Water poured into leaky container

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Homework Help Overview

The problem involves a mathematical model of water mass in a leaky container, expressed as a function of time. Participants are tasked with determining the time at which the water mass is greatest, the maximum mass, and the rate of mass change at specific times, with conversions between units required for some parts.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss their attempts to find the derivative of the mass function and evaluate it at specific times. There are questions about significant figures and unit conversions, particularly for parts c and d.

Discussion Status

Some participants have shared their calculations and results, while others express uncertainty about the correctness of their answers. There is a mix of agreement and differing opinions on the significance of figures and the appropriate units for the final answers. The discussion remains open with no explicit consensus reached.

Contextual Notes

Participants mention the potential for issues related to significant figures and the specific requirements of their online homework platform. There are indications that the problem may have ambiguities or errors, prompting some to consider reaching out to their professor for clarification.

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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 = 5.3t0.8 - 3.2t + 22, with t ≥ 0, m in grams, and t in seconds. (a) At what time is the water mass greatest, and (b) what is that greatest mass? What is the rate of mass change at (c) t = 1.6 s and (d) t = 4.9 s? Change g/s to kg/min for part c and d.

Homework Equations





The Attempt at a Solution


I already got part a and part b, but I'm having trouble with parts c and d.
This is what I got for the derivative: dm/dt=4.24t^-0.2-3.2
Part C: I plugged in 1.6 into my derivative and got 0.6595961104. then I multiplied by .06 to convert to kg/min and got 0.0395757666. But that's not the right answer.
 
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I get the same answer. Do you know what the correct answer for C is?
 
No I don't, is it maybe a problem with significant figures? My online homework is kind of picky.
 
jdawg said:
No I don't, is it maybe a problem with significant figures? My online homework is kind of picky.
If it's significant figures (which it may very well be) have you tried 0.040?
 
Thank you so much! It worked :)
 
Maybe they want the answer in grams/sec?
 
Now I'm having trouble with part D. I plugged in 4.9 into the derivative and got -0.1144922571 g/s and then converted it to kg/min and got -0.0068695354. I tried putting in -0.007 for significant figures, but that didn't work.
 
@Chestermiller No, they want it in km/min.
 
For D, try -0.01. Depending on the exact precision rules it could also be 0.
 
  • #10
I tried both and neither one worked :(
 
  • #11
Then it must be -0.069.
(I checked you calculation - it's right)
 
  • #12
That one didn't work either, maybe there's something wrong with that problem. I'll try e-mailing my professor about it. Thanks so much for your help!
 
  • #13
jdawg said:
That one didn't work either, maybe there's something wrong with that problem. I'll try e-mailing my professor about it. Thanks so much for your help!
I get -0.0069 rather than -0.069
 
  • #14
Thanks Chester, that one worked!
 

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