# Antiderivatives word problem!

• Nov 28th 2010, 12:30 PM
yess
Antiderivatives word problem!
The population of a bacterial colony t hours after observation begins is found to be changing at the rate dP/dt = 200e^0.1t+150e^-0.03t. If the population was 200,000 bacteria when observations began, what will the population be 12 hours later?
what i did:
(1/12-0)[antiderivative 200e^0.1t+150e^-0.03t]
(1/12)[(200/1.1)e^1.1t + (150/0.97)e^0.97t]
(1/12)[((200/1.1)e^13.2 + (150/0.97)e^11.64) - ((200/1.1)e^0 + (150/0.97)e^0)]
9650594.820

is this right?? seems like a big number!
• Nov 28th 2010, 12:49 PM
pickslides
$\displaystyle \displaystyle \int 200e^{0.1t}+150e^{-0.03t}~dt = 2000e^{0.1t}-\frac{100}{3}\times 150e^{-0.03t}+C$
• Nov 28th 2010, 12:50 PM
e^(i*pi)
Quote:

Originally Posted by yess
The population of a bacterial colony t hours after observation begins is found to be changing at the rate dP/dt = 200e^0.1t+150e^-0.03t. If the population was 200,000 bacteria when observations began, what will the population be 12 hours later?
what i did:
(1/12-0)[antiderivative 200e^0.1t+150e^-0.03t]
(1/12)[(200/1.1)e^1.1t + (150/0.97)e^0.97t]
(1/12)[((200/1.1)e^13.2 + (150/0.97)e^11.64) - ((200/1.1)e^0 + (150/0.97)e^0)]
9650594.820

is this right?? seems like a big number!

$\displaystyle \dfrac{dP}{dt} = 200e^{0.1t} + 150e^{-0.03t}$

I don't know where you get the 1 and 12 from, if you assume the time observations began is time 0 your limits are 12 and 0.

You've also got the integration wrong, when talking about exponentials we do not add 1 to the power: $\displaystyle \int e^{ax} = \dfrac{1}{a}e^{ax} + C$ which can be shown by taking the derivative using the chain rule

$\displaystyle \displaystyle \int^{12}_0 200e^{0.1t}+150e^{-0.03t}~dt = \left[2000e^{0.1t}-\frac{100}{3}\times 150e^{-0.03t}\right]^{12}_0$
• Nov 28th 2010, 06:04 PM
yess
Quote:

Originally Posted by e^(i*pi)
$\displaystyle \dfrac{dP}{dt} = 200e^{0.1t} + 150e^{-0.03t}$

I don't know where you get the 1 and 12 from, if you assume the time observations began is time 0 your limits are 12 and 0.

You've also got the integration wrong, when talking about exponentials we do not add 1 to the power: $\displaystyle \int e^{ax} = \dfrac{1}{a}e^{ax} + C$ which can be shown by taking the derivative using the chain rule

$\displaystyle \displaystyle \int^{12}_0 200e^{0.1t}+150e^{-0.03t}~dt = \left[2000e^{0.1t}-\frac{100}{3}\times 150e^{-0.03t}\right]^{12}_0$

so using your equation I sub in 12 for t and then I sub in 0 for t and subtract that from the value where t = 12? This gives me 6151.852215 so does this mean the population 12 hours later is about 6151.85 bacteria?
• Nov 28th 2010, 06:09 PM
skeeter
Quote:

Originally Posted by yess
so using your equation I sub in 12 for t and then I sub in 0 for t and subtract that from the value where t = 12? This gives me 6151.852215 so does this mean the population 12 hours later is about 6151.85 bacteria?

add the original 200000 ... the definite integral calculates only the change in the population.
• Nov 28th 2010, 06:41 PM
yess
Quote:

Originally Posted by skeeter
add the original 200000 ... the definite integral calculates only the change in the population.

ok thanks =]