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??

To explain how the values 25 and 81 are obtained, let's go through the solution step by step.

We start with the differential equation:

dh/dt = k/h^3

where k is an unknown constant.

To solve this equation, we can rewrite it as follows:

h^3 dh = k dt

Now, we integrate both sides with respect to their respective variables:

∫h^3 dh = ∫k dt

Evaluating the integrals:

h^4/4 + c = kt

Here, c is the constant of integration.

At t = 0, the initial height of the pile is given as ho. Plugging these values into the equation, we have:

ho^4/4 + c = 0

Simplifying for c:

c = -ho^4/4

Substituting c back into the equation, we get:

h^4/4 - ho^4/4 = kt

At t = T, the height of the pile has doubled, meaning h = 2ho. Substituting these values into the equation:

(2ho)^4/4 - ho^4/4 = kT

Simplifying:

16ho^4/4 - ho^4/4 = kT

15ho^4/4 = kT

Now, to determine the value of k, we know that k = (1/T)(15ho^4/4).

Substituting this value back into the equation, we have:

h^4 - ho^4 = 25ho^4(t/T)

Now, we want to find when h = 3ho. Substituting this value into the equation:

(3ho)^4 - ho^4 = 25ho^4(t/T)

Simplifying:

81ho^4 - ho^4 = 25ho^4(t/T)

Equating the coefficients:

80 = 25(t/T)

Solving for t:

t = (80T/25) = (16T/5)

Thus, the time after which the height of the pile is 3ho is t = 16T/5.

So, the values 25 and 81 in the solution arise from the evaluation of integrals, substitution of given values, and equating coefficients during the problem-solving process.