30 volts
Its trigonometry - big brain time.
Is this even calculus? Seems like simple geometry.
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.
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
(This is supposed tk be formatted like a triangle but it looks janky. You get the idea.)
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.
Some confusing notation here that buries the assumption that the rate of change is constant (which is true in this case). For conceptual clarity I would explain it as:
Let y(t) be the boy’s position at time t, and x(t) the girl’s position. The distance between them is S = sqrt(x^(2) + y^(2)). The distance is changing at a rate of dS/dt = dS/dx dx/dt + dS/dy dy/dt = (xdx/dt + ydy/dt)/sqrt(x^(2) + y^(2)). We are given dy/dt = 5 and dx/dt = 1, and we can determine that at t=5 we have y = 25 and x = 5. Therefore dS/dt = 130/sqrt(650) = sqrt(26) ~= 5.1.
This is a better generalised solution, yeah. Though I do think that as far as conceptual clarity goes, doing it geometrically is a bit more transparent than using the chain rule, even if it’s sort of constrained to constant speed in this case
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.
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
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.
Do we need to compensate for the curvature of the earth?
What about the local topology? What if she’s walking uphill and he’s running downhill?
That’s where the calculus comes in, otherwise it’s a basic Pythagorean theorem. I’ve compiled a list of possible latitudes (with questionable assumptions), knock yourself out
No but if they’re in Mexico you gotta keep it’s sink rate in mind
thats what the unannounced bonus points are for
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.
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
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.
Maths/Physics with imperial units seem like a crime.
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)
This problem would be exactly the same if they used 5 m/s
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
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.
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.
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
Nah it’s public opinion. People don’t want to change. The costs are annoying, but we can budget for them. We tried to metricate at the same time as Canada. They had less money per person than us but succeeded because they didn’t have a large mass of people unwilling to learn and change.
Well depends on what you mean by „success“
Comparatively though…
The unit of measurement in this instance is completely irrelevant to the math involved.
it’s not physics, it’s anthropology
Yes mph is one thing but fps!
i am running at 5 frames per second
5.099 frames per second, but allowing for rounding you could argue they are separating at 5 fps
It depends on the size of their feet.
5ft/sec?? Jesus, that’s almost 6ft/sec!
Good thing that ain’t an english teacher
Reminds me that one exurb1a pinecone video.
a^2 + b^2 = c^2
He walks 25ft north, she walks 5ft east. The distance between them is D.
25^2 + 5^2 = D^2 625 + 25 = D^2 D = √650 = 25.5ft
Read th question again. It’s not about how far apart they are after 5 seconds. It’s about how fast they are separating from each other, so it’s about a rate, not distance.
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 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.
The implication that the boy is strafing while maintaining a straight vertical line to the girl is hilarious
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.
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/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 kmHere 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 for polar regions.
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 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.
