Find curvature of r(t)?

Find curvature of r(t)=3ti+4sintj+4costk

Please show work for this problem

1 Answer

Relevance
  • 7 years ago
    Favorite Answer

    The curvature for a parametrically defined curve

    r(t) = x(t)∙i + y(t)∙j + z(t)∙k

    is given by

    κ(t) = || r'(t)×r''(t || / || r'(t) ||³

    r(t) nad r''(t) are the first and second derivative of r(t) with respect to t:

    r'(t) = x'(t)∙i + y'(t)∙j + z'(t)∙k

    r''(t) = x''(t)∙i + y''(t)∙j + z''(t)∙k

    For this problem

    r(t) = 3∙t∙i + 4∙sin(t)∙j + 4∙cos(t)∙k

    =>

    r'(t) = 3∙i + 4∙cos(t)∙j - 4∙sin(t)∙k

    r'''(t) = 0∙i - 4∙sin(t)∙j - 4∙cos(t)∙k

    The cross product of first and second derivative is:

    r'(t)×r''(t) = (3∙i + 4∙cos(t)∙j - 4∙sin(t)∙k) × (0∙i - 4∙sin(t)∙j - 4∙cos(t)∙k

    = [4∙cos(t)∙(-4∙cos(t) - (-4∙sin(t))∙(-4∙sin(t)]∙i + [(-4∙sin(t))∙0 - 3∙(-4∙cos(t)]∙j + [3∙(-4∙sin(t)) - 4∙cos(t)∙0]∙k

    = [- 16∙cos²(t) - 16∙sin²(t)]∙i + 12∙cos(t)∙j - 12∙sin(t)∙k

    = [- 16∙(sin²(t) + cos²(t)]∙i + 12∙cos(t)∙j - 12∙sin(t)∙k

    = - 16∙i + 12∙cos(t)∙j - 12∙sin(t)∙k

    So

    || r'(t)×r''(t || = √[ (-16)² + (12∙cos(t))² + (-12∙sin(t))²]

    = √[ 256² + 144∙(sin²(t) + cos²(t))]

    = √400

    = 20

    and

    || r'(t) || = √[ (3)² + (4∙cos(t))²∙ + (-4∙sin(t))² ]

    = √[ 9 + 16∙(sin²(t) + cos²(t) ]

    = √25

    = 5

    Hence,

    κ(t) = || r'(t)×r''(t || / || r'(t) ||³ = 20 / 5³ = 4/25 = 0.16

Still have questions? Get your answers by asking now.