Recent content by dimitri151

Goodness gracious, now I think it IS correct again! Aaarg.. Stop me where I go wrong: ##(p\rightarrow q)\rightarrow r## ##=(\neg p\vee q)\rightarrow r## ##=\neg(\neg p\vee q)\vee r## ##=(p\wedge \neg q)\vee r## If ##p=a\geq b## ##q=a\geq c## ##r=b\geq c## Then ##(a\geq b\rightarrow a\geq...

Ah, it's a tautology in the new form by PeroK: ##(a>b\Rightarrow a>c)\Rightarrow(b\geq c)## ##=[\neg (a>b)\vee (a>c)\Rightarrow(b\geq c)## ##=(a\leq b\vee a>c)\Rightarrow b\geq c ## ##=\neg(a\leq b\vee a>c)\vee b\geq c## ## (a>b\wedge a\leq c)\vee (b\geq c)## ##=b<c \vee b\geq c## Thanks for...

Summary:: To prove a conditional statement on a pair of inequalitites. Mentor note: Moved from technical forum section, so the post is missing the usual fields. I feel it should be possible to prove this but I keep getting lost in the symbolic manipulation. Theorem: If a>b implies a>c then...

I'll recapitulate. My original question was about this in Fundamentals of Astrodynamics, by Bate (pg15): "in general ##\boldsymbol{a\cdot\dot{a}}=a \; \dot{a}##." Here's a picture of the page: ************************************************************************************...

I think it's not the length of the time derivative of ##a##, but the time derivative of the length of ##a##.

Yes, r, ##\theta## functions of t. See first line of equations in #10. I believe that solves it.

Yes, was missing parentheses. I'm taking the length of ##\vec a## and multiplying by the derivative of the length of ##\vec a##.

Ok, got it. ##\boldsymbol{a\cdot\dot{a}}=a\dot{a}## means ##\boldsymbol{a\cdot\dot{a}}=\vert a\vert \dot{\vert a \vert}## not ##\boldsymbol{a\cdot\dot{a}}=\vert a\vert \vert \dot{a} \vert## Then if: ##\boldsymbol{a}=x \cdot \hat{i} +y \cdot \hat{j}##, ##\boldsymbol{\dot{a}}=\frac{dx}{dt} \cdot...

To me its clear ## \vec u \cdot \vec v =\vert u\vert\vert v\vert## implies ##\theta=0##. None the less, I can't find the hole in the proof in #7 which says ##\vec{r}\cdot\frac{d\vec{r}}{dt}=r \frac{dr}{dt}## for a general vector ##\vec{r}=r \cos\theta \cdot \hat{i}+r \sin\theta \cdot...

I thought this as well, but look at this. Let ##\vec{r}=r \cos\theta \cdot \hat{i}+r \sin\theta \cdot\hat{j}##. Then ##\frac{d\vec{r}}{dt}=(\frac{dr}{dt} \cos \theta -r \sin\theta \frac{d\theta}{dt})\hat{i}+(\frac{dr}{dt}\sin\theta +r \cos\theta \frac{d\theta}{dt})\hat{j}##. And ##...

Ok, solved. If you let ## \vec{r}=r \cos\theta \cdot \hat{i}+r sin\theta \cdot\hat{j}##, then calculate ##\vec{r}\cdot \frac{d \vec{r}}{dt}## you do indeed get ##r\cdot \frac{dr}{dt}## (scalar).

A more complete quote of the passage: "1. Dot multiplying equation (1.3-4) by ##\dot{r}## ##\dot{r}\cdot\ddot{r}+\dot{r}\cdot \frac{\mu}{r^3} r=0## 2. Since in general ##\boldsymbol{a\cdot\dot{a}}=a\dot{a}##, ##\boldsymbol{v=\dot{r}}##, and ##\boldsymbol{\dot{v}=\ddot{r}}##, then...

They're not including the cos of theta. ?

I'm reading Fundamentals of Astordynamics by Bate, Mueller, White and having trouble with this passage (pg15): "2. Since in general a⋅a' = a a'..." I don't think that this is the case. For instance in uniform circular motion r⋅r' = 0. Would appreciate if anyone has some insight into this.

How do I begin proving: sum(k>=1)8/(k^4+4)=pi*coth(pi)-1? I got this from Mathematica. Thanks in advance for any help.