A type of algae grows continuously such that its population doubles after 3 days. What's the population after 10 days? The 2019 Stack Overflow Developer Survey Results Are InSentence such that the universe of a structure has exactly two membersWhat's the error in this argument that Fin$le_m$InfWhat's the need for a pair type $A times B$ in Homotopy Type Theory?What's the meaning of an element that belongs to the same element?Finding a type such that $X + 1 notsimeq X$ and $X+2simeq X$Understanding types and the proof that every type is realized in an elementary extension.Show that a sequence of elements each realizing an isolated type over the previous realizes an isolated typeProof that propositional logic remains consistent after the addition of a single axiomShow that computation of all such expressions will result in the same element of $X$Do we need 'such that' after qualifiers?
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A type of algae grows continuously such that its population doubles after 3 days. What's the population after 10 days?
The 2019 Stack Overflow Developer Survey Results Are InSentence such that the universe of a structure has exactly two membersWhat's the error in this argument that Fin$le_m$InfWhat's the need for a pair type $A times B$ in Homotopy Type Theory?What's the meaning of an element that belongs to the same element?Finding a type such that $X + 1 notsimeq X$ and $X+2simeq X$Understanding types and the proof that every type is realized in an elementary extension.Show that a sequence of elements each realizing an isolated type over the previous realizes an isolated typeProof that propositional logic remains consistent after the addition of a single axiomShow that computation of all such expressions will result in the same element of $X$Do we need 'such that' after qualifiers?
$begingroup$
A type of algae grows continuously so that its population doubles in 3
days. Given a beginning population of 100 algae cells per milliliter
of water, to the nearest whole number, how many algae cells would you
expect at the end of 10 days?
I start off by dividing 10 by 3 giving me 3 doubles in population:
$$100rightarrow200rightarrow400rightarrow800$$
Now I have to deal with the extra day. I get $800+frac13cdot800approx1067$ cells per milliliter. This is incorrect. Where is my logic or math wrong? How should I solve this problem?
Thanks! Your help is appreciated!
Max0815
logic
asked Mar 23 at 18:17
Max0815Max0815
81418
$endgroup$
add a comment |
$begingroup$
A type of algae grows continuously so that its population doubles in 3
days. Given a beginning population of 100 algae cells per milliliter
of water, to the nearest whole number, how many algae cells would you
expect at the end of 10 days?
I start off by dividing 10 by 3 giving me 3 doubles in population:
$$100rightarrow200rightarrow400rightarrow800$$
Now I have to deal with the extra day. I get $800+frac13cdot800approx1067$ cells per milliliter. This is incorrect. Where is my logic or math wrong? How should I solve this problem?
Thanks! Your help is appreciated!
Max0815
logic
asked Mar 23 at 18:17
Max0815Max0815
81418
$endgroup$
$begingroup$
If the algae doubles in three days then the equation is $y(t)=y(0)cdot 2^fract3$, where $y(0)$ is the initial population. For instance, if $t=3$ the equation becomes $y(3)=y(0)cdot 2^frac33=y(0)cdot 2$,
$endgroup$
– callculus
Mar 23 at 18:37
add a comment |
$begingroup$
A type of algae grows continuously so that its population doubles in 3
days. Given a beginning population of 100 algae cells per milliliter
of water, to the nearest whole number, how many algae cells would you
expect at the end of 10 days?
I start off by dividing 10 by 3 giving me 3 doubles in population:
$$100rightarrow200rightarrow400rightarrow800$$
Now I have to deal with the extra day. I get $800+frac13cdot800approx1067$ cells per milliliter. This is incorrect. Where is my logic or math wrong? How should I solve this problem?
Thanks! Your help is appreciated!
Max0815
logic
asked Mar 23 at 18:17
Max0815Max0815
81418
$endgroup$
A type of algae grows continuously so that its population doubles in 3
days. Given a beginning population of 100 algae cells per milliliter
of water, to the nearest whole number, how many algae cells would you
expect at the end of 10 days?
I start off by dividing 10 by 3 giving me 3 doubles in population:
$$100rightarrow200rightarrow400rightarrow800$$
Now I have to deal with the extra day. I get $800+frac13cdot800approx1067$ cells per milliliter. This is incorrect. Where is my logic or math wrong? How should I solve this problem?
Thanks! Your help is appreciated!
Max0815
logic
logic
asked Mar 23 at 18:17
Max0815Max0815
81418
asked Mar 23 at 18:17
Max0815Max0815
81418
asked Mar 23 at 18:17
Max0815Max0815
81418
asked Mar 23 at 18:17
Max0815Max0815
81418
asked Mar 23 at 18:17
Max0815Max0815
81418
81418
$begingroup$
If the algae doubles in three days then the equation is $y(t)=y(0)cdot 2^fract3$, where $y(0)$ is the initial population. For instance, if $t=3$ the equation becomes $y(3)=y(0)cdot 2^frac33=y(0)cdot 2$,
$endgroup$
– callculus
Mar 23 at 18:37
add a comment |
$begingroup$
If the algae doubles in three days then the equation is $y(t)=y(0)cdot 2^fract3$, where $y(0)$ is the initial population. For instance, if $t=3$ the equation becomes $y(3)=y(0)cdot 2^frac33=y(0)cdot 2$,
$endgroup$
– callculus
Mar 23 at 18:37
$begingroup$
If the algae doubles in three days then the equation is $y(t)=y(0)cdot 2^fract3$, where $y(0)$ is the initial population. For instance, if $t=3$ the equation becomes $y(3)=y(0)cdot 2^frac33=y(0)cdot 2$,
$endgroup$
– callculus
Mar 23 at 18:37
$begingroup$
If the algae doubles in three days then the equation is $y(t)=y(0)cdot 2^fract3$, where $y(0)$ is the initial population. For instance, if $t=3$ the equation becomes $y(3)=y(0)cdot 2^frac33=y(0)cdot 2$,
$endgroup$
– callculus
Mar 23 at 18:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You are right that after $9$ days you've had $3$ doublings. You have one day, or one third of a doubling period to worry about. Your linear answer is wrong. This is a problem in exponential growth. The growth factor is the same for all time intervals of the same length. You know that factor is $2$ for $3$ day intervals so it is $z = sqrt[3]2$, the cube root of $2$, for a one day interval. So to answer your question, find the cube root $z$ of $2$ accurately enough to know $800z$ to the nearest whole number.
If you know about logarithms you can work with a formula. The number $A$ of organisms at time $t$ days starting from $100$ at $t=0$ is
$$
A = 100 times 2^t/3.
$$
Substitute $t = 10$ and solve the equation for $A$, then round to the nearest integer.
You can tell that will give the same as the first method since
$$
2^10/3 = 2^3 2^1/3 = 8 sqrt[3]2.
$$
answered Mar 23 at 18:40
Ethan BolkerEthan Bolker
46k553120
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Max0815
Mar 23 at 19:58
add a comment |
$begingroup$
As John douma has mentioned in the comments you have to use the continuous model: $P'(t)=kcdot P(t)$, where $P(t)$ is the population of the algae cells.
First of all we solve this differential equation by the method of Separation of variables.
$P'(t)=kcdot P(t)$
$fracdPdt=kcdot P(t)$
Dividing the equation by $P(t)$ and multiplying the equation by $dt$.
$frac1P(t) dP=k dt$
Integrating both sides
$int frac1P(t) dP=int k dt$
$ln(P)=kcdot t+c$
To obtain $P$ on the LHS we take both sides as an exponent of the number $e$.
$e^ln(P)=e^kcdot t+c$
$P=e^kcdot tcdot e^c$
Replacing $e^c$ by $C$
$P=Ccdot e^kcdot tquad (*)$
To determine the constants C and k we use the following information:
- Given a beginning population of $100$ algae cells per milliliter of water ...
The equation is $P(0)=100Rightarrow P(0)=Ccdot e^kcdot 0=Ccdot 1=100Rightarrow C=100$
- ... so that its population doubles in 3 days
That means after 3 days the population is $200$: $P(3)=200$
$P(3)=100cdot e^kcdot 3=200$
$e^kcdot 3=frac12$
Taking $ln()$ on both sides.
$3cdot k=lnleft( frac12right)$
$k=fraclnleft( frac12right)3$
Now we can use (*) to obtain $P(t)$
$P(t)=100cdot e^ln(2)cdot fract3=100cdot left(e^ln(2)right)^ fract3=100cdot 2^fract3$
answered Mar 23 at 19:59
callculuscallculus
18.7k31428
$endgroup$
$begingroup$
I see. Thank you for this method!
$endgroup$
– Max0815
Mar 23 at 20:05
$begingroup$
@Max0815 It is nice to hear that the answer was not for the wastebasket. You´re welcome. My intention was to show the whole calculation which is needed to obtain the equation.
$endgroup$
– callculus
Mar 23 at 20:10
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are right that after $9$ days you've had $3$ doublings. You have one day, or one third of a doubling period to worry about. Your linear answer is wrong. This is a problem in exponential growth. The growth factor is the same for all time intervals of the same length. You know that factor is $2$ for $3$ day intervals so it is $z = sqrt[3]2$, the cube root of $2$, for a one day interval. So to answer your question, find the cube root $z$ of $2$ accurately enough to know $800z$ to the nearest whole number.
If you know about logarithms you can work with a formula. The number $A$ of organisms at time $t$ days starting from $100$ at $t=0$ is
$$
A = 100 times 2^t/3.
$$
Substitute $t = 10$ and solve the equation for $A$, then round to the nearest integer.
You can tell that will give the same as the first method since
$$
2^10/3 = 2^3 2^1/3 = 8 sqrt[3]2.
$$
answered Mar 23 at 18:40
Ethan BolkerEthan Bolker
46k553120
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Max0815
Mar 23 at 19:58
add a comment |
$begingroup$
You are right that after $9$ days you've had $3$ doublings. You have one day, or one third of a doubling period to worry about. Your linear answer is wrong. This is a problem in exponential growth. The growth factor is the same for all time intervals of the same length. You know that factor is $2$ for $3$ day intervals so it is $z = sqrt[3]2$, the cube root of $2$, for a one day interval. So to answer your question, find the cube root $z$ of $2$ accurately enough to know $800z$ to the nearest whole number.
If you know about logarithms you can work with a formula. The number $A$ of organisms at time $t$ days starting from $100$ at $t=0$ is
$$
A = 100 times 2^t/3.
$$
Substitute $t = 10$ and solve the equation for $A$, then round to the nearest integer.
You can tell that will give the same as the first method since
$$
2^10/3 = 2^3 2^1/3 = 8 sqrt[3]2.
$$
answered Mar 23 at 18:40
Ethan BolkerEthan Bolker
46k553120
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Max0815
Mar 23 at 19:58
add a comment |
$begingroup$
You are right that after $9$ days you've had $3$ doublings. You have one day, or one third of a doubling period to worry about. Your linear answer is wrong. This is a problem in exponential growth. The growth factor is the same for all time intervals of the same length. You know that factor is $2$ for $3$ day intervals so it is $z = sqrt[3]2$, the cube root of $2$, for a one day interval. So to answer your question, find the cube root $z$ of $2$ accurately enough to know $800z$ to the nearest whole number.
If you know about logarithms you can work with a formula. The number $A$ of organisms at time $t$ days starting from $100$ at $t=0$ is
$$
A = 100 times 2^t/3.
$$
Substitute $t = 10$ and solve the equation for $A$, then round to the nearest integer.
You can tell that will give the same as the first method since
$$
2^10/3 = 2^3 2^1/3 = 8 sqrt[3]2.
$$
answered Mar 23 at 18:40
Ethan BolkerEthan Bolker
46k553120
$endgroup$
You are right that after $9$ days you've had $3$ doublings. You have one day, or one third of a doubling period to worry about. Your linear answer is wrong. This is a problem in exponential growth. The growth factor is the same for all time intervals of the same length. You know that factor is $2$ for $3$ day intervals so it is $z = sqrt[3]2$, the cube root of $2$, for a one day interval. So to answer your question, find the cube root $z$ of $2$ accurately enough to know $800z$ to the nearest whole number.
If you know about logarithms you can work with a formula. The number $A$ of organisms at time $t$ days starting from $100$ at $t=0$ is
$$
A = 100 times 2^t/3.
$$
Substitute $t = 10$ and solve the equation for $A$, then round to the nearest integer.
You can tell that will give the same as the first method since
$$
2^10/3 = 2^3 2^1/3 = 8 sqrt[3]2.
$$
answered Mar 23 at 18:40
Ethan BolkerEthan Bolker
46k553120
answered Mar 23 at 18:40
Ethan BolkerEthan Bolker
46k553120
answered Mar 23 at 18:40
Ethan BolkerEthan Bolker
46k553120
answered Mar 23 at 18:40
Ethan BolkerEthan Bolker
46k553120
46k553120
$begingroup$
Thank you very much!
$endgroup$
– Max0815
Mar 23 at 19:58
add a comment |
$begingroup$
Thank you very much!
$endgroup$
– Max0815
Mar 23 at 19:58
$begingroup$
Thank you very much!
$endgroup$
– Max0815
Mar 23 at 19:58
$begingroup$
Thank you very much!
$endgroup$
– Max0815
Mar 23 at 19:58
add a comment |
$begingroup$
As John douma has mentioned in the comments you have to use the continuous model: $P'(t)=kcdot P(t)$, where $P(t)$ is the population of the algae cells.
First of all we solve this differential equation by the method of Separation of variables.
$P'(t)=kcdot P(t)$
$fracdPdt=kcdot P(t)$
Dividing the equation by $P(t)$ and multiplying the equation by $dt$.
$frac1P(t) dP=k dt$
Integrating both sides
$int frac1P(t) dP=int k dt$
$ln(P)=kcdot t+c$
To obtain $P$ on the LHS we take both sides as an exponent of the number $e$.
$e^ln(P)=e^kcdot t+c$
$P=e^kcdot tcdot e^c$
Replacing $e^c$ by $C$
$P=Ccdot e^kcdot tquad (*)$
To determine the constants C and k we use the following information:
- Given a beginning population of $100$ algae cells per milliliter of water ...
The equation is $P(0)=100Rightarrow P(0)=Ccdot e^kcdot 0=Ccdot 1=100Rightarrow C=100$
- ... so that its population doubles in 3 days
That means after 3 days the population is $200$: $P(3)=200$
$P(3)=100cdot e^kcdot 3=200$
$e^kcdot 3=frac12$
Taking $ln()$ on both sides.
$3cdot k=lnleft( frac12right)$
$k=fraclnleft( frac12right)3$
Now we can use (*) to obtain $P(t)$
$P(t)=100cdot e^ln(2)cdot fract3=100cdot left(e^ln(2)right)^ fract3=100cdot 2^fract3$
answered Mar 23 at 19:59
callculuscallculus
18.7k31428
$endgroup$
$begingroup$
I see. Thank you for this method!
$endgroup$
– Max0815
Mar 23 at 20:05
$begingroup$
@Max0815 It is nice to hear that the answer was not for the wastebasket. You´re welcome. My intention was to show the whole calculation which is needed to obtain the equation.
$endgroup$
– callculus
Mar 23 at 20:10
add a comment |
$begingroup$
As John douma has mentioned in the comments you have to use the continuous model: $P'(t)=kcdot P(t)$, where $P(t)$ is the population of the algae cells.
First of all we solve this differential equation by the method of Separation of variables.
$P'(t)=kcdot P(t)$
$fracdPdt=kcdot P(t)$
Dividing the equation by $P(t)$ and multiplying the equation by $dt$.
$frac1P(t) dP=k dt$
Integrating both sides
$int frac1P(t) dP=int k dt$
$ln(P)=kcdot t+c$
To obtain $P$ on the LHS we take both sides as an exponent of the number $e$.
$e^ln(P)=e^kcdot t+c$
$P=e^kcdot tcdot e^c$
Replacing $e^c$ by $C$
$P=Ccdot e^kcdot tquad (*)$
To determine the constants C and k we use the following information:
- Given a beginning population of $100$ algae cells per milliliter of water ...
The equation is $P(0)=100Rightarrow P(0)=Ccdot e^kcdot 0=Ccdot 1=100Rightarrow C=100$
- ... so that its population doubles in 3 days
That means after 3 days the population is $200$: $P(3)=200$
$P(3)=100cdot e^kcdot 3=200$
$e^kcdot 3=frac12$
Taking $ln()$ on both sides.
$3cdot k=lnleft( frac12right)$
$k=fraclnleft( frac12right)3$
Now we can use (*) to obtain $P(t)$
$P(t)=100cdot e^ln(2)cdot fract3=100cdot left(e^ln(2)right)^ fract3=100cdot 2^fract3$
answered Mar 23 at 19:59
callculuscallculus
18.7k31428
$endgroup$
$begingroup$
I see. Thank you for this method!
$endgroup$
– Max0815
Mar 23 at 20:05
$begingroup$
@Max0815 It is nice to hear that the answer was not for the wastebasket. You´re welcome. My intention was to show the whole calculation which is needed to obtain the equation.
$endgroup$
– callculus
Mar 23 at 20:10
add a comment |
$begingroup$
As John douma has mentioned in the comments you have to use the continuous model: $P'(t)=kcdot P(t)$, where $P(t)$ is the population of the algae cells.
First of all we solve this differential equation by the method of Separation of variables.
$P'(t)=kcdot P(t)$
$fracdPdt=kcdot P(t)$
Dividing the equation by $P(t)$ and multiplying the equation by $dt$.
$frac1P(t) dP=k dt$
Integrating both sides
$int frac1P(t) dP=int k dt$
$ln(P)=kcdot t+c$
To obtain $P$ on the LHS we take both sides as an exponent of the number $e$.
$e^ln(P)=e^kcdot t+c$
$P=e^kcdot tcdot e^c$
Replacing $e^c$ by $C$
$P=Ccdot e^kcdot tquad (*)$
To determine the constants C and k we use the following information:
- Given a beginning population of $100$ algae cells per milliliter of water ...
The equation is $P(0)=100Rightarrow P(0)=Ccdot e^kcdot 0=Ccdot 1=100Rightarrow C=100$
- ... so that its population doubles in 3 days
That means after 3 days the population is $200$: $P(3)=200$
$P(3)=100cdot e^kcdot 3=200$
$e^kcdot 3=frac12$
Taking $ln()$ on both sides.
$3cdot k=lnleft( frac12right)$
$k=fraclnleft( frac12right)3$
Now we can use (*) to obtain $P(t)$
$P(t)=100cdot e^ln(2)cdot fract3=100cdot left(e^ln(2)right)^ fract3=100cdot 2^fract3$
answered Mar 23 at 19:59
callculuscallculus
18.7k31428
$endgroup$
As John douma has mentioned in the comments you have to use the continuous model: $P'(t)=kcdot P(t)$, where $P(t)$ is the population of the algae cells.
First of all we solve this differential equation by the method of Separation of variables.
$P'(t)=kcdot P(t)$
$fracdPdt=kcdot P(t)$
Dividing the equation by $P(t)$ and multiplying the equation by $dt$.
$frac1P(t) dP=k dt$
Integrating both sides
$int frac1P(t) dP=int k dt$
$ln(P)=kcdot t+c$
To obtain $P$ on the LHS we take both sides as an exponent of the number $e$.
$e^ln(P)=e^kcdot t+c$
$P=e^kcdot tcdot e^c$
Replacing $e^c$ by $C$
$P=Ccdot e^kcdot tquad (*)$
To determine the constants C and k we use the following information:
- Given a beginning population of $100$ algae cells per milliliter of water ...
The equation is $P(0)=100Rightarrow P(0)=Ccdot e^kcdot 0=Ccdot 1=100Rightarrow C=100$
- ... so that its population doubles in 3 days
That means after 3 days the population is $200$: $P(3)=200$
$P(3)=100cdot e^kcdot 3=200$
$e^kcdot 3=frac12$
Taking $ln()$ on both sides.
$3cdot k=lnleft( frac12right)$
$k=fraclnleft( frac12right)3$
Now we can use (*) to obtain $P(t)$
$P(t)=100cdot e^ln(2)cdot fract3=100cdot left(e^ln(2)right)^ fract3=100cdot 2^fract3$
answered Mar 23 at 19:59
callculuscallculus
18.7k31428
edited Mar 23 at 20:04
answered Mar 23 at 19:59
callculuscallculus
18.7k31428
answered Mar 23 at 19:59
callculuscallculus
18.7k31428
answered Mar 23 at 19:59
callculuscallculus
18.7k31428
18.7k31428
$begingroup$
I see. Thank you for this method!
$endgroup$
– Max0815
Mar 23 at 20:05
$begingroup$
@Max0815 It is nice to hear that the answer was not for the wastebasket. You´re welcome. My intention was to show the whole calculation which is needed to obtain the equation.
$endgroup$
– callculus
Mar 23 at 20:10
add a comment |
$begingroup$
I see. Thank you for this method!
$endgroup$
– Max0815
Mar 23 at 20:05
$begingroup$
@Max0815 It is nice to hear that the answer was not for the wastebasket. You´re welcome. My intention was to show the whole calculation which is needed to obtain the equation.
$endgroup$
– callculus
Mar 23 at 20:10
$begingroup$
I see. Thank you for this method!
$endgroup$
– Max0815
Mar 23 at 20:05
$begingroup$
I see. Thank you for this method!
$endgroup$
– Max0815
Mar 23 at 20:05
$begingroup$
@Max0815 It is nice to hear that the answer was not for the wastebasket. You´re welcome. My intention was to show the whole calculation which is needed to obtain the equation.
$endgroup$
– callculus
Mar 23 at 20:10
$begingroup$
@Max0815 It is nice to hear that the answer was not for the wastebasket. You´re welcome. My intention was to show the whole calculation which is needed to obtain the equation.
$endgroup$
– callculus
Mar 23 at 20:10
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$begingroup$
If the algae doubles in three days then the equation is $y(t)=y(0)cdot 2^fract3$, where $y(0)$ is the initial population. For instance, if $t=3$ the equation becomes $y(3)=y(0)cdot 2^frac33=y(0)cdot 2$,
$endgroup$
– callculus
Mar 23 at 18:37