Water poured into leaky container

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The discussion revolves around solving a problem involving the mass of water in a leaky container, expressed as a function of time. The user successfully determined the time at which the water mass is greatest and the maximum mass itself but faced challenges with the rate of mass change at specific times. For part c, after calculating the derivative at t = 1.6 s, the user converted the rate to kg/min and initially struggled with significant figures, ultimately finding a suitable answer. In part d, the user calculated the derivative at t = 4.9 s and encountered difficulties with the conversion and significant figures, leading to confusion over the correct answer. The user plans to contact their professor for clarification on the discrepancies encountered in part d.
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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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