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
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 you 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
Sure! To solve this problem, we can use differential equations. Let's break it down step by step.
Step 1: Formulating the differential equation
From the problem, we know that the rate at which the height is increasing is inversely proportional to h^3. In other words, we can express it as:
dh/dt = -k/h^3
Where dh/dt represents the rate of change of height with respect to time, and k is a constant of proportionality. The negative sign indicates that the height is decreasing over time.
Step 2: Separating variables
To solve the differential equation, we need to separate variables. We can rewrite the equation as:
h^3 * dh = -k * dt
Step 3: Integrating both sides
Now, we integrate both sides of the equation:
∫h^3 * dh = ∫-k * dt
This gives us:
(1/4) * h^4 = -k * t + C
Where C is the constant of integration.
Step 4: Solving for the constant
We know that the initial height of the pile is ho. So, at t = 0, h = ho. Substituting these values into the equation, we get:
(1/4) * ho^4 = C
Step 5: Finding the time at which the pile height is 3ho
We are given that the height of the pile doubles after time T. In other words, when t = T, h = 2ho.
Substituting these values into the equation from step 4, we have:
(1/4) * (2ho)^4 = C
Simplifying this expression, we get:
(1/4) * 16ho^4 = C
4ho^4 = C
So, we now know the value of the constant C.
Step 6: Finding the time when the height is 3ho
To find the time T1 after which the height of the pile is 3ho, we substitute these values into the equation:
(1/4) * (3ho)^4 = -k * T1 + 4ho^4
Expanding and simplifying, we get:
81ho^4/4 = -k * T1 + 4ho^4
Rearranging the equation to solve for T1, we get:
T1 = (81ho^4/4 - 4ho^4) / -k
Step 7: Final simplification
We can further simplify the equation by factoring out ho^4:
T1 = ho^4 * (81/4 - 4) / -k
T1 = ho^4 * (81/4 - 16/4) / -k
T1 = ho^4 * (65/4) / -k
So, in terms of T, the time after which the height of the pile is 3ho is given by:
T1 = - (65/4) * ho^4 / k
And this is the solution to the problem in terms of T.