Hello,
I was not certain as to whether I should begin a new thread or not, but I am still working on the integral found in the post http://ift.tt/1gQDxjB
If a moderator deems this should be merged with the linked thread, please do so. I felt as though I should begin a new thread, because the other starting becoming littered; and also, I have a lot more information I need to provide.
Okay, so I have just revisited this integral, and I still can't seem to get the correct answer. So, here is my work, starting form the beginning.
Here is the problem:
An outdoor decorative pond in the shape of a hemispherical tank is to be filled with water pumped into the tank through an inlet in its bottom. Suppose that the radius of the tank is R10 ft, that water is pumped in at a rate of pi ft^3/min, and that the tank is initially empty. See Figure 3.2.6. As the tank fills, it loses water through evaporation. Assume that the rate of evaporation is proportional to the area A of the surface of the water and that the constant of proportionality is k = 0.01
(a) The rate of change dV/dt of the volume of the water at time tis a net rate. Use this net rate to determine a differential equation for the height h of the water at time t.The volume of the water shown in the figure is [itex]V = \pi R [h(t)]^2 - \frac{1}{3} \pi [h(t)]^3[/itex], where R=10. Express the area of the surface of the water A=pi*r^2 terms of h.
(b) Solve the differential equation in part (a). Graph the
solution.
(c) If there were no evaporation, how long would it take
the tank to fill?
(d) With evaporation, what is the depth of the water at
the time found in part (c)? Will the tank ever be
filled? Prove your assertion.
-----------------------------------------------------------------------------------------------
We are given the volume as a function of the height: [itex]V = \pi R [h(t)]^2 - \frac{1}{3} \pi [h(t)]^3[/itex], where R=10, the radius of the tank. I shall denote h(t) as h, for convenience. Differentiating this with respect to time, I get
[itex]\frac{dV}{dt} = 2 \pi R h \frac{dh}{dt} - \pi h^2 \frac{dh}{dt}[/itex]
We are given that [itex]\frac{dV}{dt} = \pi - kA[/itex], where A=pi*r^2. So, the differential equation becomes,
[itex]\pi - k \pi r^2 = 2 \pi R h \frac{dh}{dt} - \pi h^2 \frac{dh}{dt}[/itex]
Dividing by pi, and solving for [itex]\frac{dh}{dt}[/itex],
[itex]1 - kr^2 = (2 R h - h^2 )\frac{dh}{dt}[/itex]
Dividing both sides by [itex]2 R h - h^2 [/itex],
[itex]\frac{1 - kr^2}{2 R h - h^2} = \frac{dh}{dt} \implies[/itex]
[itex]\frac{dh}{dt} = \frac{1 - kr^2}{2 R h - h^2} [/itex]
Using right-triangle trigonometry, I can relate the height of the water to the radius of the water at any given time, which gives me
[itex]R^2 = \left( R^2 - h \right)^2 + r^2 \implies[/itex]
[itex]r^2 = 2Rh - h^2 [/itex]
Therefore, the differential equation becomes
[itex]\frac{dh}{dt} = \frac{1 - k(2Rh - h^2)}{2 R h - h^2}[/itex]
Substituting in k=1/100 and R=10,
[itex]\frac{dh}{dt} = \frac{1 - \frac{1}{100}(20h - h^2)}{20 h - h^2}[/itex].
Multiplying the numerator and denominator by 100,
[itex]\frac{dh}{dt} = \frac{100 - 20h + h^2}{100(20 h - h^2)}[/itex].
Factoring the numerator,
[itex]\frac{dh}{dt} = \frac{(h-10)^2}{100(20 h - h^2)}[/itex]
Multiplying both sides of the equation by the reciprocal of the right-hand side of the equation, we get
[itex]\frac{100(20 h - h^2)}{(h-10)^2} \frac{dh}{dt} = 1[/itex] (1)
----------------------------------------------------------------
Now, according the answer key, the numerator should actually be [itex]100h(h-20)[/itex]. I have tried, and tried--and tried!--but I can not reproduce what the answer has. Somewhere I must be missing a negative sign, and I can not find it.
I am pretty certain that I am wrong, because when I went to go answer parts (c) and (d), after having used separation of variables on (1), which gave me a solution, I got wrong answers.
For part (c), I was able to calculate the time 666.7, which is correct according to the answer key; however, when I went to calculate the height at t=666.7, using the incorrect equation I derived from equation (1), I got a height of around 25. The reason why this is wrong is because you are supposed to show in part (d) that the height of the water in the tank can never exceed 10.
Could someone possibly help me?
I was not certain as to whether I should begin a new thread or not, but I am still working on the integral found in the post http://ift.tt/1gQDxjB
If a moderator deems this should be merged with the linked thread, please do so. I felt as though I should begin a new thread, because the other starting becoming littered; and also, I have a lot more information I need to provide.
Okay, so I have just revisited this integral, and I still can't seem to get the correct answer. So, here is my work, starting form the beginning.
Here is the problem:
An outdoor decorative pond in the shape of a hemispherical tank is to be filled with water pumped into the tank through an inlet in its bottom. Suppose that the radius of the tank is R10 ft, that water is pumped in at a rate of pi ft^3/min, and that the tank is initially empty. See Figure 3.2.6. As the tank fills, it loses water through evaporation. Assume that the rate of evaporation is proportional to the area A of the surface of the water and that the constant of proportionality is k = 0.01
(a) The rate of change dV/dt of the volume of the water at time tis a net rate. Use this net rate to determine a differential equation for the height h of the water at time t.The volume of the water shown in the figure is [itex]V = \pi R [h(t)]^2 - \frac{1}{3} \pi [h(t)]^3[/itex], where R=10. Express the area of the surface of the water A=pi*r^2 terms of h.
(b) Solve the differential equation in part (a). Graph the
solution.
(c) If there were no evaporation, how long would it take
the tank to fill?
(d) With evaporation, what is the depth of the water at
the time found in part (c)? Will the tank ever be
filled? Prove your assertion.
-----------------------------------------------------------------------------------------------
We are given the volume as a function of the height: [itex]V = \pi R [h(t)]^2 - \frac{1}{3} \pi [h(t)]^3[/itex], where R=10, the radius of the tank. I shall denote h(t) as h, for convenience. Differentiating this with respect to time, I get
[itex]\frac{dV}{dt} = 2 \pi R h \frac{dh}{dt} - \pi h^2 \frac{dh}{dt}[/itex]
We are given that [itex]\frac{dV}{dt} = \pi - kA[/itex], where A=pi*r^2. So, the differential equation becomes,
[itex]\pi - k \pi r^2 = 2 \pi R h \frac{dh}{dt} - \pi h^2 \frac{dh}{dt}[/itex]
Dividing by pi, and solving for [itex]\frac{dh}{dt}[/itex],
[itex]1 - kr^2 = (2 R h - h^2 )\frac{dh}{dt}[/itex]
Dividing both sides by [itex]2 R h - h^2 [/itex],
[itex]\frac{1 - kr^2}{2 R h - h^2} = \frac{dh}{dt} \implies[/itex]
[itex]\frac{dh}{dt} = \frac{1 - kr^2}{2 R h - h^2} [/itex]
Using right-triangle trigonometry, I can relate the height of the water to the radius of the water at any given time, which gives me
[itex]R^2 = \left( R^2 - h \right)^2 + r^2 \implies[/itex]
[itex]r^2 = 2Rh - h^2 [/itex]
Therefore, the differential equation becomes
[itex]\frac{dh}{dt} = \frac{1 - k(2Rh - h^2)}{2 R h - h^2}[/itex]
Substituting in k=1/100 and R=10,
[itex]\frac{dh}{dt} = \frac{1 - \frac{1}{100}(20h - h^2)}{20 h - h^2}[/itex].
Multiplying the numerator and denominator by 100,
[itex]\frac{dh}{dt} = \frac{100 - 20h + h^2}{100(20 h - h^2)}[/itex].
Factoring the numerator,
[itex]\frac{dh}{dt} = \frac{(h-10)^2}{100(20 h - h^2)}[/itex]
Multiplying both sides of the equation by the reciprocal of the right-hand side of the equation, we get
[itex]\frac{100(20 h - h^2)}{(h-10)^2} \frac{dh}{dt} = 1[/itex] (1)
----------------------------------------------------------------
Now, according the answer key, the numerator should actually be [itex]100h(h-20)[/itex]. I have tried, and tried--and tried!--but I can not reproduce what the answer has. Somewhere I must be missing a negative sign, and I can not find it.
I am pretty certain that I am wrong, because when I went to go answer parts (c) and (d), after having used separation of variables on (1), which gave me a solution, I got wrong answers.
For part (c), I was able to calculate the time 666.7, which is correct according to the answer key; however, when I went to calculate the height at t=666.7, using the incorrect equation I derived from equation (1), I got a height of around 25. The reason why this is wrong is because you are supposed to show in part (d) that the height of the water in the tank can never exceed 10.
Could someone possibly help me?
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