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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ArbitraryValue@sh.itjust.works 1 day 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.
andros_rex@lemmy.world 1 day ago
Rivalarrival@lemmy.today 20 hours 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.
kogasa@programming.dev 18 hours ago
The implication that the boy is strafing while maintaining a straight vertical line to the girl is hilarious
Rivalarrival@lemmy.today 16 hours ago
Technically, his starting location is defined contradictorily, being both the same position as hers, as well as “due north”. We can only proceed if a point can be considered “due north” of itself.
Technically, his initial direction of travel isn’t actually defined, nor is that direction specified as constant. Only his initial position and “5ft/sec” speed has actually been defined. He could maintain a constant distance to her, orbiting her as she travels eastward.
In five seconds, he could be a maximum of 25 feet away from the starting point. That puts him up to 30 feet west of her, or 20 feet east of her. His possible position will be defined by an ellipse with foci at the origin and her position 5 feet to the east, with a major axis length of 50 feet. If his direction is fixed, he will be on the circumference of the ellipse. If not fixed, somewhere within the ellipse.
If his direction is not fixed, and he elects to minimize his distance from her, as his distance from her approaches zero, his revolutions around her in a given time will approach infinite, and we will have to consider relativistic effects. His body will be ripped apart into a pink mist, which he will experience for all eternity. Poetic, I suppose.
rustydrd@sh.itjust.works 1 day 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.
Mirodir@discuss.tchncs.de 1 day 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.
ByteJunk@lemmy.world 1 day ago
What a convoluted way of asking for the teacher to spill the beans. I like it.
ChaoticNeutralCzech@feddit.org 1 day 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.
Mirodir@discuss.tchncs.de 1 day ago
I love you.