Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

elica85

  • 4 years ago

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

  • This Question is Closed
  1. elica85
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  2. Euler271
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 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?

  3. elica85
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 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'|

  4. elica85
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 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

  5. Euler271
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 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 :)

  6. elica85
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    thx anyway

  7. Jemurray3
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  8. Jemurray3
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  9. elica85
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    right

  10. elica85
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  13. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy