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Math Help - Length of Curve with multivariable

  1. #1
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    Length of Curve with multivariable

    I tried to evaluate the length of this curve but cannot figure out what to do with the derivative of the log(cost). The question is:

    Find the length of the curve r(t) = <1, log(cost)>
    t having the domain [0, pi/4]

    I did the integral from 0 to pi/4 of the square root of the sum of the derivative of the components and im having trouble integrating it. Can anyone help me solve this?
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  2. #2
    Super Member Random Variable's Avatar
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    \text{length} = \int^{\pi /4}_{0} ||r'(t)|| \ dt

     r'(t) = \Big(0, \text{-}\tan t\Big)

     length = \int^{\pi /4}_{0} \sqrt{0 + \tan^2t} \ dt

    and since  \tan t \ge 0 \ \text{for} \ 0 \le t \le \pi /4

     = \int^{\pi /4}_{0} \tan t \ dt

    Can you integrate that?
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  3. #3
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    how did you derive log(cost) to be -tant

    did you just change it to ln?
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  4. #4
    Super Member Random Variable's Avatar
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    Quote Originally Posted by sleepiiee View Post
    how did you derive log(cost) to be -tant

    did you just change it to ln?
    I assumed it was log base e. Is it log base 10?
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  5. #5
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    its blank, would that mean its base is 10 or e?
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  6. #6
    Super Member Random Variable's Avatar
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    Quote Originally Posted by sleepiiee View Post
    its blank, would that mean its base is 10 or e?
    most likely base e
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  7. #7
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    i got ln|secx + tanx| evaluated at 0 and pi/4

    and got the answer ln|(2)^1/2|

    can anyone confirm that the answer is correct?
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  8. #8
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    nvm i got ln[(rad 2) + 1]
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  9. #9
    Super Member Random Variable's Avatar
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    ln |secx +tanx| is the antiderivative of secx

    since the derivative of ln(cosx) is -tanx, the antiderivative of tanx is -ln(cosx)
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