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January 13

Coit tower

i am doing a project on coit tower and i need to know the exact dimensions. i need the diameter at the least. i already know the height.

thank you a student —Preceding unsigned comment added by 99.117.206.186 (talk) 02:28, 13 January 2011 (UTC)[reply]

Measure the proportions of the Coit Tower on the picture http://en.structurae.de/photos/index.cfm?JS=27571 and calculate the diameter. Bo Jacoby (talk) 07:45, 13 January 2011 (UTC).[reply]

Valid operations on cylindrical and spherical vectors

For typical rectangular vectors in R3, several operations are defined, particularly the dot and cross products. Generally, in Rn, the operations of vector addition and scalar multiplication, as well as multiplication by matrices are intuitive. For vectors defined in terms of cylindrical and spherical unit vectors, are the same operations valid? Scalar multiplication can be done by multiplying the length component, but a way to add cylindrical and spherical vectors without reverting to rectangular vectors seems lacking. Likewise, the dot and cross product, if performed with the same mechanics, don't seem to have the same physical meaning. —Preceding unsigned comment added by 68.40.57.1 (talk) 05:25, 13 January 2011 (UTC)[reply]

You would have to convert back to Cartesian coordinates first to use the normal formulas for all these operations on vectors. Of course, you could find complicated formulas for how to, say, add two vectors written in spherical coordinates, but I'm sure the result would be very complicated. To find the formulas, you would convert to Cartesian, perform the operation, and then convert back again. 82.120.58.206 (talk) 05:46, 13 January 2011 (UTC)[reply]

Time dilation limit

Resolved

So I was fiddling around with $\lim _{x\rightarrow b^{-}}f(x)={\frac {a}{\sqrt {1-{\frac {x^{2}}{b^{2}}}}}}$ on Wolfram Alpha by sticking in different numbers for $a$ and $b$ (e.g. here) and, empirically, it would appear that the function always tends towards infinity. However, how would one go about proving this limit, rather than sticking in numbers and seeing what happens? It Is Me Here ^{t / c} 22:28, 13 January 2011 (UTC)[reply]

As x approaches b,

{\frac {x^{2}}{b^{2}}}\rightarrow 1

, so the denominator goes to 0. Isn't that enough? --Tardis (talk) 00:09, 14 January 2011 (UTC)[reply]

Tardis is right, and his reasoning is what any working mathematician/physicist would use. However, it takes some training and experience to know just which arguments of this kind are actually valid. When in doubt, it always works to fall back on the definition. Confusingly, there's no actual definition in the one-sided limit article, but we have one at limit of a function#One-sided limits. Even that one does not consider the limit being infinity, but coupled with limit of a function#Limits involving infinity, we can define

\lim _{x\to b^{-}}f(x)=\infty

if for any

N

there is a

k<b

such that

f(x)>N

whenever

k<x<b

.

Proving that this definition applies is made a little easier in our case because our f is strictly increasing for x<b (which can be seen directly, or by differentiating it) -- we then only have to prove, given any N, that there is some

k<b

such that

f(k)>N

. One easily finds such an

k

in this case by assuming that N is "large enough" and solving

f(k)=N

for

k

. –Henning Makholm (talk) 00:30, 14 January 2011 (UTC)[reply]

Try first the special case a=1, b=1. Substituting $y={\sqrt {1-x^{2}}}$ simplifies your problem to $\lim _{y\rightarrow 0{+}}{\frac {1}{y}}$ . Bo Jacoby (talk) 06:21, 14 January 2011 (UTC).[reply]

Ah, OK. It's just that some limits don't seem obvious (to me, anyway) - e.g. 2.7 does not jump out at me when looking at

\lim _{x\to \infty }\left(1+{\frac {1}{x}}\right)^{x}

. It Is Me Here ^{t / c} 10:32, 14 January 2011 (UTC)[reply]

These non-obvious cases are usually ones in which the variable appears in more than one place. –Henning Makholm (talk) 10:37, 14 January 2011 (UTC)[reply]

- but in most cases the trick is to simplify to obvious limits ! Bo Jacoby (talk) 13:27, 14 January 2011 (UTC).[reply]

Thanks, all! It Is Me Here ^{t / c} 22:26, 14 January 2011 (UTC)[reply]