math please help
posted by lucy on .
i posted this
can you explain to me how to do this question i know i have to use differential equations but im not sure how to form an equation for inversely proportional
question states
grain is being poured at a steady rate to form a pile with height h, the rate at which the height is increasing is inversely proportional to h^3.
the initial height of the pile is ho and the height doubles after time T.
find in terms of T, the time after which the height of the pile is 3ho
if you could explain it as much as possible
thanks
Damon replied
dh/dt = k/h^3 where k is unknown constant
h^3 dh = k dt
h^4/4 + c = kt
at t = 0, h = ho
ho^4/4 + c = 0
c = ho^4/4
so
h^4/4 ho^4/4 = k t
at t = T, h = 2 ho
16 ho^4/4 ho^4/4 = k T
15 ho^4/4 = k T
k = (1/T)(15 ho^4/4)
so
h^4  ho^4 = 25 ho^4 (t/T)
when is h = 3 ho ?
81 ho^4  ho^4 = 25 ho^4 (t/T)
80 = 25 (t/T)
t = 80T/25 = 16 T/5
i follow it up to this point
h^4  ho^4 = 25 ho^4 (t/T)
when is h = 3 ho ?
81 ho^4  ho^4 = 25 ho^4 (t/T)
80 = 25 (t/T)
t = 80T/25 = 16 T/5
where do you get 25 from and 81
could someone help??

81 = 3^4
The "25" comes from previous step, canceling out ho^4.
81 ho^4  ho^4 = 25 ho^4 (t/T)
80 = 25 (t/T)