57. logistic growth the logistic growth model
$p(t) = \frac{0.8}{1 + 1.67e^{-0.16t}}$
represents the proportion of new cars with a global positioning system (gps). let $t = 0$ represent 2006, $t = 1$ represent 2007, and so on.
(a) what proportion of new cars in 2006 had a gps?
(b) determine the maximum proportion of new cars that have a gps.
(c) using a graphing utility, graph $p = p(t)$.
(d) when will 75% of new cars have a gps?

Question

Accepted Answer

### Part (a)
# Explanation:
## Step1: Identify $ t $ for 2006
Since $ t = 0 $ represents 2006, we substitute $ t = 0 $ into the logistic growth model $ P(t)=\frac{0.8}{1 + 1.67e^{-0.16t}} $.
## Step2: Substitute $ t = 0 $
When $ t = 0 $, $ e^{-0.16\times0}=e^{0} = 1 $. So $ P(0)=\frac{0.8}{1+1.67\times1}=\frac{0.8}{2.67}\approx0.2996 $.
# Answer:
Approximately $ 0.30 $ (or $ 30\% $) of new cars in 2006 had a GPS.

### Part (b)
# Explanation:
## Step1: Recall logistic growth limit
For a logistic growth model $ P(t)=\frac{L}{1 + be^{-kt}} $, the maximum proportion (carrying capacity) is $ L $.
## Step2: Identify $ L $ in the given model
In the model $ P(t)=\frac{0.8}{1 + 1.67e^{-0.16t}} $, we can see that $ L = 0.8 $.
# Answer:
The maximum proportion of new cars that have a GPS is $ 0.8 $ (or $ 80\% $).

### Part (c)
# Explanation:
To graph $ P(t)=\frac{0.8}{1 + 1.67e^{-0.16t}} $:
1. **Domain and Range**: The domain is $ t\geq0 $ (since $ t $ represents years starting from 2006). The range is $ 0 < P(t)<0.8 $ (from logistic growth properties, it approaches 0.8 as $ t\to\infty $ and starts at $ P(0)\approx0.3 $).
2. **Key Points**:
   - At $ t = 0 $, $ P(0)\approx0.3 $.
   - As $ t\to\infty $, $ P(t)\to0.8 $.
   - The function is increasing (since the exponent has a negative coefficient but the overall function's derivative for logistic growth is positive, indicating growth towards the carrying capacity).
3. **Using a Graphing Utility**:
   - Input the function $ y=\frac{0.8}{1 + 1.67e^{-0.16x}} $ (using $ x $ for $ t $) into a graphing calculator or software like Desmos, GeoGebra, or TI - 84.
   - Adjust the window settings: For $ x $ (t - axis), use a range like $ x:0$ to $ 30 $ (to see the approach to the carrying capacity). For $ y $ (P - axis), use $ y:0$ to $ 1 $ (since $ P(t) $ is a proportion).
# Answer:
The graph of $ P(t)=\frac{0.8}{1 + 1.67e^{-0.16t}} $ is a logistic curve that starts at $ (0, \approx0.3) $, increases, and approaches $ y = 0.8 $ as $ t$ increases. It can be plotted using a graphing utility by entering the function and adjusting the viewing window appropriately.

### Part (d)
# Explanation:
## Step1: Set $ P(t)=0.75 $
We want to find $ t $ when $ P(t)=0.75 $. So we set up the equation $ 0.75=\frac{0.8}{1 + 1.67e^{-0.16t}} $.
## Step2: Solve for $ t $
First, cross - multiply: $ 0.75(1 + 1.67e^{-0.16t})=0.8 $
Expand: $ 0.75+1.2525e^{-0.16t}=0.8 $
Subtract 0.75 from both sides: $ 1.2525e^{-0.16t}=0.8 - 0.75 = 0.05 $
Divide both sides by 1.2525: $ e^{-0.16t}=\frac{0.05}{1.2525}\approx0.0399 $
Take the natural logarithm of both sides: $ \ln(e^{-0.16t})=\ln(0.0399) $
Using the property $ \ln(e^{x}) = x $, we get: $ - 0.16t=\ln(0.0399)\approx - 3.203 $
Solve for $ t $: $ t=\frac{-3.203}{-0.16}\approx20.02 $
Since $ t = 0 $ is 2006, $ t\approx20 $ corresponds to $ 2006 + 20=2026 $.
# Answer:
Approximately in the year $ 2026 $ (when $ t\approx20 $) will 75% of new cars have a GPS.

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QUESTION IMAGE

Question

Explanation:

Part (a)

Step1: Identify \( t \) for 2006

Step2: Substitute \( t = 0 \)

Step1: Recall logistic growth limit

Step2: Identify \( L \) in the given model

Answer:

Part (b)