Rate of change question (pretty confusing)

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Homework Help Overview

The problem involves determining the rate of change of water level in a tank, described by the volume function V(h) = (0.2h^3 + 3h) m³, where h is the depth of water. Water is being added at a constant rate of 0.4 m³ per second, and the specific question is about the rate at which the water level rises when the depth is 2.1 meters.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss differentiating the volume function and applying the chain rule to relate the rates of change. There are attempts to express dH/dt in terms of known quantities, with some participants suggesting different formulations of the relationships between the rates.

Discussion Status

Several participants have shared their calculated rates of change, with some expressing uncertainty about their results. There is an ongoing dialogue about the correct application of differentiation and the relationships between the variables involved. No explicit consensus has been reached regarding the final answer.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information available for discussion. There are also informal exchanges that diverge from the mathematical focus, indicating a mix of serious inquiry and light-hearted banter.

Sirsh
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3. The volume of a tank used to provide water for animals at an animal enclosure at the local show is given by V(h) = (0.2h^3+3h)m^3 where h metres is the depth of the water in the tank at any time t seconds. Water is being added to the tank at a constant rate of 0.4 m3 per second. Find the rate at which the water level is rising when the depth of water in the tank is 2.1 metres.

So i differentiated the equation: v(h) = (0.2h^3+3h) m^3, which is v'(h) = (0.6h^2+3) m^3. But because the limit is 2.1m, and the amount of liquid going into the object is going in at 0.4m^3 a second. so if i use v'(2.1)= (0.6h^2+3) m^3 and solve for h.. i can divide h by 0.4 and find out the time it took to reach 2.1m?

Thank you.
 
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U need to find dH/dt, i.e the rate at which water level is rising,,
lets say, dV/dH=(dV/dt).(dt/dH).
write dV/dH expression, from the relation given at h=2.1 m.
since know what dV/dt is i.e, o.4 m^3/sec, u can find out dH/dt!
 
What was the answer you got? I got 0.071 m/s.
 
i got 0.053
 
Then I'll confirm that Anakin_k is correct.
 
I think it should be dV/dT = dH/dT * dV/dH rather than what you had, Mandeep.
 
its all the same! u cn write it whichever way u wish too
u have dV/dt, dV/dh... its not a big deal to get dH/dt
 
dude m really sorry... i messd up the calc..
its 0.071 only!
 
Mandeep Deka said:
dude m really sorry... i messd up the calc..
its 0.071 only!

A piece of writing advice. You don't really need to put exclamation marks in everything you write. At some point it starts to lose its meaning. In fact, in a math homework forum I dare say an exclamation point would probably never be necessary.
 
  • #10
deltapapazulu said:
In fact, in a math homework forum I dare say an exclamation point would probably never be necessary.

But if I couldn't use exclamation marks then what good does does it this equality? 3+3=3!

:smile:
 
  • #11
I think you must be really excited about 3+3=3. I, on the other hand, know that 3+3!=3. :p
 
  • #12
Mindscrape said:
I think you must be really excited about 3+3=3.
I never claimed this. Please, quote specifically where I ever claimed such an invalid response in any of my posts here.

I, on the other hand, know that 3+3!=3. :p
You don't know very much then do you :-p
 
  • #13
Mentallic said:
But if I couldn't use exclamation marks then what good does does it this equality? 3+3=3!

:smile:

Mindscrape said:
I think you must be really excited about 3+3=3. I, on the other hand, know that 3+3!=3. :p

A few spaces would make it more clear that you meant 3 + 3 != 3, or better yet, 3 + 3 \neq 3.
 

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