Math / Calculus

QuestionSection C(One Questions) (15 Marks) Let the rate of increase of bacteria is proportional to its size it is found that a culture of 300 bacteria has increased to 6 hundrod affer 90 minutes. (a) Express the population after $t$ hours as a function of $f$ . (b) How many bacteria are there in 5 hours? (c) How many hours and minutes will it take the culture to reach 20,000 bacteria. ( 3 marks

Studdy Solution

STEP 1

1. The rate of increase of bacteria is proportional to its size.
2. Initially, there are 300 bacteria.
3. After 90 minutes (or 1.5 hours), the population increases to 600 bacteria.
4. We need to express the population as a function of time, find the population after 5 hours, and determine the time to reach 20,000 bacteria.

STEP 2

1. Set up the differential equation for exponential growth.
2. Solve the differential equation to find the function of population over time.
3. Use the function to find the population after 5 hours.
4. Use the function to find the time required to reach 20,000 bacteria.

STEP 3

The rate of increase of bacteria is proportional to its size, which can be expressed as the differential equation: $\frac{dP}{dt} = kP$ where $P$ is the population size and $k$ is the proportionality constant.

STEP 4

Solve the differential equation: $\frac{dP}{dt} = kP$ This is a separable differential equation. Separate variables and integrate: $\int \frac{1}{P} \, dP = \int k \, dt$ $\ln |P| = kt + C$ Exponentiate both sides to solve for $P$ : $P(t) = Ce^{kt}$ where $C$ is the constant of integration.

STEP 5

Use the initial condition $P(0) = 300$ to find $C$ : $300 = Ce^{k \cdot 0} \Rightarrow C = 300$ Thus, the equation becomes: $P(t) = 300e^{kt}$

STEP 6

Use the condition $P(1.5) = 600$ to find $k$ : $600 = 300e^{1.5k}$ $2 = e^{1.5k}$ Take the natural logarithm of both sides: $\ln 2 = 1.5k$ $k = \frac{\ln 2}{1.5}$

STEP 7

Substitute $k$ back into the population function: $P(t) = 300e^{\left(\frac{\ln 2}{1.5}\right)t}$ To find the population after 5 hours: $P(5) = 300e^{\left(\frac{\ln 2}{1.5}\right)5}$

STEP 8

Calculate $P(5)$ : $P(5) = 300 \cdot 2^{\frac{5}{1.5}}$ $P(5) \approx 300 \cdot 8$ $P(5) \approx 2400$

STEP 9

To find the time $t$ when $P(t) = 20000$ : $20000 = 300e^{\left(\frac{\ln 2}{1.5}\right)t}$ $\frac{20000}{300} = e^{\left(\frac{\ln 2}{1.5}\right)t}$ $\frac{200}{3} = e^{\left(\frac{\ln 2}{1.5}\right)t}$ Take the natural logarithm of both sides: $\ln \left(\frac{200}{3}\right) = \left(\frac{\ln 2}{1.5}\right)t$ $t = \frac{1.5 \ln \left(\frac{200}{3}\right)}{\ln 2}$

STEP 10

Calculate $t$ : $t \approx \frac{1.5 \times 4.3219}{0.6931}$ $t \approx 9.35 \text{ hours}$ Convert 0.35 hours to minutes: $0.35 \times 60 \approx 21 \text{ minutes}$
The population after 5 hours is approximately 2400 bacteria, and it will take approximately 9 hours and 21 minutes for the culture to reach 20,000 bacteria.

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Studdy Solution

STEP 1

STEP 2

1. Set up the differential equation for exponential growth.
2. Solve the differential equation to find the function of population over time.
3. Use the function to find the population after 5 hours.
4. Use the function to find the time required to reach 20,000 bacteria.

STEP 3

The rate of increase of bacteria is proportional to its size, which can be expressed as the differential equation: $\frac{dP}{dt} = kP$ where $P$ is the population size and $k$ is the proportionality constant.

STEP 4

Solve the differential equation: $\frac{dP}{dt} = kP$ This is a separable differential equation. Separate variables and integrate: $\int \frac{1}{P} \, dP = \int k \, dt$ $\ln |P| = kt + C$ Exponentiate both sides to solve for $P$ : $P(t) = Ce^{kt}$ where $C$ is the constant of integration.

STEP 5

Use the initial condition $P(0) = 300$ to find $C$ : $300 = Ce^{k \cdot 0} \Rightarrow C = 300$ Thus, the equation becomes: $P(t) = 300e^{kt}$

STEP 6

Use the condition $P(1.5) = 600$ to find $k$ : $600 = 300e^{1.5k}$ $2 = e^{1.5k}$ Take the natural logarithm of both sides: $\ln 2 = 1.5k$ $k = \frac{\ln 2}{1.5}$

STEP 7

Substitute $k$ back into the population function: $P(t) = 300e^{\left(\frac{\ln 2}{1.5}\right)t}$ To find the population after 5 hours: $P(5) = 300e^{\left(\frac{\ln 2}{1.5}\right)5}$

STEP 8

Calculate $P(5)$ : $P(5) = 300 \cdot 2^{\frac{5}{1.5}}$ $P(5) \approx 300 \cdot 8$ $P(5) \approx 2400$

STEP 9

STEP 10

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Studdy Solution

STEP 1

STEP 2

1. Set up the differential equation for exponential growth.2. Solve the differential equation to find the function of population over time.3. Use the function to find the population after 5 hours.4. Use the function to find the time required to reach 20,000 bacteria.

STEP 3

The rate of increase of bacteria is proportional to its size, which can be expressed as the differential equation: dPdt=kP\frac{dP}{dt} = kPdtdP​=kP where P P P is the population size and k k k is the proportionality constant.

STEP 4

STEP 5

Use the initial condition P(0)=300 P(0) = 300 P(0)=300 to find C C C: 300=Cek⋅0⇒C=300300 = Ce^{k \cdot 0} \Rightarrow C = 300300=Cek⋅0⇒C=300 Thus, the equation becomes: P(t)=300ektP(t) = 300e^{kt}P(t)=300ekt

STEP 6

Use the condition P(1.5)=600 P(1.5) = 600 P(1.5)=600 to find k k k: 600=300e1.5k600 = 300e^{1.5k}600=300e1.5k 2=e1.5k2 = e^{1.5k}2=e1.5k Take the natural logarithm of both sides: ln⁡2=1.5k\ln 2 = 1.5kln2=1.5k k=ln⁡21.5k = \frac{\ln 2}{1.5}k=1.5ln2​

STEP 7

Substitute k k k back into the population function: P(t)=300e(ln⁡21.5)tP(t) = 300e^{\left(\frac{\ln 2}{1.5}\right)t}P(t)=300e(1.5ln2​)t To find the population after 5 hours: P(5)=300e(ln⁡21.5)5P(5) = 300e^{\left(\frac{\ln 2}{1.5}\right)5}P(5)=300e(1.5ln2​)5

STEP 8

Calculate P(5) P(5) P(5): P(5)=300⋅251.5P(5) = 300 \cdot 2^{\frac{5}{1.5}}P(5)=300⋅21.55​ P(5)≈300⋅8P(5) \approx 300 \cdot 8P(5)≈300⋅8 P(5)≈2400P(5) \approx 2400P(5)≈2400

STEP 9

STEP 10

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1. Set up the differential equation for exponential growth.
2. Solve the differential equation to find the function of population over time.
3. Use the function to find the population after 5 hours.
4. Use the function to find the time required to reach 20,000 bacteria.

The rate of increase of bacteria is proportional to its size, which can be expressed as the differential equation: $\frac{dP}{dt} = kP$ where $P$ is the population size and $k$ is the proportionality constant.

Use the initial condition $P(0) = 300$ to find $C$ : $300 = Ce^{k \cdot 0} \Rightarrow C = 300$ Thus, the equation becomes: $P(t) = 300e^{kt}$

Use the condition $P(1.5) = 600$ to find $k$ : $600 = 300e^{1.5k}$ $2 = e^{1.5k}$ Take the natural logarithm of both sides: $\ln 2 = 1.5k$ $k = \frac{\ln 2}{1.5}$

Substitute $k$ back into the population function: $P(t) = 300e^{\left(\frac{\ln 2}{1.5}\right)t}$ To find the population after 5 hours: $P(5) = 300e^{\left(\frac{\ln 2}{1.5}\right)5}$

Calculate $P(5)$ : $P(5) = 300 \cdot 2^{\frac{5}{1.5}}$ $P(5) \approx 300 \cdot 8$ $P(5) \approx 2400$