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anonymous

  • 5 years ago

how is x=tanθ when using trig sub

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  1. anonymous
    • 5 years ago
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    For (a^2+x^2), you substitute x = a tan (theta) because when you replace the equation a^2 + x^2 with a^2 + (a tan (theta))^2, you get a^2 ( 1 + tan^2 (theta)) which becomes a^2 (sec^2 (theta)).

  2. anonymous
    • 5 years ago
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    I'm trying to figure out when solving for arc length r=e^θ from (1,0) to origin, you have to convert the limits in terms of theta. I found that someone used tanθ = x to solve the problem. I'm wondering how the person got the equation.

  3. anonymous
    • 5 years ago
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    For arc length, can you use the equation \[\int\limits_(\sqrt(1+(f'(x))^2) \] from a to b?

  4. anonymous
    • 5 years ago
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    ∫ab sqrt (r^2 + (dr/dθ)^2) dθ is what i used

  5. anonymous
    • 5 years ago
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    Are you doing it in polar?

  6. anonymous
    • 5 years ago
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    yes

  7. anonymous
    • 5 years ago
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    You are integrating dθ - so you need the limits in the integral to be in terms of theta.

  8. anonymous
    • 5 years ago
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    http://tutorial.math.lamar.edu/Classes/CalcII/PolarArcLength.aspx i'm confused about the trig sub that he used to convert the limits

  9. anonymous
    • 5 years ago
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    Well you'd still use a a tan \[\theta\] because it will simplify if you use trig identities.

  10. anonymous
    • 5 years ago
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    \[a \tan(\theta)\]

  11. anonymous
    • 5 years ago
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    but how is tanθ = x or y

  12. anonymous
    • 5 years ago
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    isnt it tanθ = y/x?

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