the function at that point. When you have an infinite sum like this, you use an integral!

The picture you listed there is the fundamental theorem of calculus. There's more to integrals than just finding areas. The fundamental theorem of calculus relates derivatives and integrals together by saying that for a function f, the integral of it f's derivative is just f. So similar to how functions have inverse functions, if you take the integral of a derivative, you get back out the original function. And this is incredibly useful, because often you'll be faced with solving differential equations (equations involving derivatives, where you have to find what the function is), and to do this, you're going to end up having to integrate (take the integral) at some point to get out your function you want.

This is a completely non-rigorous explanation of calculus, but I hope it shed some light on what calculus is all about. I personally use calculus in physics daily–it's that essential. So learn it!

Posted on 25 February 2017 - 12:02 AM

idk r u

Posted on 25 February 2017 - 04:49 AM

What's 6 to the power of 300?

Posted on 25 February 2017 - 06:30 AM

big number

Last edited on 25 February 2017 - 01:37 PM by Legendcaleb

What are the applications of polar form of complex number (z) ? I get how to do them I just don't know the application.

Posted on 25 February 2017 - 02:58 PM

https://gyazo.com/b07ee73cc62e1fa88984dc49284976ca help me

Last edited on 25 February 2017 - 03:04 PM by AlbertEinstein

It's a lot easier to manipulate complex numbers in polar form.

For example, which do you think is easier: multiplying (34-(59/32)i)(25+(3/2)i) or 3e^(i*pi) 4e^(i(pi/4)) (These are not equivalent, but I'm making a point with this example)?

I'd say the second one, because the moduli multiply together quickly to get 12 and then you just add the exponents. If you meant rcos(theta) + i rsin(theta) = z, then that's the equivalent statement of writing z = re^(itheta)–not sure if you've seen the latter.

Exponential functions and trigonometric functions are innately related by that relationship. In fact, you can always write trig functions in terms of exponentials:

sin(x) = (e^(ix) - e^(-ix))/2i; cos(x) = (e^(ix) + e^(-ix))/2.

Again, this makes computations a lot simpler since you don't have to deal with products of trig functions–just simple exponentials.

Another reason I can think of polar form being useful is because of the geometric information it immediately gives you. Right of the bat, you know the modulus (length) of the complex number and how it's rotated in the complex (Argand?) plane. This gives a geometric interpretation to, for example, multiplying complex numbers: you rotate it and make it longer or shorter.

Having the exponential e^(ix) (often called a phase) in there with the modulus r is useful in quantum mechanics. Quantum mechanics necessitates complex numbers. Physical meaning rarely comes from absolute phases, but rather from from relative phases and the modulus. Having it written in this form, like I said earlier, makes computations easier and gives better intuition than just Cartesian coordinate mappings of x+iy.

Posted on 25 February 2017 - 03:12 PM

I have to stretch myself to recall my geometry knowledge.

Two sets of congruent angles that come to mind are alternate interior angles and vertical angles (correct me if I gave the wrong names to these angles).

Alternate interior angles are the ones made by a line intersecting two parallel lines. In this case, I see that 5 and 9 are in this category, and as a result, 22 is as well.

Vertical angles are angles that are formed by intersecting lines. In this case, the vertical angle to 1 is 3 and is hence congruent. But we can apply that same logic to any of the other angles we found to be congruent to 1 by transitivity of congruence. Hence, 7, 11, and 19 are also congruent to 1.

I think line E is just there to throw you off for this problem. Again, correct me if I'm wrong.

Last edited on 26 February 2017 - 02:00 PM by ThatOneCombo

https://cdn.discordapp.com/attachments/242290205585965057/285485923062382593/20170226_135846.jpg

if you could, could you work this out on paper cuz its kinda hard to read it on the forums with messy notation

Posted on 26 February 2017 - 04:43 PM

This doesn't have a lot of wacky notation, so I'm just gonna type it out and hope for the best.

This is an ideal application of using Newton's Second Law for rotations, i.e., for torques.

a.) Assuming that there's no tilt to the biceps muscle (I'm assuming there's not because there's no angles or vertical distances given), we'll say that it's a completely vertical force going up (call it B for biceps).

(Note: there's also a normal force from the interaction at the elbow joint between the humerus and radius/ulna, but it doesn't contribute anything to the rotational dynamics of the system since it has a zero lever arm.)

We define counterclockwise inducing torques to be positive and clockwise inducing torques to be negative. Thus, there's a negative torque for both the center of mass of the forearm and for the mass.

Since we're in equilibrium, the sum of the torques must be zero. Then,

sum(torques) = 0 = B(0.05m) - (12N)(0.16m) - (6.0kg)(9.81 m/s^2)(0.35m). Solving for B, we have B = 450 N vertically.

b.) This is the same problem, but now we aren't in equilibrium (calling new force on biceps B'):

sum(torques) = I(alpha) = (0.80 kg*m^2)(5.0 rad/s^2) = B'(0.05m) - (12N)(0.16m) - (6.0kg)(9.81 m/s^2)(0.35m).

Notice that we made alpha positive since we defined counterclockwise (up) to be positive.

Solving for B, we have B = 530 N vertically. So the process of raising rotating the arm actually increases the stress on the biceps in this model.

Last edited on 28 February 2017 - 08:11 PM by Soulero

all of these are graphing inequalities

A farmer wants to create a rectangular pen for his hogs. The fence should be at least 60 feet long and the distance around it should be no more than 310 feet. What are the possible dimensions?

Steve is currently looking to buy audio books and songs for his iPod. Steve wants at least one, but no more than 4 audio books on his iPod. He also wants more than 10 songs. Each audio book costs $14 and each song costs $1.29. Steve has $63 to spend on books and music. What are the possible combinations of audio books and songs Steve can buy?

Joe needs at least 30 hot dogs and 30 buns for a cookout. Packages of hot dogs cost 3.20 and contain 10 hot dogs. Packages of buns cost 2.50 and contain 8 buns. The maximum amount of money Joe can spend on hot dogs and buns is $38. What are the possible combinations of packages that Joe could buy?

@AlbertEinstein

Posted on 28 February 2017 - 11:58 PM

I'll write out the inequalities for each one, but I'm not going to graph them. Remember that strict inequalities are dotted lines, and the others are normal lines.

1.) Let l be the length of the fence, and let w be its width. Of course, l,w > 0. Now, I'm actually not sure how to interpret this problem. The wording makes it seem as if how "long" it should be is different from the "distance around it." I'll present both of my interpretations of this.

If it means long as in length of the fence, then 310 >= l >= 60 and l >= w (length vs. width), and 2(l+w) <= 310 (perimeter of a rectangle).

If it means long as in perimeter, then 60 <= 2(l+w) <= 310 (this is two inequalities–graph both) along with l >= w and l,w > 0.

2.) Call audio books B and songs S. Immediately, we have 0 <= B <= 4 and S > 10. We can find the cost inequality by multiplying B and S by their respective prices: 14B + 1.29S <= 63. Graph these inequalities and look at the integer quantities in the shaded region (no half books or quarter songs) for the solutions.

3.) Call PACKAGES of hot dogs h and PACKAGES of buns b. Since each h has 10 hot dogs, and each b has 8 buns, we have

10h >= 30 and 8b >= 30.

This problem is a bit more interesting, because now we can't purchase individual units of hot dogs and buns–they come in packs. That's why instead of individual hot dogs and buns as our variables, we're using packages.

The cost function of our system in inequalities is 3.20h + 2.50b <= 38.

Now, ideally, we'd like to have the same number of hot dogs as buns, but I'm not sure if the problem wants this much thought.

If we wanted to find equal numbers of hot dogs as buns, after graphing and finding our possible solutions, we simply require

h/10 = b/8, or 4h/5 = b (h,b must be integers!), since this equates the number of hot dogs to buns. Whichever points from our graph satisfy this relationship will give us the ideal situation where we have equal numbers of hot dogs and buns.

Graphically, this is just looking for integer solutions on the line b = .8h that lie in our shaded region.

Posted on 01 March 2017 - 01:18 PM

Automatically Deleted

Last edited on 02 March 2017 - 01:57 AM by AlbertEinstein

It's not often that I have to deal with number theory problems, so I had to get a refresher on some useful inequalities.

One site led me to a useful manipulation the Cauchy-Schwarz inequality (useful in quantum mechanics actually): http://math.stackexchange.com/questions/130773/find-bound-for-sum-of-square-roots.

Disclaimer: These proofs are in no way formal.

I'll do b.) first

Using Cauchy-Schwarz with the second vector as (1,1,…,1) and the first as (sqrt(x1),…sqrt(xn)) with xi = Ci, we obtain

[sum_{i=1}^{n} (sqrt(x_i))]^2 <= (x1+ … + xn)(1+…+1) = n(x1 + … + xn).

Square root both sides and call the sum on the left (the one we're interested in) S:

S <= sqrt[n(x1 + … + xn)].

The binomial theorem gives a quick method to calculate the sum of binomial coefficients:

sum_{i=0}^n (n i) 1^n 1^(n-i) = 1 + sum_{i=1}^n (n i) = (1 + 1)^n = 2^n.

Thus, x1 + … + xn = 2^n - 1.

Finally, S <= sqrt[n(2^n-1)], as desired. QED.

For the first part, let's start at the result of the last part:

Let's start back at S <= [n(x1+ … +xn)].

We'll use Young's inequality for p = 2, q = 2, which can be given intuitively by the following:

Suppose for a,b positive integers, we will have (a-b)^2 >= 0. Then a^2 - 2ab + b^2 >= 0 and thus, ab <= (a^2 + b^2)/2.

Thus, after writing our sum for x1 + … + xn as 2^n - 1, we have

S <= (n + 2^n - 1)/2 = 2^(n-1) + (n-1)/2, as desired. QED.

Last edited on 02 March 2017 - 03:05 AM by Prisoners

So you're more than 100 years old? Good one retard, 19th century is 1800's.

Also is your life this shit you have to do resort to doing maths problems for people on minecraft?

Last edited on 02 March 2017 - 03:24 AM by AlbertEinstein

Hello, Prisoners.

You missed the point of that. That was a bit of humor on the fact that Albert Einstein (Ever heard of him?) was born in the 1800s, or the 19th century. I thought most people would understand that, but I'm glad I could clear it up.

And that's an interesting take on it, indeed. Doing "maths problems" is entertaining for me, and I enjoy doing it. And given the high activity on this post, it's apparent that others are glad this thread is here.

No need to be hostile, little guy.



Looking back, you have actually commented on this post before.

Are you upset that I didn't do your midterm or something?

Posted on 02 March 2017 - 03:26 AM

nah bro i aced it all g :)