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anilorap

  • 3 years ago

let r(t)=(sint, cost,t) find k(t)

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  1. anilorap
    • 3 years ago
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    k(t)=|r'(t)xr''(t)| / |r'(t)|^3

  2. anilorap
    • 3 years ago
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    r'(t)= (cost,-sint,1) r''(t)=(-sint,-cost,0) |r'(t)|= square root of 2

  3. ash2326
    • 3 years ago
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    r'(t)r''(t)=(-sint cost, sint cost, 0) |r'(t)r''(t)|=root(sin^2t cos^2t+sin^2t cos^2+0) |r'(t)r''(t)|=root(2 sint^2t cos^2t)=sint.cost root(2) |r't|^3=2root (2) k(t)=(sint*cos t)/2

  4. anilorap
    • 3 years ago
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    mm,,, in the cross product i got (-cost, sint,-1)

  5. him1618
    • 3 years ago
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    |dT/dt| x 1/|v|

  6. him1618
    • 3 years ago
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    T = v(t)/|v(t)|

  7. mathmate
    • 3 years ago
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    r(t)=(sint, cost,t) find k(t) r'(t)=<cos(t),-sin(t),1> r"(t)=<-sin(t),-cos(t),0> r'(t)xr"(t) = <cos(t),-sin(t),-1> |r'(t)xr"(t)| = sqrt(2) |r'(t)| = sqrt(2) |r'(t)xr"(t)| / |r'(t)|^3 = sqrt(2)/sqrt(2)^3 = 1/2

  8. anilorap
    • 3 years ago
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    yea thats what i have

  9. mathmate
    • 3 years ago
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    Great!

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