Comment on iHave a Lovesick Teacher
Hozerkiller@lemmy.ca 2 weeks agoIt’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.
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.
kogasa@programming.dev 2 weeks ago
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.
davetortoise@reddthat.com 2 weeks ago
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