Solved on Dec 15, 2023

Insect population of $\mathrm{P}(t) = 600e^{0.02t}$ at time $t$ days. Find: (a) population at $t=0$ , (b) growth rate, (c) graph, (d) population after 10 days, (e) when population reaches 900, (f) when population doubles.

STEP 1

Assumptions
1. The insect population size $\mathrm{P}$ at time $\mathrm{t}$ (in days) is given by the function $\mathrm{P}(\mathrm{t})=600 \mathrm{e}^{0.02 \mathrm{t}}$ .
2. The function is continuous and differentiable for all $\mathrm{t} \geq 0$ .
3. The base of the natural logarithm $\mathrm{e}$ is approximately 2.71828.
4. The growth rate refers to the derivative of the population function with respect to time.

STEP 2

To determine the number of insects at $t=0$ days, we substitute $t=0$ into the population function $\mathrm{P}(\mathrm{t})$ .
$\mathrm{P}(0) = 600 \mathrm{e}^{0.02 \cdot 0}$

STEP 3

Since $\mathrm{e}^{0} = 1$ , we can simplify the expression.
$\mathrm{P}(0) = 600 \cdot 1$

STEP 4

Calculate the population at $t=0$ days.
$\mathrm{P}(0) = 600$
The number of insects at $t=0$ days is 600.

STEP 5

To find the growth rate of the insect population, we need to differentiate the population function $\mathrm{P}(\mathrm{t})$ with respect to $\mathrm{t}$ .
$\frac{d\mathrm{P}}{d\mathrm{t}} = \frac{d}{d\mathrm{t}} (600 \mathrm{e}^{0.02 \mathrm{t}})$

STEP 6

Apply the chain rule to differentiate the exponential function.
$\frac{d\mathrm{P}}{d\mathrm{t}} = 600 \cdot 0.02 \mathrm{e}^{0.02 \mathrm{t}}$

STEP 7

The growth rate of the insect population is the derivative we just found.
$Growth\, rate = 600 \cdot 0.02 \mathrm{e}^{0.02 \mathrm{t}}$

STEP 8

To graph the function using a graphing utility, input the function $\mathrm{P}(\mathrm{t})=600 \mathrm{e}^{0.02 \mathrm{t}}$ into the utility and plot it for a range of $\mathrm{t}$ values, typically from $t=0$ to a reasonable upper limit to observe the growth trend.

STEP 9

To determine the population after 10 days, substitute $t=10$ into the population function $\mathrm{P}(\mathrm{t})$ .
$\mathrm{P}(10) = 600 \mathrm{e}^{0.02 \cdot 10}$

STEP 10

Calculate the value of the exponential term.
$\mathrm{e}^{0.02 \cdot 10} = \mathrm{e}^{0.2}$

STEP 11

Use a calculator to find the approximate value of $\mathrm{e}^{0.2}$ .
$\mathrm{e}^{0.2} \approx 1.22140$

STEP 12

Calculate the population at $t=10$ days.
$\mathrm{P}(10) = 600 \cdot 1.22140$

STEP 13

Multiply to find the population after 10 days.
$\mathrm{P}(10) \approx 600 \cdot 1.22140 \approx 732.84$
The population after 10 days is approximately 733 insects (rounded to the nearest whole number).

STEP 14

To find when the insect population will reach 900, we set $\mathrm{P}(\mathrm{t})$ equal to 900 and solve for $\mathrm{t}$ .
$900 = 600 \mathrm{e}^{0.02 \mathrm{t}}$

STEP 15

Divide both sides of the equation by 600.
$\frac{900}{600} = \mathrm{e}^{0.02 \mathrm{t}}$

STEP 16

Simplify the fraction.
$1.5 = \mathrm{e}^{0.02 \mathrm{t}}$

STEP 17

Take the natural logarithm of both sides to solve for $\mathrm{t}$ .
$\ln(1.5) = \ln(\mathrm{e}^{0.02 \mathrm{t}})$

STEP 18

Apply the property of logarithms $\ln(\mathrm{e}^x) = x$ .
$\ln(1.5) = 0.02 \mathrm{t}$

STEP 19

Divide both sides by 0.02 to isolate $\mathrm{t}$ .
$\mathrm{t} = \frac{\ln(1.5)}{0.02}$

STEP 20

Use a calculator to find the approximate value of $\mathrm{t}$ .
$\mathrm{t} \approx \frac{\ln(1.5)}{0.02} \approx 20.79$
The insect population will reach 900 after approximately 20.79 days.

STEP 21

To determine when the insect population will double, we set $\mathrm{P}(\mathrm{t})$ equal to twice the initial population of 600.
$2 \cdot 600 = 600 \mathrm{e}^{0.02 \mathrm{t}}$

STEP 22

Divide both sides of the equation by 600.
$2 = \mathrm{e}^{0.02 \mathrm{t}}$

STEP 23

Take the natural logarithm of both sides to solve for $\mathrm{t}$ .
$\ln(2) = \ln(\mathrm{e}^{0.02 \mathrm{t}})$

STEP 24

Apply the property of logarithms $\ln(\mathrm{e}^x) = x$ .
$\ln(2) = 0.02 \mathrm{t}$

STEP 25

Divide both sides by 0.02 to isolate $\mathrm{t}$ .
$\mathrm{t} = \frac{\ln(2)}{0.02}$

STEP 26

Use a calculator to find the approximate value of $\mathrm{t}$ .
$\mathrm{t} \approx \frac{\ln(2)}{0.02} \approx 34.66$
The insect population will double after approximately 34.66 days.

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Insect population of P(t)=600e0.02t\mathrm{P}(t) = 600e^{0.02t}P(t)=600e0.02t at time ttt days. Find: (a) population at t=0t=0t=0, (b) growth rate, (c) graph, (d) population after 10 days, (e) when population reaches 900, (f) when population doubles.

STEP 1

STEP 2

To determine the number of insects at t=0t=0t=0 days, we substitute t=0t=0t=0 into the population function P(t)\mathrm{P}(\mathrm{t})P(t).P(0)=600e0.02⋅0\mathrm{P}(0) = 600 \mathrm{e}^{0.02 \cdot 0}P(0)=600e0.02⋅0

STEP 3

Since e0=1\mathrm{e}^{0} = 1e0=1, we can simplify the expression.P(0)=600⋅1\mathrm{P}(0) = 600 \cdot 1P(0)=600⋅1

STEP 4

Calculate the population at t=0t=0t=0 days.P(0)=600\mathrm{P}(0) = 600P(0)=600The number of insects at t=0t=0t=0 days is 600.

STEP 5

To find the growth rate of the insect population, we need to differentiate the population function P(t)\mathrm{P}(\mathrm{t})P(t) with respect to t\mathrm{t}t.dPdt=ddt(600e0.02t)\frac{d\mathrm{P}}{d\mathrm{t}} = \frac{d}{d\mathrm{t}} (600 \mathrm{e}^{0.02 \mathrm{t}})dtdP​=dtd​(600e0.02t)

STEP 6

Apply the chain rule to differentiate the exponential function.dPdt=600⋅0.02e0.02t\frac{d\mathrm{P}}{d\mathrm{t}} = 600 \cdot 0.02 \mathrm{e}^{0.02 \mathrm{t}}dtdP​=600⋅0.02e0.02t

STEP 7

The growth rate of the insect population is the derivative we just found.Growth rate=600⋅0.02e0.02tGrowth\, rate = 600 \cdot 0.02 \mathrm{e}^{0.02 \mathrm{t}}Growthrate=600⋅0.02e0.02t

STEP 8

To graph the function using a graphing utility, input the function P(t)=600e0.02t\mathrm{P}(\mathrm{t})=600 \mathrm{e}^{0.02 \mathrm{t}}P(t)=600e0.02t into the utility and plot it for a range of t\mathrm{t}t values, typically from t=0t=0t=0 to a reasonable upper limit to observe the growth trend.

STEP 9

To determine the population after 10 days, substitute t=10t=10t=10 into the population function P(t)\mathrm{P}(\mathrm{t})P(t).P(10)=600e0.02⋅10\mathrm{P}(10) = 600 \mathrm{e}^{0.02 \cdot 10}P(10)=600e0.02⋅10

STEP 10

Calculate the value of the exponential term.e0.02⋅10=e0.2\mathrm{e}^{0.02 \cdot 10} = \mathrm{e}^{0.2}e0.02⋅10=e0.2

STEP 11

Use a calculator to find the approximate value of e0.2\mathrm{e}^{0.2}e0.2.e0.2≈1.22140\mathrm{e}^{0.2} \approx 1.22140e0.2≈1.22140

STEP 12

Calculate the population at t=10t=10t=10 days.P(10)=600⋅1.22140\mathrm{P}(10) = 600 \cdot 1.22140P(10)=600⋅1.22140

STEP 13

Multiply to find the population after 10 days.P(10)≈600⋅1.22140≈732.84\mathrm{P}(10) \approx 600 \cdot 1.22140 \approx 732.84P(10)≈600⋅1.22140≈732.84The population after 10 days is approximately 733 insects (rounded to the nearest whole number).

STEP 14

To find when the insect population will reach 900, we set P(t)\mathrm{P}(\mathrm{t})P(t) equal to 900 and solve for t\mathrm{t}t.900=600e0.02t900 = 600 \mathrm{e}^{0.02 \mathrm{t}}900=600e0.02t

STEP 15

Divide both sides of the equation by 600.900600=e0.02t\frac{900}{600} = \mathrm{e}^{0.02 \mathrm{t}}600900​=e0.02t

STEP 16

Simplify the fraction.1.5=e0.02t1.5 = \mathrm{e}^{0.02 \mathrm{t}}1.5=e0.02t

STEP 17

Take the natural logarithm of both sides to solve for t\mathrm{t}t.ln⁡(1.5)=ln⁡(e0.02t)\ln(1.5) = \ln(\mathrm{e}^{0.02 \mathrm{t}})ln(1.5)=ln(e0.02t)

STEP 18

Apply the property of logarithms ln⁡(ex)=x\ln(\mathrm{e}^x) = xln(ex)=x.ln⁡(1.5)=0.02t\ln(1.5) = 0.02 \mathrm{t}ln(1.5)=0.02t

STEP 19

Divide both sides by 0.02 to isolate t\mathrm{t}t.t=ln⁡(1.5)0.02\mathrm{t} = \frac{\ln(1.5)}{0.02}t=0.02ln(1.5)​

STEP 20

Use a calculator to find the approximate value of t\mathrm{t}t.t≈ln⁡(1.5)0.02≈20.79\mathrm{t} \approx \frac{\ln(1.5)}{0.02} \approx 20.79t≈0.02ln(1.5)​≈20.79The insect population will reach 900 after approximately 20.79 days.

STEP 21

To determine when the insect population will double, we set P(t)\mathrm{P}(\mathrm{t})P(t) equal to twice the initial population of 600.2⋅600=600e0.02t2 \cdot 600 = 600 \mathrm{e}^{0.02 \mathrm{t}}2⋅600=600e0.02t

STEP 22

Divide both sides of the equation by 600.2=e0.02t2 = \mathrm{e}^{0.02 \mathrm{t}}2=e0.02t

STEP 23

Take the natural logarithm of both sides to solve for t\mathrm{t}t.ln⁡(2)=ln⁡(e0.02t)\ln(2) = \ln(\mathrm{e}^{0.02 \mathrm{t}})ln(2)=ln(e0.02t)

STEP 24

Apply the property of logarithms ln⁡(ex)=x\ln(\mathrm{e}^x) = xln(ex)=x.ln⁡(2)=0.02t\ln(2) = 0.02 \mathrm{t}ln(2)=0.02t

STEP 25

Divide both sides by 0.02 to isolate t\mathrm{t}t.t=ln⁡(2)0.02\mathrm{t} = \frac{\ln(2)}{0.02}t=0.02ln(2)​

STEP 26

Use a calculator to find the approximate value of t\mathrm{t}t.t≈ln⁡(2)0.02≈34.66\mathrm{t} \approx \frac{\ln(2)}{0.02} \approx 34.66t≈0.02ln(2)​≈34.66The insect population will double after approximately 34.66 days.

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Insect population of $\mathrm{P}(t) = 600e^{0.02t}$ at time $t$ days. Find: (a) population at $t=0$ , (b) growth rate, (c) graph, (d) population after 10 days, (e) when population reaches 900, (f) when population doubles.

To determine the number of insects at $t=0$ days, we substitute $t=0$ into the population function $\mathrm{P}(\mathrm{t})$ .
$\mathrm{P}(0) = 600 \mathrm{e}^{0.02 \cdot 0}$

Since $\mathrm{e}^{0} = 1$ , we can simplify the expression.
$\mathrm{P}(0) = 600 \cdot 1$

Calculate the population at $t=0$ days.
$\mathrm{P}(0) = 600$
The number of insects at $t=0$ days is 600.

To find the growth rate of the insect population, we need to differentiate the population function $\mathrm{P}(\mathrm{t})$ with respect to $\mathrm{t}$ .
$\frac{d\mathrm{P}}{d\mathrm{t}} = \frac{d}{d\mathrm{t}} (600 \mathrm{e}^{0.02 \mathrm{t}})$

Apply the chain rule to differentiate the exponential function.
$\frac{d\mathrm{P}}{d\mathrm{t}} = 600 \cdot 0.02 \mathrm{e}^{0.02 \mathrm{t}}$

The growth rate of the insect population is the derivative we just found.
$Growth\, rate = 600 \cdot 0.02 \mathrm{e}^{0.02 \mathrm{t}}$

To graph the function using a graphing utility, input the function $\mathrm{P}(\mathrm{t})=600 \mathrm{e}^{0.02 \mathrm{t}}$ into the utility and plot it for a range of $\mathrm{t}$ values, typically from $t=0$ to a reasonable upper limit to observe the growth trend.

To determine the population after 10 days, substitute $t=10$ into the population function $\mathrm{P}(\mathrm{t})$ .
$\mathrm{P}(10) = 600 \mathrm{e}^{0.02 \cdot 10}$

Calculate the value of the exponential term.
$\mathrm{e}^{0.02 \cdot 10} = \mathrm{e}^{0.2}$

Use a calculator to find the approximate value of $\mathrm{e}^{0.2}$ .
$\mathrm{e}^{0.2} \approx 1.22140$

Calculate the population at $t=10$ days.
$\mathrm{P}(10) = 600 \cdot 1.22140$

Multiply to find the population after 10 days.
$\mathrm{P}(10) \approx 600 \cdot 1.22140 \approx 732.84$
The population after 10 days is approximately 733 insects (rounded to the nearest whole number).

To find when the insect population will reach 900, we set $\mathrm{P}(\mathrm{t})$ equal to 900 and solve for $\mathrm{t}$ .
$900 = 600 \mathrm{e}^{0.02 \mathrm{t}}$

Divide both sides of the equation by 600.
$\frac{900}{600} = \mathrm{e}^{0.02 \mathrm{t}}$

Simplify the fraction.
$1.5 = \mathrm{e}^{0.02 \mathrm{t}}$

Take the natural logarithm of both sides to solve for $\mathrm{t}$ .
$\ln(1.5) = \ln(\mathrm{e}^{0.02 \mathrm{t}})$

Apply the property of logarithms $\ln(\mathrm{e}^x) = x$ .
$\ln(1.5) = 0.02 \mathrm{t}$

Divide both sides by 0.02 to isolate $\mathrm{t}$ .
$\mathrm{t} = \frac{\ln(1.5)}{0.02}$

Use a calculator to find the approximate value of $\mathrm{t}$ .
$\mathrm{t} \approx \frac{\ln(1.5)}{0.02} \approx 20.79$
The insect population will reach 900 after approximately 20.79 days.

To determine when the insect population will double, we set $\mathrm{P}(\mathrm{t})$ equal to twice the initial population of 600.
$2 \cdot 600 = 600 \mathrm{e}^{0.02 \mathrm{t}}$

Divide both sides of the equation by 600.
$2 = \mathrm{e}^{0.02 \mathrm{t}}$

Take the natural logarithm of both sides to solve for $\mathrm{t}$ .
$\ln(2) = \ln(\mathrm{e}^{0.02 \mathrm{t}})$

Apply the property of logarithms $\ln(\mathrm{e}^x) = x$ .
$\ln(2) = 0.02 \mathrm{t}$

Divide both sides by 0.02 to isolate $\mathrm{t}$ .
$\mathrm{t} = \frac{\ln(2)}{0.02}$

Use a calculator to find the approximate value of $\mathrm{t}$ .
$\mathrm{t} \approx \frac{\ln(2)}{0.02} \approx 34.66$
The insect population will double after approximately 34.66 days.