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February 27, 2015

February 27, 2015

Posted by **lucy** on Tuesday, February 2, 2010 at 8:03am.

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??

- math please help -
**drwls**, Tuesday, February 2, 2010 at 9:30am81 = 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)

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