find the curvature of the plane curve y=x^n, n>0

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find the curvature of the plane curve y=x^n, n>0

Mathematics
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bonus question from an old test so it's suppose to be tricky...
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?
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'|

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the difference is this asks to find all that in a space curve and this bonus question asks to find in a plane curve
lol, never mind, i thought it was basic algebra, not advanced linear algebra :P. i can't help at all with that, my bad :)
thx anyway
It's neither, its calculus. A plane curve is nothing but a space curve with no zero component. You could parameterize it like \[\vec{r} = \] or \[\vec{r} = \] if it would make you more comfortable.
I'm sorry, with no z-component :)
right
ok so if i use that, i should be able to get it..feel free to solve so i can check later, thx!!
So the tangent vector \[T = \frac{\vec{r'}}{|r'|} = \frac{<1,nx^{n-1}>}{\sqrt{1+(nx^{n-1})^2}}\] 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 \[\vec{r} = \] for some parameter t. We're actually using x itself as the parameter, so x' = 1 and x'' = 0. This yields \[\kappa = \frac{|n(n-1)x^{n-2}|}{(1 + n^2x^{2n-2})^\frac{3}{2} } \]
your way seems much faster, easier, and cleaner. mine is getting too messy and it's getting too late

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