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iHave a Lovesick Teacher

⁨714⁩ ⁨likes⁩

Submitted ⁨⁨2⁩ ⁨weeks⁩ ago⁩ by ⁨nutbutter@discuss.tchncs.de⁩ to ⁨[deleted]⁩

https://discuss.tchncs.de/pictrs/image/dd1f8577-6bbc-477e-a0d8-b43f55c66232.jpeg

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  • trxxruraxvr@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

    They probably would have stayed together if they had just had the sense to use SI units.

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  • polle@feddit.org ⁨2⁩ ⁨weeks⁩ ago

    Maths/Physics with imperial units seem like a crime.

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    • ColeSloth@discuss.tchncs.de ⁨2⁩ ⁨weeks⁩ ago

      The unit of measurement in this instance is completely irrelevant to the math involved.

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    • Soulg@ani.social ⁨2⁩ ⁨weeks⁩ ago

      This problem would be exactly the same if they used 5 m/s

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      • ChaoticNeutralCzech@feddit.org ⁨2⁩ ⁨weeks⁩ ago

        Except now the walking speed of 1 m/s (3.6 km/h) and running speed of 5 m/s (18 km/h) are realistic

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    • Atlusb@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

      Yes mph is one thing but fps!

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      • general_kitten@sopuli.xyz ⁨2⁩ ⁨weeks⁩ ago

        i am running at 5 frames per second

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    • captainlezbian@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

      Yeah tbh I think that’s a good next step for metrication. Currently it’s all in US customary until college with a brief lesson on metric every few years. With some high school teachers doing a bit more in metric. If you swap it these people will just instinctively use and understand metric.

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      • MajorasTerribleFate@lemmy.zip ⁨2⁩ ⁨weeks⁩ ago

        I’m good with intuiting 1-3 meters, and I have a decent estimate for 1 centimeter. Beyond that, about all I got (yes I know this is all very rough estimates) is “a kilogram is a couple pounds, a kilometer is a short mile, a liter is basically a quart”. But I guess I don’t even have a good intuition for a quart because that in my head rests at the crossroads of “4 cups or ¼ gallon”.

        Make metric meaningful early in life, or actively make referencing real metric measures a part of your life, if you want it to stick.

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      • Soulg@ani.social ⁨2⁩ ⁨weeks⁩ ago

        The biggest issue with swapping to metric I’m aware of isn’t the people getting used to it as much as the massive monetary and labor cost of replacing signs/notices/etc around the whole country

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    • HeyThisIsntTheYMCA@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

      I am on my phone so sorry for terseness but maths breakup problem trapeze lady twelve feet per second south human cannonball man two hundred yards per second every direction but south how far apart after two burritos and a threesome (human cannonball not involved)

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    • Alberat@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

      it’s not physics, it’s anthropology

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  • ArbitraryValue@sh.itjust.works ⁨2⁩ ⁨weeks⁩ ago

    You don’t need calculus to do this. Neither one is accelerating, so just calculate the velocity of one in the reference frame of the other by subtracting the vectors: from the point of view of the boy, the girl’s velocity vector has a component of -5 ft/sec north and 1 ft/sec east, so the magnitude is 26^0.5 ft/sec.

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    • Mirodir@discuss.tchncs.de ⁨2⁩ ⁨weeks⁩ ago

      The “5 seconds after they started moving” is relevant. If we assume this takes place on Earth (i.e. on the surface of a sphere with a set pair of north/south poles), the angle between the two vectors changes depending on their current position.

      If it’s not on the equator, it’s also slightly up to interpretation if “Due East” means they’ll turn to stay on the same latitude, always adjusting to stay moving east forever or if they’ll do a great circle. In the former case, the north moving one will eventually get stuck at the north-pole too instead of continuing their circle around the globe. Most likely not within 5 seconds though, unless the place they started was within 25 feet of the north-pole.

      To actually do the math we’ll need to know (or somehow deduce) where “the place where everything about them began” is though.

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      • ByteJunk@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

        What a convoluted way of asking for the teacher to spill the beans. I like it.

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      • ChaoticNeutralCzech@feddit.org ⁨2⁩ ⁨weeks⁩ ago

        You actually only need to know the latitude for that… except the local terrain will play a larger role anyway, unless they started very close to a pole and follow rhumb lines (in this case a meridian/circle of latitude) as opposed to great circles, so better just ask for full coordinates.

        We don’t have many data points in the question so let’s extrapolate into the past. There is the hint that they met 8 years earlier at the same spot, during which he’d have gone 1 262 304 000 ft or 384 750.2592 km, completing 9.617 polar circumferences of the Earth (40 007.863 km each).

        Huh, that’s not a whole number. In some languages, “eight long years” might mean “a little over 8 years” so let’s assume he finished exactly 10 polar circumnavigations, which took 8 years and about 116 days. Her walked distance over that time is 5x smaller, 2 polar circumnavigations’ worth or 80 015.726 km. This is only exactly 2 great circles if they are polar, but we know that it’s impossible to go due east from either pole. Therefore, she must have gone around a circle of latitude. To end up in the same spot, she must have not-quite-circumnavigated-but-enough-for-Phileas-Fogg the Earth (aka crossed every meridian but not the equator) an integer number of times. After a simple conversion, we can construct a table of the options.

        To calcuate latitude from circle-of-latitude circumference (colc), we’ll be using geodetic ↔ ECEF conversion equations (except those with the perverse prime vertical radius of curvature 𝑁 of course) and their notation (simplified with 𝑦 = 0, 𝜆 = 0, ℎ = 0 to ignore longitude and elevation) with values of the WGS-84 ellipsoid. The relationship we’re seeking is between colc/2𝛑 = 𝑝, circle-of-latitude radius, which is at zero longitude equal to ECEF 𝑋, and 𝜙 (latitude). See also Wikipedia on Earth radius by location but remember to skip anything with 𝑁, we’re not doing that.

        The geocentric radius (𝑅) is related to 𝜙 (latitude) like this but we only need the distance to axis of rotation 𝑝.

        (𝑍/𝑝)(cot 𝜙) = (1 − 𝑒²) → (𝑏²/𝑎²)(𝑍/𝑝) = 1/(cot 𝜙) = tan 𝜙 → 𝜙 = atan((𝑏²/𝑎²)(𝑍/𝑝))
        (using 𝑒² = 1 −  𝑏²/𝑎²)

        Since sin² 𝛂 = 1 − cos² 𝛂 and we can normalize 𝑍 and 𝑝 to the unit circle with ellipsoid radii 𝑏 and 𝑎 respectively:
        𝑍²/𝑏² = (𝑍/𝑏)² = 1 − (𝑝/𝑎)² = 1 − 𝑝²/𝑎², therefore 𝑍 = √(𝑏²(1 − 𝑝²/𝑎²)).

        All in all, 𝑝 → 𝜙 conversion is:
        𝜙 = atan((𝑏²/𝑎²)(√(𝑏²(1 - 𝑝²/𝑎²))/𝑝))

        (Presumably, this could be simpified further but since I can just put this into a calculator so idc)
        Per WGS-84:
        𝑎 = 6378.137 km
        𝑏 = 6352.752 km

        Here are the results. Finding appropriate meeting locations at some of the 25+ possible latitudes on either hemisphere is left as an exercise to the reader.

        nqcbefPFs colc = 2𝛑𝑝 [km] Latitude [°N/°S]
        1 too big N/A
        2 6 367.449 3.277975
        3 4 244.966 47.934779
        4 3 183.724 59.758044
        5 2 546.979 66.211738
        6 2 122.483 70.346611
        7 1 819.271 73.238734
        8 1 591.862 75.380740
        9 1 414.988 77.033209
        10 1 273.489 78.347789
        11 1 157.718 79.419029
        12 1 061.241 80.309059
        13 979.607 81.060439
        14 909.635 81.703329
        15 848.993 82.259706
        16 795.931 82.745975
        17 749.111 83.174629
        18 707.494 83.555355
        19 670.257 83.895779
        20 636.744 84.201988
        21 606.423 84.478901
        22 578.859 84.730536
        23 553.691 84.960206
        24 530.620 85.170671
        25 509.395 85.364245
        26 489.803 85.542885
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    • andros_rex@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

      I mean, there’s quite a few calculus problems that you can do without calculus. Pretty much 80% of the optimization problems/max’s and minimums in a typical Calc 1 class can be done if you remember that -b/2a is the vertex of a quadratic.

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    • rustydrd@sh.itjust.works ⁨2⁩ ⁨weeks⁩ ago

      I guess the calculus portion of this is to write the separation as a function of time, s = √26*t, and then realize that the “speed” of separation is the same regardless of time, because the first derivative is a constant.

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    • Rivalarrival@lemmy.today ⁨2⁩ ⁨weeks⁩ ago

      The boy’s speed is given as 5 ft/sec, but the question is ambiguous as to whether his position remains due north of the starting point, or due north of the girl. Your approach assumes the former, but his 5ft/sec speed may include the girl’s 1ft/sec eastward component.

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      • kogasa@programming.dev ⁨2⁩ ⁨weeks⁩ ago

        The implication that the boy is strafing while maintaining a straight vertical line to the girl is hilarious

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  • ooterness@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

    At any given time T, the coordinates form a right triangle with legs of length 5T and T. Therefore the distance D is given by D^2 = (5T)^2 + T^2 = 26T^2. This simplifies to D = T * sqrt(26). Therefore the rate of separation is sqrt(26) ft/sec regardless of time

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    • Ziglin@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

      You forgot about the curvature of the earth!

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      • baggachipz@sh.itjust.works ⁨2⁩ ⁨weeks⁩ ago

        It’s flat, idiot

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      • ByteJunk@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

        Also, where’s the topographic map of the region? How can you expect us to come up with something remotely accurate without knowing this, is the third dimension a joke to you, are we all dots of ink on a paper?

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  • osanna@lemmy.vg ⁨2⁩ ⁨weeks⁩ ago

    5ft/sec?? Jesus, that’s almost 6ft/sec!

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  • Caketaco@lemmy.dbzer0.com ⁨2⁩ ⁨weeks⁩ ago

    A girl was excited for her sweet 16, and she asked her boyfriend to buy her a car. He said yes. The night of the party, he didn’t come. She was very sad. Then she found out he’d died trying to drive two cars at once to her party. Like this if you love your boyfriend.

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    • Wispy2891@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

      i have the feeling that the kind of person that buys a car for his 16yo gf, is like the one that’s leading the united states

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      • bitjunkie@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

        Image

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    • HeyThisIsntTheYMCA@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

      What if I love your boyfriend?

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  • spicytuna62@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

    Do we need to compensate for the curvature of the earth?

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    • ChaoticNeutralCzech@feddit.org ⁨2⁩ ⁨weeks⁩ ago

      That’s where the calculus comes in, otherwise it’s a basic Pythagorean theorem. I’ve compiled a list of possible latitudes, knock yourself out

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    • Agent641@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

      What about the local topology? What if she’s walking uphill and he’s running downhill?

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    • ThunderComplex@lemmy.today ⁨2⁩ ⁨weeks⁩ ago

      No but if they’re in Mexico you gotta keep it’s sink rate in mind

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    • mrgoosmoos@lemmy.ca ⁨2⁩ ⁨weeks⁩ ago

      thats what the unannounced bonus points are for

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  • TigerAce@lemmy.dbzer0.com ⁨2⁩ ⁨weeks⁩ ago

    It depends on the size of their feet.

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  • hardcoreufo@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

    Is this even calculus? Seems like simple geometry.

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    • Hozerkiller@lemmy.ca ⁨2⁩ ⁨weeks⁩ ago

      It’s not how far apart they are It’s how fast they are seperating. I’m not sure but the rate they are seperating might change as they get further apart due to the triangle or something? Otherwise yeah just a triangle.

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      • davetortoise@reddthat.com ⁨2⁩ ⁨weeks⁩ ago

        Rigourously overcomplicating the problem: Let dy = distance boy travels north in time dt, and dx = distance girl travels east in time dt. We know that dy = 5dx from the question, hence:

           |
           |
        

        dy | | |__ dx

        And the distance they separate dS in time dt is clearly the hypoteneuse. So we can write:

        dS = sqrt(dx^2 + dy^2)

        And divide through by dt:

        dS/dt = sqrt( (dx/dt)^2 + (dy/dt)^2 )

        Simply gives the rate of separation dS/dt as 5.1 feet per second.

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  • farmgineer@nord.pub ⁨2⁩ ⁨weeks⁩ ago

    Shouldn’t it be ‘after having been together’?

    What is ‘at the same time’ referring to in that sentence? They wanted to break up at the same time (as in both had the idea)? They wanted to break up at the same time on the clock to continue the theme of things being same-y?

    The boy is due north of what? The place? The girl? Also, the girl should be wondering about her decision, I think.

    (I don’t even speak English every day anymore, so I could be wrong).

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    • Bonsoir@lemmy.ca ⁨2⁩ ⁨weeks⁩ ago

      They said goodbye at a given position and are then leaving each in a different direction. They are starting to move at the same time from the same point.

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      • NoSpotOfGround@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

        Wait, we know their position exactly? That means we have no idea what their velocities are!

        Actually, their velocities are specified precisely in the problem description.

        What? Velocity too? Now we know nothing!!

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      • Rivalarrival@lemmy.today ⁨2⁩ ⁨weeks⁩ ago

        Is he due north of the starting point, or of the girl? If the former, he is traveling due north. If the latter, his 5ft/sec velocity has a 1ft/sec eastward component, and we need to calculate the northern component.

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    • misericordiae@literature.cafe ⁨2⁩ ⁨weeks⁩ ago

      (I don’t even speak English every day anymore, so I could be wrong).

      You’re not wrong. I think some of it is the difference between casual speech and formal writing (people are more likely to say “after being” but write “after having been”, especially in published work)**, but some of it is also just poorly phrased. It makes enough sense to a native speaker to get what the problem is asking, though.

      __

      ** I think the first may be correct in some cases, but idk the rule.

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      • farmgineer@nord.pub ⁨2⁩ ⁨weeks⁩ ago

        It only bothered me because I saw that it was a school assignment and I thought it would be to a higher standard. In casual speech, I don’t really care unless the meaning is unclear.

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  • FinjaminPoach@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

    Sigh i miss high school maths. Even i’m lovesick now.

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  • CileTheSane@lemmy.ca ⁨2⁩ ⁨weeks⁩ ago

    This is the type of question my Calc teacher would put on a test. He liked doing weird questions probably because we’re doing fucking calculus here, it doesn’t make sense in “every day word problems” (I have never once needed to know how long it will take me to fill a funnel while I let water pour out it at the same time, and where the water line will be after exactly 7 seconds) so trying to make a word problem out of it is already an exercise in nonsense.

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  • ITGuyLevi@programming.dev ⁨2⁩ ⁨weeks⁩ ago

    Really depends on which way the dude is going. It said that he is due North and running at 5 feet per second, is he running east? If so they would be like 4 feet apart, unless he started 4 feet further West than she did.

    Really need some more info for this dumb word problem.

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    • ChaoticNeutralCzech@feddit.org ⁨2⁩ ⁨weeks⁩ ago

      Instructions unclear, let’s assume he started 1,312,593,975 feet further back (see my other comment)

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  • Grail@multiverse.soulism.net ⁨2⁩ ⁨weeks⁩ ago

    Good thing that ain’t an english teacher

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  • mandolrain@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

    One of my teachers in uni always used sex potion and leather wear examples. He had been reported for sexual abuse a year prior but continued in his assignments

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    • ProfessorHoover@infosec.pub ⁨2⁩ ⁨weeks⁩ ago

      One of my game development lecturers always used giving handjobs to stakeholders as an example of balancing priorities and keeping everyone happy while being pulled in multiple directions. I’m assuming someone complained though because when I had him for a different class he’d stopped.

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      • Leather@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

        The truth is always suppressed.

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  • Jankatarch@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

    Reminds me that one exurb1a pinecone video.

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  • FreshParsnip@lemmy.ca ⁨2⁩ ⁨weeks⁩ ago

    30 volts

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  • dhruv3006@lemmy.world ⁨2⁩ ⁨weeks⁩ ago

    Its trigonometry - big brain time.

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