Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

elica85 Group TitleBest ResponseYou've already chosen the best response.1
bonus question from an old test so it's suppose to be tricky...
 2 years ago

Euler271 Group TitleBest ResponseYou've already chosen the best response.0
well, this question seems flawed to me. the curve x^n can not really be defined. although, all curves of the form x^n, n>0 look the same, and that is exponential. The only point these have in common is (1,1). Do you know what your teacher meant by: find?
 2 years ago

elica85 Group TitleBest ResponseYou've already chosen the best response.1
the question before this one asks to find the unit tangent vector T, curvature k, and unit normal vector N and i know that T=r'/r', k=(dT/dt)/r', and N=T'/T'
 2 years ago

elica85 Group TitleBest ResponseYou've already chosen the best response.1
the difference is this asks to find all that in a space curve and this bonus question asks to find in a plane curve
 2 years ago

Euler271 Group TitleBest ResponseYou've already chosen the best response.0
lol, never mind, i thought it was basic algebra, not advanced linear algebra :P. i can't help at all with that, my bad :)
 2 years ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
It's neither, its calculus. A plane curve is nothing but a space curve with no zero component. You could parameterize it like \[\vec{r} = <x,x^n>\] or \[\vec{r} = <x,x^n,0> \] if it would make you more comfortable.
 2 years ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
I'm sorry, with no zcomponent :)
 2 years ago

elica85 Group TitleBest ResponseYou've already chosen the best response.1
ok so if i use that, i should be able to get it..feel free to solve so i can check later, thx!!
 2 years ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
So the tangent vector \[T = \frac{\vec{r'}}{r'} = \frac{<1,nx^{n1}>}{\sqrt{1+(nx^{n1})^2}}\] the curvature can also be written \[\kappa = \frac{x' y'' + x'' y'}{(x'^2+y'^2)^\frac{3}{2}}\] where the above variables are defined as \[\vec{r} = <x(t),y(t)> \] for some parameter t. We're actually using x itself as the parameter, so x' = 1 and x'' = 0. This yields \[\kappa = \frac{n(n1)x^{n2}}{(1 + n^2x^{2n2})^\frac{3}{2} } \]
 2 years ago

elica85 Group TitleBest ResponseYou've already chosen the best response.1
your way seems much faster, easier, and cleaner. mine is getting too messy and it's getting too late
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.