Math / Calculus

QuestionThe flow diagram below shows the volume of fluid in a chemostat (a growth chamber for bacteria).
Poll 1: The ODE is: A. $\frac{d V(t)}{d t}=f-d$ B. $\frac{d V(t)}{d t}=f V(t)$

Studdy Solution

STEP 1

What is this asking? We're looking at how the volume of liquid in a bacteria tank changes over time, given a constant inflow and outflow. Watch out! Don't mix up the rates of flow with the volume itself!

STEP 2

1. Analyze the Flow
2. Set up the ODE
3. Solve the ODE

STEP 3

Alright, so we've got this bacteria tank, right?
It's called a chemostat, which sounds super fancy!
Liquid is flowing in at a rate of $f$ and flowing out at a rate of $d$ .
So, the change in volume is simply the difference between these two rates.

STEP 4

Think of it like a bathtub!
If you turn on the faucet, the water level rises.
If you pull the plug, the water level drops.
The rate of change of the water level depends on both the inflow and outflow.

STEP 5

Now, let's translate this into math!
The rate of change of the volume, written as $\frac{d V(t)}{d t}$ , is equal to the inflow rate $f$ minus the outflow rate $d$ .
So, our Ordinary Differential Equation (ODE) is: $\frac{d V(t)}{d t} = f - d$

STEP 6

Notice that the correct answer to Poll 1 is A!
The rate of change of the volume depends only on the difference between $f$ and $d$ , not on the volume $V(t)$ itself.

STEP 7

This ODE is pretty straightforward to solve!
Since $f$ and $d$ are constants, $f - d$ is also a constant.
Let's call this constant $c$ , so $c = f - d$ .
Now our ODE looks like this: $\frac{d V(t)}{d t} = c$

STEP 8

To solve this, we just integrate both sides with respect to $t$ : $\int \frac{d V(t)}{d t} \, dt = \int c \, dt$ $V(t) = ct + k$ where $k$ is the constant of integration.

STEP 9

We're given that the initial volume is $V(0) = 250 \, \text{ml}$ .
Let's plug this in to find $k$ : $V(0) = c \cdot 0 + k = 250$ So, $k = \textbf{250}$ .

STEP 10

Now we can write the complete solution: $V(t) = ct + 250$ Remembering that $c = f - d$ , we get our final equation: $V(t) = (f - d)t + 250$

STEP 11

The volume of fluid in the chemostat at any time $t$ is given by $V(t) = (f - d)t + 250$ .

Was this helpful?

QuestionThe flow diagram below shows the volume of fluid in a chemostat (a growth chamber for bacteria).
Poll 1: The ODE is: A. $\frac{d V(t)}{d t}=f-d$ B. $\frac{d V(t)}{d t}=f V(t)$

Studdy Solution

STEP 1

What is this asking? We're looking at how the volume of liquid in a bacteria tank changes over time, given a constant inflow and outflow. Watch out! Don't mix up the rates of flow with the volume itself!

STEP 2

1. Analyze the Flow
2. Set up the ODE
3. Solve the ODE

STEP 3

Alright, so we've got this bacteria tank, right?
It's called a chemostat, which sounds super fancy!
Liquid is flowing in at a rate of $f$ and flowing out at a rate of $d$ .
So, the change in volume is simply the difference between these two rates.

STEP 4

Think of it like a bathtub!
If you turn on the faucet, the water level rises.
If you pull the plug, the water level drops.
The rate of change of the water level depends on both the inflow and outflow.

STEP 5

Now, let's translate this into math!
The rate of change of the volume, written as $\frac{d V(t)}{d t}$ , is equal to the inflow rate $f$ minus the outflow rate $d$ .
So, our Ordinary Differential Equation (ODE) is: $\frac{d V(t)}{d t} = f - d$

STEP 6

Notice that the correct answer to Poll 1 is A!
The rate of change of the volume depends only on the difference between $f$ and $d$ , not on the volume $V(t)$ itself.

STEP 7

This ODE is pretty straightforward to solve!
Since $f$ and $d$ are constants, $f - d$ is also a constant.
Let's call this constant $c$ , so $c = f - d$ .
Now our ODE looks like this: $\frac{d V(t)}{d t} = c$

STEP 8

To solve this, we just integrate both sides with respect to $t$ : $\int \frac{d V(t)}{d t} \, dt = \int c \, dt$ $V(t) = ct + k$ where $k$ is the constant of integration.

STEP 9

We're given that the initial volume is $V(0) = 250 \, \text{ml}$ .
Let's plug this in to find $k$ : $V(0) = c \cdot 0 + k = 250$ So, $k = \textbf{250}$ .

STEP 10

Now we can write the complete solution: $V(t) = ct + 250$ Remembering that $c = f - d$ , we get our final equation: $V(t) = (f - d)t + 250$

STEP 11

The volume of fluid in the chemostat at any time $t$ is given by $V(t) = (f - d)t + 250$ .

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionThe flow diagram below shows the volume of fluid in a chemostat (a growth chamber for bacteria).Poll 1: The ODE is: A. dV(t)dt=f−d\frac{d V(t)}{d t}=f-ddtdV(t)​=f−d B. dV(t)dt=fV(t)\frac{d V(t)}{d t}=f V(t)dtdV(t)​=fV(t)

Studdy Solution

STEP 1

What is this asking? We're looking at how the volume of liquid in a bacteria tank changes over time, given a constant inflow and outflow. Watch out! Don't mix up the *rates* of flow with the *volume* itself!

STEP 2

1. Analyze the Flow2. Set up the ODE3. Solve the ODE

STEP 3

Alright, so we've got this bacteria tank, right?It's called a chemostat, which sounds super fancy!Liquid is flowing *in* at a rate of fff and flowing *out* at a rate of ddd.So, the change in volume is simply the difference between these two rates.

STEP 4

Think of it like a bathtub!If you turn on the faucet, the water level rises.If you pull the plug, the water level drops.The rate of change of the water level depends on *both* the inflow and outflow.

STEP 5

Now, let's translate this into math!The rate of change of the volume, written as dV(t)dt\frac{d V(t)}{d t}dtdV(t)​, is equal to the inflow rate fff *minus* the outflow rate ddd.So, our Ordinary Differential Equation (ODE) is: dV(t)dt=f−d \frac{d V(t)}{d t} = f - d dtdV(t)​=f−d

STEP 6

Notice that the correct answer to Poll 1 is A!The rate of change of the volume depends *only* on the difference between fff and ddd, not on the volume V(t)V(t)V(t) itself.

STEP 7

This ODE is pretty straightforward to solve!Since fff and ddd are constants, f−df - df−d is also a constant.Let's call this constant ccc, so c=f−dc = f - dc=f−d.Now our ODE looks like this: dV(t)dt=c \frac{d V(t)}{d t} = c dtdV(t)​=c

STEP 8

To solve this, we just integrate both sides with respect to ttt: ∫dV(t)dt dt=∫c dt \int \frac{d V(t)}{d t} \, dt = \int c \, dt ∫dtdV(t)​dt=∫cdt V(t)=ct+k V(t) = ct + k V(t)=ct+kwhere kkk is the constant of integration.

STEP 9

We're given that the **initial volume** is V(0)=250 mlV(0) = 250 \, \text{ml}V(0)=250ml.Let's plug this in to find kkk: V(0)=c⋅0+k=250 V(0) = c \cdot 0 + k = 250 V(0)=c⋅0+k=250 So, k=250k = \textbf{250}k=250.

STEP 10

Now we can write the complete solution: V(t)=ct+250 V(t) = ct + 250 V(t)=ct+250 Remembering that c=f−dc = f - dc=f−d, we get our **final equation**: V(t)=(f−d)t+250 V(t) = (f - d)t + 250 V(t)=(f−d)t+250

STEP 11

The volume of fluid in the chemostat at any time ttt is given by V(t)=(f−d)t+250V(t) = (f - d)t + 250V(t)=(f−d)t+250.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionThe flow diagram below shows the volume of fluid in a chemostat (a growth chamber for bacteria).
Poll 1: The ODE is: A. $\frac{d V(t)}{d t}=f-d$ B. $\frac{d V(t)}{d t}=f V(t)$

What is this asking? We're looking at how the volume of liquid in a bacteria tank changes over time, given a constant inflow and outflow. Watch out! Don't mix up the rates of flow with the volume itself!

1. Analyze the Flow
2. Set up the ODE
3. Solve the ODE

Alright, so we've got this bacteria tank, right?
It's called a chemostat, which sounds super fancy!
Liquid is flowing in at a rate of $f$ and flowing out at a rate of $d$ .
So, the change in volume is simply the difference between these two rates.

Think of it like a bathtub!
If you turn on the faucet, the water level rises.
If you pull the plug, the water level drops.
The rate of change of the water level depends on both the inflow and outflow.

Now, let's translate this into math!
The rate of change of the volume, written as $\frac{d V(t)}{d t}$ , is equal to the inflow rate $f$ minus the outflow rate $d$ .
So, our Ordinary Differential Equation (ODE) is: $\frac{d V(t)}{d t} = f - d$

Notice that the correct answer to Poll 1 is A!
The rate of change of the volume depends only on the difference between $f$ and $d$ , not on the volume $V(t)$ itself.

This ODE is pretty straightforward to solve!
Since $f$ and $d$ are constants, $f - d$ is also a constant.
Let's call this constant $c$ , so $c = f - d$ .
Now our ODE looks like this: $\frac{d V(t)}{d t} = c$

To solve this, we just integrate both sides with respect to $t$ : $\int \frac{d V(t)}{d t} \, dt = \int c \, dt$ $V(t) = ct + k$ where $k$ is the constant of integration.

We're given that the initial volume is $V(0) = 250 \, \text{ml}$ .
Let's plug this in to find $k$ : $V(0) = c \cdot 0 + k = 250$ So, $k = \textbf{250}$ .

Now we can write the complete solution: $V(t) = ct + 250$ Remembering that $c = f - d$ , we get our final equation: $V(t) = (f - d)t + 250$

The volume of fluid in the chemostat at any time $t$ is given by $V(t) = (f - d)t + 250$ .