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Cake day: July 10th, 2023

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  • ltxrtquq@lemmy.mltoScience Memes@mander.xyzSquare!
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    21 days ago

    Given r=f(θ), we are generally not concerned with r′=f′(θ); that describes how fast r changes with respect to θ

    You’re using the derivative of a polar equation as the basis for what a tangent line is. But as the textbook explains, that doesn’t give you a tangent line or describe the slope at that point. I never bothered defining what “tangent” means, but since this seems so important to you why don’t you try coming up with a reasonable definition?


  • ltxrtquq@lemmy.mltoScience Memes@mander.xyzSquare!
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    21 days ago

    I think we fundamentally don’t agree on what “tangent” means. You can use

    x=f(θ)cosθ, y=f(θ)sinθ to compute dydx

    as taken from the textbook, giving you a tangent line in the terms used in polar coordinates. I think your line of reasoning would lead to r=1 in polar coordinates being a line, even though it’s a circle with radius 1.


  • ltxrtquq@lemmy.mltoScience Memes@mander.xyzSquare!
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    22 days ago

    Given r=f(θ), we are generally not concerned with r′=f′(θ); that describes how fast r changes with respect to θ

    I think this part from the textbook describes what you’re talking about

    Instead, we will use x=f(θ)cosθ, y=f(θ)sinθ to compute dydx.

    And this would give you the actual tangent line, or at least the slope of that line.