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elica85
 3 years ago
find the curvature of the plane curve y=x^n, n>0
elica85
 3 years ago
find the curvature of the plane curve y=x^n, n>0

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

Euler271
 3 years ago
Best ResponseYou've already chosen the best response.0well, 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?

elica85
 3 years ago
Best ResponseYou've already chosen the best response.1the 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'

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

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

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1It'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.

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

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

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1So 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 \[\kappa = \frac{n(n1)x^{n2}}{(1 + n^2x^{2n2})^\frac{3}{2} } \]

elica85
 3 years ago
Best ResponseYou've already chosen the best response.1your way seems much faster, easier, and cleaner. mine is getting too messy and it's getting too late
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