# Water poured into leaky container

1. Jan 16, 2014

### jdawg

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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.

2. Jan 16, 2014

### .Scott

I get the same answer. Do you know what the correct answer for C is?

3. Jan 16, 2014

### jdawg

No I don't, is it maybe a problem with significant figures? My online homework is kind of picky.

4. Jan 16, 2014

### .Scott

If it's significant figures (which it may very well be) have you tried 0.040?

5. Jan 16, 2014

### jdawg

Thank you so much! It worked :)

6. Jan 16, 2014

### Staff: Mentor

Maybe they want the answer in grams/sec?

7. Jan 16, 2014

### jdawg

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.

8. Jan 16, 2014

### jdawg

@Chestermiller No, they want it in km/min.

9. Jan 16, 2014

### .Scott

For D, try -0.01. Depending on the exact precision rules it could also be 0.

10. Jan 16, 2014

### jdawg

I tried both and neither one worked :(

11. Jan 16, 2014

### .Scott

Then it must be -0.069.
(I checked you calculation - it's right)

12. Jan 16, 2014

### jdawg

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. Jan 16, 2014

### Staff: Mentor

I get -0.0069 rather than -0.069

14. Jan 16, 2014

### jdawg

Thanks Chester, that one worked!