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**jacs****Member**- Registered: 2006-01-16
- Posts: 2

Hi, my first post, and i need help with some applications of calculus to physics

A tank contains 100 litres of brine whose concentration is 3 grams/litres. Three litres of brine whose concentration is 2 grams/litre flow into the tank each minute and at the same time 3 litres of mixture flow out each minute.

Show that the quantity of salt, Q gram, in the tank at any time t is given by:

Q = 200 + 100e[sup]-0.03t[/sup]

i have managed to get Q = 200 - 100e[sup]-0.03t[/sup]

and cannot figure out why the minus is there, i have included the pdf of the working i have done so far, any help appreciated

thanks

jacs

*Last edited by jacs (2006-01-16 02:05:35)*

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

You use a different method for solving differential equations than I'm used to, and quite an awkward one at that. See if you can follow this, and if you want, I can try to find the error in your steps.

Integrate both sides

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,555

Nice work, Ricky!

jacs wrote:

Show that the quantity of salt, Q gram, in the tank at any time t is given by:

Q = 200 + 100e[sup]-0.03t[/sup]

And, of course, the 100e[sup]-0.03t[/sup] term decreases with t, so it *should* be added.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**jacs****Member**- Registered: 2006-01-16
- Posts: 2

thank you thankyou thankyou I see it now.... i had the negative log and so cheated assuming that the 3Q was 300 instead of 900 (which i secretly knew was wrong...lol....but coudnt get rid of negative log so sort fudged the results)

what i needed was just the clever maniplutaion of the minus...ohhh soo easy now i see

thanks

jacs

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