Hard Differential Equations problem and Newton's Law of Cool

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lamaclass

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Oct 18, 2009
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I'm not sure what to do for either of these problems:

1. The rate of growth of a population of rats is proportional to the size of the population. P(t) represents the population of time t. Starting with 500 rats, after 10 days, the population is growing at a rate of 25 new rats each day. Find P(t) in terms of t.

2. At 1 PM, one afternoon there is a power failure in your home in northern Wisconsin and your heat does not work without electricity. When the power goes out it is 64 degrees Farenheit in your house. At 10 PM, it is 48 degrees Farenheit in the house. At that that same time, you hear that the temperature is 10 degrees Farenheit and will be all night. Assuming the temperature T in your house follows Newton's law of cooling, what will the tempertaure be at 7 AM the next morning? When would your pipes be in danger of freezing (nearest minute)?
 
lamaclass said:
I'm not sure what to do for either of these problems:

1. The rate of growth of a population of rats is proportional to the size of the population. P(t) represents the population of time t. Starting with 500 rats, after 10 days, the population is growing at a rate of 25 new rats each day. Find P(t) in terms of t.

2. At 1 PM, one afternoon there is a power failure in your home in northern Wisconsin and your heat does not work without electricity. When the power goes out it is 64 degrees Farenheit in your house. At 10 PM, it is 48 degrees Farenheit in the house. At that that same time, you hear that the temperature is 10 degrees Farenheit and will be all night. Assuming the temperature T in your house follows Newton's law of cooling, what will the tempertaure be at 7 AM the next morning? When would your pipes be in danger of freezing (nearest minute)?
Click to expand...

Can you begin by starting with naming variables and taking a stab at deriving the ODE?

Then we will worry about solving it.
 
1. The rate of growth of a population of rats is proportional to the size of the population. P(t) represents the population of time t. Starting with 500 rats, after 10 days, the population is growing at a rate of 25 new rats each day. Find P(t) in terms of t.

I believe for this one, the variables would be set up as:

P(t)/dP(t) = Ce[sup:2ofe6abs]kt[/sup:2ofe6abs]

So would C be 500 as the initial point and then we'd have to start out solving for k?

2. At 1 PM, one afternoon there is a power failure in your home in northern Wisconsin and your heat does not work without electricity. When the power goes out it is 64 degrees Farenheit in your house. At 10 PM, it is 48 degrees Farenheit in the house. At that that same time, you hear that the temperature is 10 degrees Farenheit and will be all night. Assuming the temperature T in your house follows Newton's law of cooling, what will the tempertaure be at 7 AM the next morning? When would your pipes be in danger of freezing (nearest minute)?

Hmm it looks like for this one, 64 degrees would be the starting point but I'm not sure if the same formula would be used as it was for the first one. :?
 
lamaclass said:
1. The rate of growth of a population of rats is proportional to the size of the population. P(t) represents the population of time t. Starting with 500 rats, after 10 days, the population is growing at a rate of 25 new rats each day. Find P(t) in terms of t.

I believe for this one, the variables would be set up as:

P(t)/dP(t) = Ce[sup:2x04ljve]kt[/sup:2x04ljve]

You are mixing up "Differential equation" with final form.

You are given:The rate of growth of a population of rats is proportional to the size of the population

So:

dP/dt = kP

dP/P = kdt

P = C*e[sup:2x04ljve]kt[/sup:2x04ljve]

As you had said befre C = 500

P = 500e[sup:2x04ljve]kt[/sup:2x04ljve]

we still need to find 'k'. For that we use:

after 10 days, the population is growing at a rate of 25 new rats each day

dP/dt = k* 500* e[sup:2x04ljve]kt[/sup:2x04ljve]

then

25 = k * 500 * e[sup:2x04ljve]k*10[/sup:2x04ljve]

Find 'k' from above - put it all together - and you are done...

So would C be 500 as the initial point and then we'd have to start out solving for k?

2. At 1 PM, one afternoon there is a power failure in your home in northern Wisconsin and your heat does not work without electricity. When the power goes out it is 64 degrees Farenheit in your house. At 10 PM, it is 48 degrees Farenheit in the house. At that that same time, you hear that the temperature is 10 degrees Farenheit and will be all night. Assuming the temperature T in your house follows Newton's law of cooling, what will the tempertaure be at 7 AM the next morning? When would your pipes be in danger of freezing (nearest minute)?

Hmm it looks like for this one, 64 degrees would be the starting point but I'm not sure if the same formula would be used as it was for the first one. :?

For this one - the DE would be:

dT/dt = -k(T-10)

and gofrom there
Click to expand...
 
Thank you for your help!

For the first one, here's what I got:

25=500e[sup:liohjfib]10k[/sup:liohjfib]

1/20=e[sup:liohjfib]10k[/sup:liohjfib]

ln 1/20 = 10k

ln 1/20/10 = k

=500e[sup:liohjfib]ln 1/20/10t[/sup:liohjfib]

=370.57 min

Is this correct for the first one?
 
lamaclass said:
Thank you for your help!

For the first one, here's what I got:

25=500e[sup:1z7lxumh]10k[/sup:1z7lxumh] <<< Where did you get that?

1/20=e[sup:1z7lxumh]10k[/sup:1z7lxumh]

ln 1/20 = 10k

ln 1/20/10 = k

=500e[sup:1z7lxumh]ln 1/20/10t[/sup:1z7lxumh]

=370.57 min

Is this correct for the first one?
Click to expand...
 
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