Question

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

Answer 1

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

Answer 2

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?

Answer 3

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'|

Answer 4

the difference is this asks to find all that in a space curve and this bonus question asks to find in a plane curve

Answer 5

lol, never mind, i thought it was basic algebra, not advanced linear algebra :P.

i can't help at all with that, my bad :)

Answer 6

thx anyway

Answer 7

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.

Answer 8

I'm sorry, with no z-component :)

Answer 9

right

Answer 10

ok so if i use that, i should be able to get it..feel free to solve so i can check later, thx!!

Answer 11

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
\[\vec{r} = <x(t),y(t)> \]
for some parameter t. We're actually using x itself as the parameter, so x' = 1 and x'' = 0. This yields
\[\kappa = \frac{|n(n-1)x^{n-2}|}{(1 + n^2x^{2n-2})^\frac{3}{2} } \]

Answer 12

your way seems much faster, easier, and cleaner. mine is getting too messy and it's getting too late