Rates of Change/Newtons Law of Cooling (1 Viewer)

[tommykins]

I've covered this but I totally forgot how to do it, if someone can provide step by step solutions with explanations, love will be given.

1 ) A population of bacteria is decreasing as it is attacked by an antibody at a rate given by dp/dt = -k(P - 32000). If the number of bacteria decreases from 4.9*10^7 to 2.1*10^5 after 3 hours, find -

a) how many bacteria will be left after 5 hours
b) the rate at which the number of bacteria will be decreasing after 5 hours
c) the eventual number of bacteria left after the antibody finishes its work.


I know I need to find the formulas, but no idea how to start.

Thank you in advance

 

[lyounamu]

I've covered this but I totally forgot how to do it, if someone can provide step by step solutions with explanations, love will be given.

1 ) A population of bacteria is decreasing as it is attacked by an antibody at a rate given by dp/dt = -k(P - 32000). If the number of bacteria decreases from 4.9*10^7 to 2.1*10^5 after 3 hours, find -

a) how many bacteria will be left after 5 hours
b) the rate at which the number of bacteria will be decreasing after 5 hours
c) the eventual number of bacteria left after the antibody finishes its work.


I know I need to find the formulas, but no idea how to start.

Thank you in advance

Is it dp/dt or dP/dt? (I think it is dP/dt in this case)

 

[tommykins]

回复: Re: Rates of Change/Newtons Law of Cooling

dP/dt , sorry didn't think it'd be that big a deal.

 

[Zephyrio]

Re: 回复: Re: Rates of Change/Newtons Law of Cooling

Population = P + Ae^(-kt)
That's the standard formula you'd use for this kind of question.

Population = 32000 + Ae^(-kt)
When t = 0, P = 4.9*10^7
From that, find k.

Once you've found k, plug t = 5 into the question to find how much bacteria is there after 5 hours.

 

[tommykins]

回复: Re: 回复: Re: Rates of Change/Newtons Law of Cooling

Ahh yes I've figured it out, thanks alot

EDIT : Can't get b) out.

 

[lyounamu]

I've covered this but I totally forgot how to do it, if someone can provide step by step solutions with explanations, love will be given.

1 ) A population of bacteria is decreasing as it is attacked by an antibody at a rate given by dp/dt = -k(P - 32000). If the number of bacteria decreases from 4.9*10^7 to 2.1*10^5 after 3 hours, find -

a) how many bacteria will be left after 5 hours
b) the rate at which the number of bacteria will be decreasing after 5 hours
c) the eventual number of bacteria left after the antibody finishes its work.


I know I need to find the formulas, but no idea how to start.

Thank you in advance

dP/dt = -k(P-32000)
dt/dP = 1/(-k(P-32000) (because you cannot integrate until you have same variable)
t = 1/-k . ln(P-32000) + c
t - c = 1/-k . ln(P-32000)
-k(t-c) = ln (P-32000)
P-32000 = e^(-k(t-c))
P-32000 = e^-kt . e^kc (where e^kc is a constant, i.e. A)
P-32000 = Ae^-kt
P = Ae^-kt + 32000

When t=0, P = 4.9 . 10^7. So, A=4.9 . 10^7 - 32000 = 48968000
When t=3, P = 2.1 . 10^5
Therefore, 2.1 . 10^5 = 48968000e^-3k + 32000
178000/48968000 = e^-3k
-3k = ln(178000/4896800)
k = ln(178000/4896800)/-3
= 1.1048512...

Now, P = 4896800e^-5 . 1.1048512... + 32000
= 51532.5266...

dP/dt = -kAe^-kt + 32000
= -1.1048512... . 4896800 . e^-1.1048512... . t + 32000
When t = 5, dP/dt = -1.1048512 ... . 4896900 . 3^-1.1-48512... . 5 + 32000
= ... (cannot be bothered to find it... lost all the data of k & other stuffs in my calculator)

 

[tommykins]

回复: Re: Rates of Change/Newtons Law of Cooling

dP/dt = -k(P-32000)
dt/dP = 1/(-k(P-32000) (because you cannot integrate until you have same variable)
t = 1/-k . ln(P-32000) + c
t - c = 1/-k . ln(P-32000)
-k(t-c) = ln (P-32000)
P-32000 = e^(-k(t-c))
P-32000 = e^-kt . e^kc (where e^kc is a constant, i.e. A)
P-32000 = Ae^-kt
P = Ae^-kt + 32000

When t=0, P = 4.9 . 10^7. So, A=4.9 . 10^7 - 32000 = 48968000
When t=3, P = 2.1 . 10^5
Therefore, 2.1 . 10^5 = 48968000e^-3k + 32000
178000/48968000 = e^-3k
-3k = ln(178000/4896800)
k = ln(178000/4896800)/-3
= 1.1048512...

Now, P = 4896800e^-5 . 1.1048512... + 32000
= 51532.5266...

dP/dt = -kAe^-kt + 32000
= -1.1048512... . 4896800 . e^-1.1048512... . t + 32000
When t = 5, dP/dt = -1.1048512 ... . 4896900 . 3^-1.1-48512... . 5 + 32000
= ... (cannot be bothered to find it... lost all the data of k & other stuffs in my calculator)

That k value is incorrect, I got the k value to be -1.8723 etc.and that got me the right answer for part a).

b) is giving me trouble though.

 

[lyounamu]

Re: 回复: Re: Rates of Change/Newtons Law of Cooling

That k value is incorrect, I got the k value to be -1.8723 etc.and that got me the right answer for part a).

b) is giving me trouble though.

I probably made a calculation error. I did it on the computer so I guess it's common to make a mistake.

 

[tommykins]

回复: Re: 回复: Re: Rates of Change/Newtons Law of Cooling

Yeah. I tried differentiating P and subbing in t = 5 but that still doesn't work.

Meh thanks anyways.