HallsofIvy

"I am a layman trying to understand above theorems. This could be a stupid question." There is no such thing as a stupid question! "Does these theorems imply that we actually cannot prove that 2+2 = 4???" Gosh, I may have to reconsider! No, Godel's theorem say that, given any set of axioms large enough to encompass the properties of the non-negative integers there must exist some theorem that can neither be prove nor disproved. It does not say that a specific theorem cannot be proved. In fact, if we were able to identify a specific theorem that can not be proved nor disproved, we can always extend the axioms, perhaps by adding that theorem itself as an axiom, so that theorem can be proved. Of course, there would then be still another theorem that cannot be proved nor disproved.

So you are arguing that NO ONE can criticize your paper because YOU did not include a "working roadmap" or "outline"?

For a "free projectile" (one that has an initial force but no continuing thrust), and ignoring air resistance or other friction, the acceleration is the constant -g, the acceleration due to gravity. Since the gravity, set into an "xy- coordinate system" with positive y upward and positive x to the right, is <0, -g>, the velocity, at any time t, is <vx, vy- gt> where "vx" is the initial velocity in the x-direction and "vy" is the initial velocity in the y-direction. Taking the initial speed to be "v" at angle $\theta$ to the horizontal, the initial velocity is $\left< v cos(\theta), v sin(\theta)\right>$ so we have the velocity at time t to be $\left<v cos(\theta), v sin(\theta)- gt\right>$ and, integrating that with respect to time, and taking the initial position to be (0, 0), the position at time t is $\left<v cos(\theta)t, v sin(\theta)t- \frac{1}{2}gt^2\right>. Now, suppose, after time t, the projectile is at point (w, h). That is, that the projectile is distance w from the initial point, horizontally, and at height h (both of which may be positive or negative). Then we must have $v cos(\theta)t= w$ and $v sin(\theta)t- \frac{1}{2}gt^2= h$. How we proceed depends upon exactly what the problem is. If we are given initial speed and angle of the projectile we can immediately calculate w and h. If we are given w and h and are asked to find the necessary angle and initial speed to achieve that, we can solve the first equation for t, $t= \frac{w}{v cos(\theta)}$ and put that into the second equation to get an equation, $w tan(\theta)- \frac{1}{2}\frac{gw^2}{v^2cos^2(\theta)}$ relating v and $\theta$ (there will be more than one correct answer).

I'm sure a silver plate would be sufficient!

And where did you read this?

When I was in college (in Cambridge, Mass. where it get cold in the winter!) I heard about (but never knew anyone who actually did it) putting fermented liquor in a plastic bottle and leaving it out over night. Since alcohol freezes at a much lower temperature than water (114 degrees C lower) in the morning you would have a block of ice with a liquid center. Use an ice pick (does any body have one of those any more?) to punch a hole to that liquid center and pour the high alcohol center into a container.

I am afraid that before anyone can answer that, you will have to say where you draw the line between "regular" and "irregular".

You include 18000 N for "dependent relatives" but I see nothing that says this man HAS any "dependent relatives"!

I see your point- but that depends on exactly what you mean by "has a derivative". I would say that f(x)= |x| certainly has a derivative. That derivative does not happen to be defined at x= 0. The original post did not say anything about the derivative being defined "for all x".

Why isn't it? The OP asked for an example of a function that is continuous but its derivative is not. y= |x| is continuous for all x and its derivative, y'(x)= 1 for x> 0, y'(x)= -1 for x< 0, y'(0) not defined, is not continuous at x= 0.

I have no idea what you are asking. Is your question about "Euclidean division (the "Euclidean algorithm") itself or about why it is named for Euclid?

Sorry, I misread the question.

I see no question here. What question are you trying to answer?

You can't! The form you give can only be used for quadratic polynomials in x and y while the polynomial $$3x^2y+ 4xy$$ is cubic. If this were simply $3x^2+ 4xy$ then you could. Treating (x, y) as the vector $\begin{pmatrix}x \\ y \end{pmatrix}$ and its transpose is $\begin{pmatrix} x & y \end{pmatrix}$ and writing Q as the generic 2 by 2 matrix $\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ and the vector a as $\begin{pmatrix}p \\ q\end{pmatrix}$, "$x^TQx+ a^Tx$" becomes $\begin{pmatrix}x & y \end{pmatrix}\begin{pmatrix}a & b \\ c & d \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}+ \begin{pmatrix}p & q \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}$$= ax^2+ (b+ c)xy+ dy^2+ px+ qy$. In order that this equal $3x^2+ 4y$ we must have a= 3, b+ c= 3, d= p= q= 0. Of course, b+ c= 3 does not tell us either b or c separately. The simplest thing to do (and what is generally done) is to require that Q be a symmetric matrix so that b= c= 3/2. Then $Q= \begin{pmatrix}3 & \frac{3}{2} \\ \frac{3}{2} & 0 \end{pmatrix}$. Then $3x^2+ 4xy= \begin{pmatrix}x & y \end{pmatrix}\begin{pmatrix}3 & \frac{3}{2} \\ \frac{3}{2} & 0 \end{pmatrix}\begin{pmatrix}x \\ y \end{pmatrix}$.

For example, if u(x, t)= x2+ t2 then u(., t) is the function that takes 3 to the function u(3, t)= 9+ t2, that takes 5 to u(5, t)= 25+ t2, etc. It maps a number to a function.

Given the x-y position of the mass you can calculate the lengths of the two springs and so the force they apply to the mass. Take the point at which the left spring is attached to the platform as (0, 0) so that the other spring is at (0.6, 0) (in meters). You say that "If the springs are coupled to each other without the mass they are at resting position". That position would be along the platform at (0.3, 0) so I will interpret that as saying that the natural length of the springs is 0.3 m. With the mass at (x, y) at rest, the length of each spring will be $\sqrt{x^2+ y^2}$ so the force due to each spring will be [tex]10(\sqrt{x^2+ y^2}- 0,3)[/tex]. At rest, the vertical component of force of the two springs must be mg= 2(9.81)= 19.62 N. The vertical component is proportional to the vertical component of the position vector, x. That is [tex]\frac{F_v}{10(\sqrt{x^2+ y^2}- 0.3}= \frac{x}{\sqrt{x^2+ y^2}}[/tex] so [tex]F_v= \frac{10x\sqrt{x^2+ y^2}- 0.3}{\sqrt{x^2+ y^2}}= 19.62[/tex]. It is clear from symmetry that the x component of the rest position is 0.3 m so that becomes [tex]\frac{3\sqrt{y^2+ 0.09}- 0.3}{\sqrt{y^2+ 0.09}}= 19.62}[/tex]. Solve that to find the y component of the initial position. That should be enough to get you started.

This is where you are misunderstanding. Force does NOT "equal a displaced object". The force is equal to the weight of the water displaced by the object. And "weight" is a force, of course. The volume of the displaced water is, of course, the same as the volume of the object. First, as I said above, it is NOT the "amount of liquid that was displaced", it is the weight of the amount of liquid that was displaced. Yes, they would, and they don't!

28 is straight forward if you know some basic facts: The midpoint of the line segment between (x_1, y_1) and (x_2, y_2) is ((x_1+ x_2)/2, (y_1+ y_2)/2). The slope of the line between those two points is m= (y_2- y_1)/(x_2- x_1). The line perpendicular to y= mx+ b has slope -1/m. The distance between points (x_1, y_1) and (x_2, y_2) is sqrt((x_2- x_1)^2+ (y_2- y_1)^2). Whoever gave you this problem expects you to know that!

"b) Consider when the "Mega-Drop" falls with acceleration. i) Compare the vertical forces acting on the passenger with those mentioned in (a)(ii)." There is still the "weight", the downward force due to gravity. If this were "freefall" there would be no upward force due to the seat but that isn't actually said. If there were some acceleration downward but not freefall the upward force due to the seat will be less than the weight of the passenger.