Water Mass in Leaking Container

In summary, the problem involves finding the maximum mass of water in a leaking container over a period of time. Using the given equation, the maximum mass is approximately 33.1 grams at t = 16.2 seconds. To find the rate of mass change, the derivative of the equation is taken and plugged in for the specified times, then converted to the proper units of kg/min. The final answers for parts (c) and (d) are 0.0857 kg/min and 0.0437 kg/min respectively.
  • #1
IsoGD

Homework Statement


Water is poured into a container that has a leak. The mass m of the water is given as a function of time t by m = 6.00t 0.8 - 2.75t + 22.00, with t
gteq.gif
0, m in grams, and t in seconds.

(a) At what time is the water mass greatest? (s)
(b) What is that greatest mass? (g)
(c) In kilograms per minute, what is the rate of mass change at t = 2.00 s? (kg/min)
(d) In kilograms per minute, what is the rate of mass change at t = 5.00 s? (kg/min)

Homework Equations


m = 6.00t 0.8 - 2.75t + 22.00

The Attempt at a Solution


I didn't really have any clue on how to start this problem at all, so I was just able to find that the greatest mass is ~33.1 grams from guessing and checking. After that, I tried to see if I could work out what the time was by plugging the mass back into the equation, but the answer I got (17.7 seconds) was showing as incorrect on Webassign. So I really just need to be explained how to even start this problem in the first place so that I can work from there.

*Edited superscripts in*
 
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  • #2
IsoGD said:
6.00t 0.8 - 2.75t + 22.00, with t
gteq-gif.gif
0, m in grams, and t in seconds.
From your answer, I deduce this is supposed to be 6.00t0.8. You can use the X2 button above the typing area to get superscripts.
How do you normally find maxima and minima of functions? Haven't you been taught a calculus method?
 
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  • #3
haruspex said:
From your answer, I deduce this is supposed to be 6.00t0.8. You can use the X2 button above the typing area to get superscripts.
How do you normally find maxima and minima of functions? Haven't you been taught a calculus method?
Sorry about that, didn't notice the mistake. And this homework was assigned on the first day, Friday the 18th, without any prior instruction, and I'm taking Calculus and this Physics class at the same time so I haven't learned any methods for this.
 
  • #4
IsoGD said:
Sorry about that, didn't notice the mistake. And this homework was assigned on the first day, Friday the 18th, without any prior instruction, and I'm taking Calculus and this Physics class at the same time so I haven't learned any methods for this.
Ok, not unusual that you need calculus methods in physics before you get taught them in maths.
Do you know how to differentiate xk with respect to x?
 
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  • #5
haruspex said:
Ok, not unusual that you need calculus methods in physics before you get taught them in maths.
Do you know how to differentiate xk with respect to x?
Well, after a bit of looking things up, I found out how to. I got the derivative (24 / (5*t1/5)) - 2.75 after some trial and error. But after I found that and tried to solve it for where the slope equaled zero (like I read I was supposed to do?) I only found a time of t = .062 seconds, which is obviously wrong. What is it that I'm doing wrong? Am I not supposed to set the derivative equal to zero and solve for this problem and I'm actually supposed to something else?
 
  • #6
How did you get 0.062? Have you tried using logarithms to solve it rather than trial and error?
 
  • #7
scottdave said:
How did you get 0.062? Have you tried using logarithms to solve it rather than trial and error?
When I looked up how to do this, it said that I should set the derivative equal to zero, and solve for t. Every time I did that I got .062. I never saw anything on logarithms before.
 
  • #8
I see what you did, now. Did you take into account that t1/5 is in the denominator? So logs are not the only way to solve it.
 
  • #9
scottdave said:
I see what you did, now. Did you take into account that t1/5 is in the denominator? So logs are not the only way to solve it.
Yes, I did now. I took a fresh look at the equation again and tried to just re-solve from the beginning of the derivative being equal to zero, as it turns it out my issue was just some bad math at that step. I was able to get t = 16.2 seconds after I tried it again and it was correct on Webassign. And to double check, I plugged that back into the original equation and got m = 33.1 grams.

Now, I'm not sure how to do parts c and b for this question. I've tried plugging t = 2 into the original equation without the constant 22, and converting to kg/min but it's obviously not that easy, but that's all I could think to do with it.
 
  • #10
What you need to know about derivatives, since they are new to you: The derivative (with respect to time) is equal to the rate of change.
So if you have a function f(t) which is equal to grams, when you plug in seconds for time, then taking the derivative (which you have now done) will be an expression in grams/second. Plug in the specified times, then convert to the proper units (from grams/second to kg/minute)
 
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  • #11
scottdave said:
What you need to know about derivatives, since they are new to you: The derivative (with respect to time) is equal to the rate of change.
So if you have a function f(t) which is equal to grams, when you plug in seconds for time, then taking the derivative (which you have now done) will be an expression in grams/second. Plug in the specified times, then convert to the proper units (from grams/second to kg/minute)
Wow, that explains a lot. Thank you! Thanks to that, I was able to get the two answers at .0857 kg/min and .0437 kg/min respectively for c and d, and they were correct on Webassign. I'm now finally done with this homework assignment because this was the last problem on it, so thanks for the help!
 
  • #12
IsoGD said:
Wow, that explains a lot. Thank you! Thanks to that, I was able to get the two answers at .0857 kg/min and .0437 kg/min respectively for c and d, and they were correct ...

Great! Do you understand how the derivative (rate) equaling zero is the place where a maximum (or minimum) will occur (for any general case)?
 
  • #13
scottdave said:
Great! Do you understand how the derivative (rate) equaling zero is the place where a maximum (or minimum) will occur (for any general case)?
Because the derivative (or rate) is essentially the slope of the equation, and when the slope (or rate) equals zero, the line is horizontal, meaning that it has hit a maximum or minimum, right?
 
  • #14
IsoGD said:
Because the derivative (or rate) is essentially the slope of the equation, and when the slope (or rate) equals zero, the line is horizontal, meaning that it has hit a maximum or minimum, right?
Yes.
 

Related to Water Mass in Leaking Container

1. How does water mass affect a leaking container?

The water mass in a leaking container can significantly alter the overall weight and distribution of the container, potentially causing it to become unstable and potentially leading to further leakage or damage.

2. How can the water mass in a leaking container be measured?

The water mass in a leaking container can be measured using various techniques such as weighing the container before and after the leak, using displacement methods, or using specialized equipment such as a hydrometer or moisture meter.

3. What factors can affect the water mass in a leaking container?

The water mass in a leaking container can be affected by various factors such as the size and shape of the container, the type of material the container is made of, the temperature and pressure of the surrounding environment, and the rate of leakage.

4. Can the water mass in a leaking container be controlled?

In some cases, the water mass in a leaking container can be controlled by adjusting the container's position or using absorbent materials to soak up the water. However, if the leak is significant or the container is compromised, it may be difficult to control the water mass.

5. How can the water mass in a leaking container be minimized?

The water mass in a leaking container can be minimized by quickly identifying and repairing the source of the leak. Additionally, implementing proper maintenance procedures and regularly inspecting the container can help prevent leaks from occurring in the first place.

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