Question

will fan and medal
Find the curvature of r(t) = (12sint,5t,12cost) at the point (0,5pi,-12)

Answer 1

i used second formula in http://tutorial.math.lamar.edu/Classes/CalcIII/Curvature_files/eq0005P.gif

Answer 2

Yeah, that should work

Answer 3

i have an answer and would like someone to verify my answer is correct

Answer 4

Sure, can you please show all your steps though.

Answer 5

my answer came out to 12

Answer 6

Well that's not very useful to me, if you want to check your steps post them.

Answer 7

i dont want to type it all out. i was hoping someone would work the problem to check

Answer 8

We're not here to do your work, but we are willing check YOUR work.

Answer 9

ok ill wait for someone who is capable of doing the problem

Answer 10

What makes you think anyone here is willing to type it all out? At the very least, you could show the tiniest bit of effort to perhaps *show* you've tried.

Answer 11

i do not anyone to type it out. Just to work the problem. i am not a very fast typer and suck with the equation button. i showed the formula i was using

Answer 12

its an exam review with no answers. I was hoping someone would work the problem and say yes that is what i got or no that is not what i got

Answer 13

Grr I'm a lil rusty on these.
What is T again?
\[\Large\rm \hat T=\frac{\vec r'}{||\vec r'||}\]Like that or something?

Answer 14

Tangent

Answer 15

Mmm ok I'll try using that one then. >.<
In case we end up with different answers, then we can maybe back up and see where we went wrong.

Answer 16

ok sounds great

Answer 17

Might take me a few minutes >.< ahhh the rust is real

Answer 18

Checking with Mathematica, the answer is not \(12\).

Answer 19

If you can show your steps, you know we can pinpoint where you made the mistake...otherwise seems you just want a direct answer.

Answer 20

that strange that i would get perfect squares in my simplification then. What was it?

Answer 21

ohh wait i got my mistake

Answer 22

\[\Large\rm \vec r=\left<12 \sin t, 5t, 12 \cos t\right>\]Then,\[\Large\rm \vec r'=\left<12 \cos t, 5, -12 \sin t\right>\]and,\[\Large\rm \vec r''=\left<-12 \sin t, 0, -12\cos t\right>\]

And you were using the second method?

Answer 23

i did not cube the bottom was it 12/169 ?

Answer 24

yes i was using the second method and that is what i got for the r values

Answer 25

@SithsAndGiggles was it 12/169

Answer 26

No, but it's certainly closer. You're off by about 0.03. Have you attempted using the formula on the left?

Answer 27

if im off by .03 then maybe mathmatica is a approximation instead of exact?
i am not using a calculator

Answer 28

Yah I hate doing that stupid cross product >.<

Answer 29

i agree lol

Answer 30

\[\begin{align*}\vec{T}(t)&=\frac{\vec{r}~'(t)}{\|\vec{r}~'(t)\|}\\\\&=\frac{\langle 12\cos t,5,-12\sin t\rangle}{\sqrt{12^2\cos^2t+25+(-12)^2\sin^2t}}\\\\&=\frac{\langle 12\cos t,5,-12\sin t\rangle}{\sqrt{144+25}}\\\\
&=\frac{1}{13}\langle 12\cos t,5,-12\sin t\rangle\end{align*}\]
Is this what you have?

Answer 31

i never calculated the tangent but yeah that looks right

Answer 32

Hmm I came up with 12/169 also :P darn

Answer 33

i think that is probably right then. did you do the first formula?

Answer 34

Ya, I kept screwing up doing that cross product, so I switched to the first one.
Wolfram seems to agree. :O
https://www.wolframalpha.com/input/?i=curvature+of+r%3D%3C12+sin+t%2C+5t%2C+12+cos+t%3E+at+t%3Dpi

Answer 35

wow thanks i did know how to type that in to wolf, that would of saved a lot of time

Answer 36

thanks for the help

Answer 37

Ya, whenever you're not quite sure how to type it, try using a lot of "words" instead of symbols. usually works :O

Answer 38

ok i tried typing the vector into it but i never typed curvature. good to know. thank as well sith

Answer 39

Aw Siff >.< you were gonna show up the magic lol

Answer 40

Doing both methods, I didn't see a time where I could plug in the t value.
Every t just fell out in the end.

So this surface has constant curvature I guess? It's some type of sphericalllll shinanigans or something?

Answer 41

Yes, we often call it an osculating sphere if k>0

Answer 42

does that look right to you astro lol

Answer 43

There might be some typo in the Mathematica file then. I'm getting \(\dfrac{12}{169}\) now. We have \(\vec{T}~'(t)=\dfrac{1}{13}\langle -12\sin t,0,-12\cos t\rangle\), which means \(\left\|\vec{T}~'(t)\right\|=\sqrt{\dfrac{(-12)^2\sin^2t+0^2+(-12)^2\cos^2t}{13^2}}=\dfrac{12}{13}\).

Then\[\begin{align*}
\kappa(t)&=\frac{\dfrac{12}{13}}{13}=\frac{12}{169}
\end{align*}\]

Answer 44

Yes it's good

Answer 45

good i glad your here to say you agree

Answer 46

Sorry to doubt you I suppose, I deal with a lot of people who just want the answer.

Answer 47

@zepdrix Yeah the curve defines a helix.

Answer 48

Hmm neato :)

Answer 49

@SithsAndGiggles @Astrophysics I dunno why you guys fuss about it so much :D
If someone doesn't care enough about their education to actually try, it's on them. It'll catch up to them eventually. They're just doing themselves a disservice.

I guess I just like to assume that people have good intentions when they're on this site, maybe that's naive though :)

Answer 50

Thanks @zepdrix I often tell myself that to, you have the right idea, they will eventually learn I guess, the way I handle it never really works or it's very rare. Thanks :)

Answer 51

this is calculus 3 as well, if i just put the answer down my prof would laugh and give me a zero even if it was correct