gah! help again please h=2.5−ce^(−0.016t) this is the equation i have from solving dh/dt=0.016(2.5-h) now when h is at max h=2.5 how long will it take to become max (where t is time and h is hieght)

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gah! help again please h=2.5−ce^(−0.016t) this is the equation i have from solving dh/dt=0.016(2.5-h) now when h is at max h=2.5 how long will it take to become max (where t is time and h is hieght)

Mathematics
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unless im missing something it seems the only t that yields h of 2.5 is as t goes to infinity so you could say lim h = 2.5 as t->infinity
yeah thats why it doesnt seem to make sense, as its wanting a time for it to reach 2.5 which seems to make no sense!
your equation for h is correct for the given dh/dt whats the context of the problem

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the filling of a lock in a canal, According to the mathematics of your solution, how long would it take for the lock to fill? which follows Solve the differential (dh/dt=0.016(2.5-h)) equation to find h in terms of t.
im guessing h needs to be 2.5 for the lock to be filled in that case the lock slowly is filled at an ever decreasing rate such that the lock is never completely full something like that would be my interpretation
yeah, the source of the questions often asks annoying questions! half the time it makes no sense!
good luck :)
cheers :)
the next part is Find how long it takes in minutes for the water to rise to within 2mm of the lock being completely full. so it still has to be rearranged to find that
id think anyway!
well for this part you can evaluate write t in terms of h h=2.5−ce^(−0.016t) ->e^(-.016t) = (2.5-h)/c ->t = -ln((2.5-h)/c)/.016 as long as h<2.5 and c > 0 t is defined plug in h = 2.5 - 2mm not sure what units the 2.5 is and of course you have to solve for c, given initial conditions do they say what the water level is at t=0 hope that helps
nope they dont ever give a value of c and thank you!

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