At what time was the yeast added?

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Homework Help Overview

The problem involves a scenario from a physics story where a girl suffocates due to rapidly rising bread dough caused by yeast. The dimensions of the kitchen and the initial volume of the dough are provided, along with the rate at which the dough expands. The main question is to determine the time at which the yeast was added, based on the information given.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss their calculated times for when the yeast was added, with differing results. Some mention using geometric series to approach the problem, while others suggest logarithmic methods for solving the volume expansion.

Discussion Status

Multiple interpretations of the problem are being explored, with participants sharing their reasoning and calculations. Some guidance has been offered regarding the mathematical setup of the volume expansion, but no consensus has been reached on the correct time.

Contextual Notes

Participants are working with the constraints of the problem as presented in a fictional context, and there is uncertainty regarding the rounding of calculations and the interpretation of the mathematical series involved.

mborn
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Hi, I had this story in my Physics book

The police was baffled by what seemed to be the prefect nurder of a girlwho had been found, apparently suffocated, in her kitchen. The girl had been making bread in her kitchen, whose dimensions were 6,10, 10 m. she had formed the dough into a ball of volume 1/6 cubic m and turned away to wash some dishes. The criminal added a special virulent strain of yeast to the bread. As a result the bread immediately started to rise in volume, triple every 4 min. before long the dough filled the room, stopping the clock at 3:48 and squashing the girl into the wall. By the time the police came the next day, the yeast worked itself and the dough returned to its normal size.

at what time was the yeast added?

My answer is 3:20 is it true?

M B
 
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I got 3:18. Maybe we rounded differently?
 
I too (like tide) get t ~ 29 min 50 sec
 
Ok!,
Since the answer is not there in my book, I say I solved it as a geometric series, am I right?
 
I am not sure what you mean when you say you "solved it as a geometric series." Please explain. :-)
 
Well,
a geometric series goes like this
a, ar, a(r^2), a(r^3), ...
General term take the following look
ar^(n-1)
now we have the initial volume (1/6 cubic meter) which is the first term, a.
the nth term is the final volume 6*10*10. therefore we need to know how mnay terms are there to reach 600 starting with 1/6
accordingly;
600=(1/6) * 3^(n-1) and solve for n-1
the time taken to reach the volume 600 is (n-1) * 4 min
since it was found that the clock stopped at 3:48
therefore, t0=3:48 - (n-1)*4

M B
 
The question reduces to how you solved that last equation for n. I think it might be clearer if you set up the problem as follows:

Since the volume triples every four minutes you can write the volume as

[tex]V = V_0 \times 3^{t/4}[/tex]

where t is the elapsed time. You can solve for t using logarithms:

[tex]t = 4 \frac {\ln V/V_0}{\ln 3}[/tex]
 
same as my answer without rounding.

Thanks

M B