Comment on iHave a Lovesick Teacher
davetortoise@reddthat.com 2 weeks agoRigourously 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