## elica85 find the curvature of the plane curve y=x^n, n>0 2 years ago 2 years ago

1. elica85

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

2. Euler271

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?

3. elica85

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'|

4. elica85

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

5. Euler271

lol, never mind, i thought it was basic algebra, not advanced linear algebra :P. i can't help at all with that, my bad :)

6. elica85

thx anyway

7. Jemurray3

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.

8. Jemurray3

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

9. elica85

right

10. elica85

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

11. Jemurray3

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 $\kappa = \frac{|n(n-1)x^{n-2}|}{(1 + n^2x^{2n-2})^\frac{3}{2} }$

12. elica85

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