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6

HINT: $$y''(t)=\frac{a}{y(t)^2}\Longleftrightarrow$$ $$y''(t)y'(t)=\frac{ay'(t)}{y(t)^2}\Longleftrightarrow$$ $$\int y''(t)y'(t)\space\text{d}t=\int\frac{ay'(t)}{y(t)^2}\space\text{d}t\Longleftrightarrow$$ $$\frac{y'(t)^2}{2}=\text{C}_1-\frac{a}{y(t)}\Longleftrightarrow$$ $$y'(t)^2=2\text{C}_1-\frac{2a}{y(t)}\Longleftrightarrow$$ $$y'(t)=\pm\sqrt{\text{C}_1-...


3

The two expressions might look different at first glance but lead to the same thing in the end. Your expression evaluates as$$ \int_{a}^{t} (t-x) \rho(x)dx=\int_{t}^{b} (x-t) \rho(x) dx $$ $$ \implies \int_{a}^{b} (t-x) \rho(x)dx - \int_{t}^{b} (t-x) \rho(x)dx =\int_{t}^{b} (x-t) \rho(x) dx $$ $$ \implies \int_{a}^{b} (t-x) \rho(x)dx = \int_{t}^{b} (t-x) \...


3

The reason we run into problems is we have not really regularized our loop integral. We are dealing with an ill-defined quantity, and it is natural that we run into contradictions. The way to go is to regularize first, which in essence defines our object of interest, and then manipulate it. We will then see that some formal operations we use are justified, ...


3

I can give a heuristic derivation of the results found numerically by @RaymondManzoni and @Arentino. I will try to make things more rigorous the next days First of all we observe the invariance of the integral under the transformation $x \leftrightarrow y$ (i relabel $u\rightarrow x, v\rightarrow y $). Therefore we can write by symmetry $$ I_n=\color{...


2

The integral can be computed with Mathematica and the result (for $n>3$) is: $$ \begin{align} c_n=&{1\over n-3} \left({4^{-1/n}\over\sqrt\pi} \Gamma\left({1\over2} - {1\over n}\right) \Gamma\left({1} + {1\over n}\right) -{}_2F_1\left({1\over n}, {n-1\over n}; {n+1\over n}; -1\right)\\ +{1\over n-2}\ {}_2F_1\left({n-2\over n}, {n-1\over n}; {2n-2\...


2

In terms of polar coordinates, $(u^n+v^n)^{-1+1/n} \leq r^{-n+1} 2^{(-n/2+1)(-1+1/n)} = r^{-n+1} 2^{n/2-3/2+1/n}$. This is attained along $\theta=\pi/4$. This makes geometric sense: the $n$ norms with $n \geq 2$ are smaller than the $2$ norm, and the place where they are smaller by the largest amount is along the lines $y=\pm x$. We're dealing with a ...


2

Jacob, you probably won't like this, but the easiest way I see to do the calculation rigorously is to use differential forms. (One reference that's somewhat accessible is my textbook Multivariable Mathematics ..., but you can find plenty of others.) If $S$ is a closed (oriented) surface in $\Bbb R^3$ not containing the origin and $S^2$ is the unit sphere, ...


2

Yes, the expression you got is for the median, while the intent (I am guessing) was to balance the "counter-clockwise" torques (the left-hand side of your last equation) against the "clockwise" ones about $t$. The explanation for the discrepancy you (wisely) ask about is that the torque about $t$ exerted on the rod by point mass $\rho(x)$ is $(x - t)\rho(x)$...


1

By writing $$\frac{d\dot x}{dt}=\frac{d\dot x}{dx}\frac{dx}{dt},$$ you have assumed that $\dot x$ can be written as a function of $x$, so that you can apply chain rule. However, in the case of the pendulum, we cannot write $\dot x$ as a function of $x$. It is easy to see that the pendulum can be either rising or dropping if we are only given the position of ...


1

From the point of view of Mathematics, the following equation can be solved on an interval $[a,b]$ with $c\in[a,b]$, for all choices of constants $A$, $B$: $$ -\frac{d}{dx}\left(p(x)\frac{df}{dx}\right) +q(x)f(x) = \lambda w(x) f(x),\\ a \le x \le b, \;\;\; f(c) = A,\; f'(c)=B. $$ You can get by with very few assumptions on the ...


1

$$a=\frac{dv}{dt}=32 \\ \int_0^t \frac{dv}{dt}\ dt = \int_0^t 32\ dt \\ v(t)-v(0)=32t \\ v(t) = 32t+v(0)$$ But $v(0)$ is $1500\ ft/s$ in the negative direction (because the bomb is initially travelling upward with the plane and you chose downward as the positive direction). So $v(t) = 32t-1500$. Then integrate again to find $s(t)$.


1

Let $M$ be the plane of the mirror, $\bar{p}$ the image of $p$ under reflection in $M$, and $\bar{r}_{i}$ the image of $r_{i}$ under reflection in $M$. For each $i$, the points $p$, $\bar{p}$, $r_{i}$, and $\bar{r}_{i}$ lie in a plane orthogonal to $M$ (because the segments $p\bar{p}$ and $r_{i}\bar{r}_{i}$ are orthogonal to $M$), and the reflected ray from ...


1

For the projectile to be moving away, you require $|r|$ to be increasing, therefore $|r|^2=r\cdot r$ to be increasing Hence $$\frac {d}{dt}(r\cdot r)=2r\cdot\dot{r}>0$$


1

Consider the statement If I look at an individual particle entering the chamber at time $0$, it is travelling $Ce^{−At}$ metres per second after $t$ seconds. The constants $C$ and $A$ are the same for every particle. If $t=0$, the speed would be $C\cdot e^{−A*0}=C\cdot e^0=C$. Therefore, $v_i=C$ (Thanks @Semiclassical for pointing this out before). So ...


1

A well-known special case too long for a comment: If we consider point charges in the plane or line charges in space, the neatest choice of $\alpha$ is 2. In that case, if the $q_i$ are integers, you will find all stationary points of your force field as zeros of the derivative of the polynomial $$ p(z) \ := \ \prod_{i=1}^m (z - P_i)^{q_i} \, , $$ which ...



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