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anonymous

  • 5 years ago

Find the curvature at the point (1,0). Let x=e^tcost, y=e^tsint. What would t=?

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  1. anonymous
    • 5 years ago
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    x'=e^tcost-e^tsint, x"=e^tcost-e^tsint-e^tsint+e^tcost y'=e^tsint+e^tcost, y"=e^tsint+e^tcost+e^tcost-e^tsint K=x'y"-y'x"/Sq. x'+y' all i need is what t equals

  2. anonymous
    • 5 years ago
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    I'll admit this one threw me for a loop too. You have to notice that the y-coordinate is 0, so plug it into the y equation. Use the zero product property to conclude that e^t = 0 or sint = 0. e^t is most definitely not going to equal 0 because 0 is not in the range of e^t (horizontal asymptote, remember?). So, sint = 0. If you solve for t, you'll find that occurs when t = 0.

  3. anonymous
    • 5 years ago
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    thanks that is what i was thinking i just wasnt sure if you could break it up like that

  4. anonymous
    • 5 years ago
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    yeah and you could even check it in the x equation. If you let t = 0, you would get x = 1.

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