Thank you for your help,

Printable View

- Dec 2nd 2009, 01:09 AMkiranizzleHelp on exponential question
Thank you for your help,

- Dec 2nd 2009, 01:46 AMCalculus26
I'll give you the answer for the growing case in i and ii

Note y = Ae^(kt)

y(0) = A

y = y(0)e^(kt) = y0 e^(kt)

if y(t1) = y1

Then y1 = y0 e^(kt1)

1/kln(y1/y0) = t1

Similarly for ii y(tn) = yn

yields 1/k ln(yn/y0) = tn

For your Lake problem You know your solution can't be correct

If P(t)=100,000e^(-t/1000) then P(0) =100,00

But if the concentration is initially .1% and there are 108 m^3

P(0) = .108m^3 - Dec 2nd 2009, 02:41 AMCalculus26lake pblm revisited
It occurred to me you mean 10^8 m^3 and 10^5 m^3/day

otherwise if it were 108 and 105 you would have a mud puddle.

so dP/dt = pollution in -pollution out

dp/dt = 0 - %pollution*flow rate

dP/dt = -(P(t)/10^8) (10^5 m^2/day)

dP/dt = -P(t)/1000 P(0) = .001* 10^8 = 10^5

dp/P = -1/1000dt

ln(P) = -t/1000 +C

e^(ln(P) = e^(-t/1000+C) = e^C e^(-t/1000)

Since e^C is a constant we can just call it C

P = C*e^(-t/1000)

P(0) = 100,000 = C

P = 100,000 e^(-t/1000)

To find when it 1/10 original solve 10,000 = 100,000 e^(-t/1000)

1/10 = e^(-t/1000)

1/1000ln(10) = t