# exponential function

• December 18th 2008, 08:23 AM
nrslmz
exponential function
1000000 viruses are reproducing exponentially with a rate of 160% in four hours. At the same time, the body is eliminating the viruses constantly at 50000 per hour. What would be the function giving the virus population at a given time?

k must be ln(1,6)/4, because
e^k*4 must be 1,6.
the differential equation for exponential growth is
dP/dt = Po * e ^(kt) * k

• December 18th 2008, 08:48 AM
tester85
This is what i thought but i maybe wrong. What do you think ?

v(t) = 1000000^1.6(t/4) - 50000t
• December 18th 2008, 08:52 AM
nrslmz
Yes, I have tried that one. I think that is incorrect because while the viruses grow exponentially, the elimination is constant. Each time the viruses reproduce it takes out 50000, changing the initial virus population. If you do that, I think (BUT NOT KNOW) multiplicating the initial virus number by e^k t times won't work.
• December 18th 2008, 09:02 AM
tester85
Ok noted. Anyway previous post thread updated due to type should be - 5000t instead of + 5000t. Will let you know if i work something out.
• December 18th 2008, 01:29 PM
chabmgph
Quote:

Originally Posted by nrslmz
1000000 viruses are reproducing exponentially with a rate of 160% in four hours. At the same time, the body is eliminating the viruses constantly at 50000 per hour. What would be the function giving the virus population at a given time?

k must be ln(1,6)/4, because
e^k*4 must be 1,6.
the differential equation for exponential growth is
dP/dt = Po * e ^(kt) * k

Hi nrslmz,

This looks like an interesting question. I am probably trying the long way, but this is what I have worked out:

When $t = 0, P_0=1000000$

When $t = 1, P_1=P_0(1.6)^{\frac{1}{4}}-50000$
$=1000000(1.6)^{\frac{1}{4}}-50000$

When $t = 2, P_2=P_1(1.6)^{\frac{2}{4}}-50000$
$=[1000000(1.6)^{\frac{1}{4}}-50000](1.6)^{\frac{2}{4}}-50000$
$=1000000(1.6)^{\frac{1}{4}+\frac{2}{4}}-50000(1.6)^{\frac{2}{4}}-50000$

When $t = 3, P_3=P_2(1.6)^{\frac{3}{4}}-50000$
$=[1000000(1.6)^{\frac{1}{4}+\frac{2}{4}}-50000(1.6)^{\frac{2}{4}}-50000](1.6)^{\frac{3}{4}}-50000$
$=1000000(1.6)^{\frac{1}{4}+\frac{2}{4}+\frac{3}{4} }-50000(1.6)^{\frac{2}{4}+\frac{3}{4}}-50000(1.6)^{\frac{3}{4}}-50000$

When $t = 4, P_4=P_3(1.6)^{\frac{4}{4}}-50000$
$=[1000000(1.6)^{\frac{1}{4}+\frac{2}{4}+\frac{3}{4}}-50000(1.6)^{\frac{2}{4}+\frac{3}{4}}-50000(1.6)^{\frac{3}{4}}-50000](1.6)^{\frac{4}{4}}-50000$
$=1000000(1.6)^{\frac{1}{4}+\frac{2}{4}+\frac{3}{4} +\frac{4}{4}}-50000(1.6)^{\frac{2}{4}+\frac{3}{4}+\frac{4}{4}}-50000(1.6)^{\frac{3}{4}+\frac{4}{4}}-50000(1.6)^{\frac{4}{4}}-50000$