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The number of bacteria in the early stages of a growing colony of bacteria can be approximated using the function: $$N(t) = 450e^{0.065t}$$ where $t$ is the time, measured in hours, and $N(t)$ is the number of bacteria in the colony at time $t$ - Leaving Cert Mathematics - Question 9 - 2020

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Question 9

The-number-of-bacteria-in-the-early-stages-of-a-growing-colony-of-bacteria-can-be-approximated-using-the-function:--$$N(t)-=-450e^{0.065t}$$--where-$t$-is-the-time,-measured-in-hours,-and-$N(t)$-is-the-number-of-bacteria-in-the-colony-at-time-$t$-Leaving Cert Mathematics-Question 9-2020.png

The number of bacteria in the early stages of a growing colony of bacteria can be approximated using the function: $$N(t) = 450e^{0.065t}$$ where $t$ is the time, ... show full transcript

Worked Solution & Example Answer:The number of bacteria in the early stages of a growing colony of bacteria can be approximated using the function: $$N(t) = 450e^{0.065t}$$ where $t$ is the time, measured in hours, and $N(t)$ is the number of bacteria in the colony at time $t$ - Leaving Cert Mathematics - Question 9 - 2020

Step 1

Find the number of bacteria in the colony after 4-5 hours.

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Answer

To find the number of bacteria after 4-5 hours, substitute t=4.5t = 4.5 into the function:

N(4.5)=450e0.065×4.5N(4.5) = 450e^{0.065 \times 4.5}

Calculating this gives:

N(4.5)=450e0.2925603N(4.5) = 450e^{0.2925} \approx 603

Thus, the number of bacteria after approximately 4.5 hours is 603.

Step 2

Find the time, in hours, that it takes the colony to grow to 790 bacteria.

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Answer

To find the time tt when N(t)=790N(t) = 790, we set up the equation:

790=450e0.065t790 = 450e^{0.065t}

Now, divide both sides by 450:

790450=e0.065t\frac{790}{450} = e^{0.065t}

Taking the natural logarithm:

ln(790450)=0.065t\ln\left(\frac{790}{450}\right) = 0.065t

Solving for tt gives:

t=ln(790450)0.0658.7t = \frac{\ln\left(\frac{790}{450}\right)}{0.065} \approx 8.7

Thus, it takes approximately 8.7 hours.

Step 3

Using the function N(t) = 450e^{0.065t}, find the average number of bacteria in the colony during the period from t = 3 to t = 12.

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Answer

To find the average number of bacteria from t=3t = 3 to t=12t = 12, we use the formula:

Average=1baabN(t)dt\text{Average} = \frac{1}{b-a} \int_a^b N(t) dt

where a=3a = 3 and b=12b = 12:

=1123312450e0.065tdt= \frac{1}{12-3} \int_3^{12} 450e^{0.065t} dt

Calculating the integral:

=19[4500.065e0.065t]312= \frac{1}{9} \left[ \frac{450}{0.065}e^{0.065t} \right]_{3}^{12}

Evaluating gives:

19(4500.065(e0.78e0.195))743\frac{1}{9} \left( \frac{450}{0.065}(e^{0.78} - e^{0.195}) \right) \approx 743

Thus, the average number of bacteria is approximately 743.

Step 4

Find the rate at which N(t) = 450e^{0.065t} is changing when t = 12.

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Answer

To find the rate of change, we first need the derivative:

N(t)=450imes0.065e0.065tN'(t) = 450 imes 0.065 e^{0.065t}

Evaluating at t=12t = 12:

N(12)=450imes0.065e0.78638.8N'(12) = 450 imes 0.065 e^{0.78} \approx 638.8

Thus, the rate at which the colony is growing at t=12t = 12 is approximately 638.8 bacteria per hour.

Step 5

After t hours, the rate of increase of N(t) is greater than 90 bacteria per hour.

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Answer

Setting up the inequality:

N(t)>90N'(t) > 90

Finding N(t)N'(t):

450imes0.065e0.065t>90450 imes 0.065 e^{0.065t} > 90

This simplifies to:

e0.065t>9029.25e^{0.065t} > \frac{90}{29.25}

Taking the natural logarithm:

0.065t>ln(9029.25)0.065t > \ln\left(\frac{90}{29.25}\right)

Solving for tt leads to:

t>ln(90)ln(29.25)0.06518t > \frac{\ln(90) - \ln(29.25)}{0.065} \approx 18

Thus, the least value of kk is 18.

Step 6

Find the time, at which the number of bacteria in both colonies will be equal.

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Answer

Setting the two equations equal:

450e0.065t=220e0.17t450e^{0.065t} = 220e^{0.17t}

Rearranging, we have:

450220=e0.17t0.065t\frac{450}{220} = e^{0.17t - 0.065t}

Taking logarithms:

ln(450220)=(0.170.065)t\ln\left(\frac{450}{220}\right) = (0.17 - 0.065)t

Solving for tt gives:

t=ln(450220)0.1057t = \frac{\ln\left(\frac{450}{220}\right)}{0.105} \approx 7

Thus, the time at which the number of bacteria in both colonies will be equal is approximately 7 hours.

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