You don’t need calculus to do this. Neither one is accelerating, so “5 seconds after they started moving” is irrelevant. 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 orthogonal components of -5 ft/sec north and 1 ft/sec east, so the magnitude is 26^0.5 ft/sec.
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.
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 and ♀︎ a circle of latitude) as opposed to great circles, so better just ask for full coordinates.
What? The teacher does not want to talk about it? Let’s find out anyway, to the best of my abilities. For now, we’ll be assuming Earth is a fully walkable ellipsoid.
We don’t have many data points in the question so let’s extrapolate their movement 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 (ellipses, really) if they are polar, but we know that it’s impossible to go due east from either pole. Therefore, we’ll use the other option you pointed out, of her having gone at a constant bearing of 090, her path being a circle of latitude (aka a “parallel”). 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.
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 I can just put this into a calculator so idc)
Per WGS-84:
𝑎 = 6378.137 km
𝑏 = 6356.752 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. Also note that “rainy days” don’t occur in some places, which is why I didn’t bother adding more rows after I got within 500 km of the pole.
nqcbefPFs
colc/2𝛑 = 𝒑 [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
Rows where the number of not-quite-circumnavigations is divisible by 2, 5, or 10 are especially interesting because then the couple would meet 3, 6 and 11 times over the 8.32-year relationship, respectively, rather than just twice.
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.
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 rate of separation is the same regardless of time, because the first derivative is a constant.
You don’t need calculus to do this. Neither one is accelerating, so “5 seconds after they started moving” is irrelevant. 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 orthogonal components of -5 ft/sec north and 1 ft/sec east, so the magnitude is 26^0.5 ft/sec.
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.
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 and ♀︎ a circle of latitude) as opposed to great circles, so better just ask for full coordinates.
What? The teacher does not want to talk about it? Let’s find out anyway, to the best of my abilities. For now, we’ll be assuming Earth is a fully walkable ellipsoid.
We don’t have many data points in the question so let’s extrapolate their movement 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 (ellipses, really) if they are polar, but we know that it’s impossible to go due east from either pole. Therefore, we’ll use the other option you pointed out, of her having gone at a constant bearing of 090, her path being a circle of latitude (aka a “parallel”). 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 I can just put this into a calculator so idc)
Per WGS-84:
𝑎 = 6378.137 km
𝑏 = 6356.752 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. Also note that “rainy days” don’t occur in some places, which is why I didn’t bother adding more rows after I got within 500 km of the pole.
Rows where the number of not-quite-circumnavigations is divisible by 2, 5, or 10 are especially interesting because then the couple would meet 3, 6 and 11 times over the 8.32-year relationship, respectively, rather than just twice.
I love you.
What a convoluted way of asking the teacher to spill the beans. I like it.
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.
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 rate of separation is the same regardless of time, because the first derivative is a constant.