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Joined 1 year ago
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Cake day: July 10th, 2023

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  • why do people have this innate ability to underestimate what we might be capable of?

    Because we can see what we’re currently capable of in terms of climate change, and the outlook is pretty bleak

    why do you think its impossible for us to become masters of our own genome?

    Because even in the best case scenario, this is dangerously close to eugenics

    not getting off this rock means our species is doomed regardless of how ‘perfect’ we keep earth.

    If we can’t keep earth livable, an entire self-regulating planet that’s been livable for hundreds of millions or billions of years, what are our chances of keeping anywhere else livable?


  • ltxrtquq@lemmy.mltoScience Memes@mander.xyzSquare!
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    1 month 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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    1 month 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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    1 month 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.


  • ltxrtquq@lemmy.mltoScience Memes@mander.xyzSquare!
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    1 month ago

    Polar Functions and dydx

    We are interested in the lines tangent a given graph, regardless of whether that graph is produced by rectangular, parametric, or polar equations. In each of these contexts, the slope of the tangent line is dydx. Given r=f(θ), we are generally not concerned with r′=f′(θ); that describes how fast r changes with respect to θ. Instead, we will use x=f(θ)cosθ, y=f(θ)sinθ to compute dydx.

    From the link above. I really don’t understand why you seem to think a tangent line in polar coordinates would be a circle.


  • ltxrtquq@lemmy.mltoScience Memes@mander.xyzSquare!
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    1 month ago

    A straight line in polar coordinates with the same tangent would be a circle.

    I’m not sure that’s true. In non-euclidean geometry it might be, but aren’t polar coordinates just an alternative way of expressing cartesian?

    Looking at a libre textbook, it seems to be showing that a tangent line in polar coordinates is still a straight line, not a circle.