They probably would have stayed together if they had just had the sense to use SI units.
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.
What a convoluted way of asking the teacher to spill the beans. I like it.
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.
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.
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
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. The problem doesn’t actually constrain him to a northerly heading, or even to a constant heading. He could choose to orbit her at a constant distance as she travels eastward.
His possible position will be up to 25ft from the origin, putting him at most 30 feet from her, if his direction is opposite hers. If his direction is constant, his minimum distance from her will be 20 feet.
If his direction is not constant, 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. From his perspective, his body will be ripped apart into a dizzying pink mist, which he will experience for all eternity. Poetic, I suppose.
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
You forgot about the curvature of the earth!
It’s flat, idiot
I didn’t know the earth was flat when using imperial, but it makes perfect sense!
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?
5ft/sec?? Jesus, that’s almost 6ft/sec!
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
What if I love your boyfriend?
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?
No but if they’re in Mexico you gotta keep it’s sink rate in mind
thats what the unannounced bonus points are for
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It depends on the size of their feet.
Sigh i miss high school maths. Even i’m lovesick now.
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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).
They said goodbye at a given position and are then leaving each in a different direction. They start to move at the same time from the same point.
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!!
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.
(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.
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** I think the first may be correct in some cases, but idk the rule.
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.
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.
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.
Good thing that ain’t an english teacher
Reminds me that one exurb1a pinecone video.
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.
The truth is always suppressed.
