Pedro asked in Science & MathematicsPhysics · 7 years ago

# Find curvature of r(t)?

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

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• 7 years ago

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