Exponential Growth and Decay

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Dean54321

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A tank in the shape of a vertical hexagonal prism with base area A is filled to a depth of 25 metres. The liquid inside is leaking through a small hole in the bottom of the tank, and it is found that the change in volume at any instant t hours after the tank starts leaking is proportional to the depth h metres, that is, dV/dt = −kh.

Show that dh/dt = − kh/A.
 
The volume of a vertical hexagonal prism is

[MATH]V = \frac{3\sqrt{3}}{2}a^2h[/MATH]
Differentiate both sides with respect to time. What do you get?
 
Please read https://www.freemathhelp.com/forum/threads/guidelines-summary.112086/#post-433156

One of the things we ask there is to say what you are studying. Another is to show your thoughts or work. You did neither. So I must GUESS that you are studying differential equations.

[MATH]dV = A \ dh \implies \int dV = \int A \ dh \implies V = Ah + c.[/MATH]
What is your next step? Do you need to find c or is c irrelevant to this problem?

EDIT: This is, in my opinion, an easier way to proceed than nasi’s, but both get you to the same place in the end.
 
Last edited:
JeffM said:
Please read https://www.freemathhelp.com/forum/threads/guidelines-summary.112086/#post-433156

One of the things we ask there is to say what you are studying. Another is to show your thoughts or work. You did neither. So I must GUESS that you are studying differential equations.

[MATH]dV = A \ dh \implies \int dV = \int A \ dh \implies V = Ah + c.[/MATH]
What is your next step? Do you need to find c or is c irrelevant to this problem?

EDIT: This is, in my opinion, an easier way to proceed than nasi’s, but both get you to the same place in the end.
Click to expand...
So I'm stuck on how to work out the equation in terms of h. Since for further rate problems, you normally only have 2 variables. (quantity and time)
It is asking for dh/dt,

I know dv/dt = -kh
so, h = Ae-kh ?
 
Dean54321 said:
So I'm stuck on how to work out the equation in terms of h. Since for further rate problems, you normally only have 2 variables. (quantity and time)
It is asking for dh/dt,

I know dv/dt = -kh
so, h = Ae-kh ?
Click to expand...
Did you read what I asked you to read? Apparently not.

Did you answer my questions? Apparently not.

Did you try what nasi suggested? Apparently not.

I am going to try one more time with you.

I gave you the equation

[MATH]V = Ah + c.[/MATH]
But c is irrelevant to this question and, relevant or not, c = 0 in this specific case.

So differentiate both sides of V = Ah, CONSIDERING BOTH AS FUNCTIONS OF TIME. If two functions are equal their derivatives are equal, right? That is what nasi was suggesting, right?

So [MATH]\dfrac{dV}{dt} = A * \dfrac{dh}{dt}.[/MATH]
Now what?
 
nasi112 said:
The volume of a vertical hexagonal prism is

[MATH]V = \frac{3\sqrt{3}}{2}a^2h[/MATH]
Differentiate both sides with respect to time. What do you get?
Click to expand...
Unnecessary complication considering the FIND and the GIVEN of the problem.

According to the problem:
Dean54321 said:
A tank in the shape of a vertical hexagonal prism with base area A is filled to a depth of 25 metres. The liquid inside is leaking through a small hole in the bottom of the tank, and it is found that the change in volume at any instant t hours after the tank starts leaking is proportional to the depth h metres, that is, dV/dt = −kh.

Show that dh/dt = − kh/A
Click to expand...

V = A * h

h = V/A

dh/dt = ?
 
Subhotosh Khan said:
Unnecessary complication considering the FIND and the GIVEN of the problem.
Click to expand...

It is not a complication at all. The OP should know that the base area of the vertical hexagonal prism is [MATH]A = \frac{3\sqrt{3}}{2}a^2[/MATH].

Then, the volume is [MATH]V = \frac{3\sqrt{3}}{2}a^2h = Ah[/MATH]
 
JeffM said:
Did you read what I asked you to read? Apparently not.

Did you answer my questions? Apparently not.

Did you try what nasi suggested? Apparently not.

I am going to try one more time with you.

I gave you the equation

[MATH]V = Ah + c.[/MATH]
But c is irrelevant to this question and, relevant or not, c = 0 in this specific case.

So differentiate both sides of V = Ah, CONSIDERING BOTH AS FUNCTIONS OF TIME. If two functions are equal their derivatives are equal, right? That is what nasi was suggesting, right?

So [MATH]\dfrac{dV}{dt} = A * \dfrac{dh}{dt}.[/MATH]
Now what?
Click to expand...
ok so V = Ah (also I'm not sure why you would add c since it is simply volume)
therefore uf dV/dt=-kh
then, d(Ah)/dv = -kh
A*(dh/dv) = -kh
dh/dv = -kh/A

Thank you very much.
 
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