Math Problem Statement

Cost of Scientific Periodicals The table below shows the average cost C, in dollars, of chemistry and physics periodicals t years after 1980.29 t= years since 1980 0 5 10 15 20 22 C= cost, in dollars 140 250 410 780 1300 1520

a. Make an exponential model of C as a function of t. b. Plot the data and the exponential model. c. What was the yearly percentage growth rate of the cost of chemistry and physics periodicals? d. If this exponential trend continues, what will be the expected average cost of physics and chem- istry periodicals in 2020? Round your answer to the nearest dollar.

Solution

Let’s solve this step by step.

Part a: Making an Exponential Model

We need to find an exponential model of the form: C(t)=C0ektC(t) = C_0 \cdot e^{kt} where C0C_0 is the initial cost, kk is the growth rate, and tt is the time in years since 1980.

Given data points:

  • t=0t = 0, C=140C = 140
  • t=5t = 5, C=250C = 250
  • t=10t = 10, C=410C = 410
  • t=15t = 15, C=780C = 780
  • t=20t = 20, C=1300C = 1300
  • t=22t = 22, C=1520C = 1520

To find the parameters C0C_0 and kk, we start with the initial value C0=140C_0 = 140, since this is the cost at t=0t = 0.

Next, we use the data points to find kk. We can select two points and solve for kk. Let’s use (t=0,C=140)(t = 0, C = 140) and (t=5,C=250)(t = 5, C = 250).

The model equation becomes: 250=140e5k250 = 140 \cdot e^{5k}

Divide both sides by 140: 250140=e5k\frac{250}{140} = e^{5k}

Calculating the left side: 1.7857e5k1.7857 \approx e^{5k}

Taking the natural logarithm of both sides: ln(1.7857)=5k\ln(1.7857) = 5k k=ln(1.7857)50.115k = \frac{\ln(1.7857)}{5} \approx 0.115

So, the exponential model is: C(t)=140e0.115tC(t) = 140 \cdot e^{0.115t}

Part b: Plotting the Data and the Exponential Model

To plot the data, we will use the original data points and also plot the exponential model C(t)=140e0.115tC(t) = 140 \cdot e^{0.115t}. This involves calculating the model values for various values of tt and comparing them with the given data points.

Part c: Yearly Percentage Growth Rate

The yearly percentage growth rate can be calculated using the formula: Growth Rate=(ek1)×100\text{Growth Rate} = (e^k - 1) \times 100 Substituting k=0.115k = 0.115: Growth Rate=(e0.1151)×10012.2%\text{Growth Rate} = (e^{0.115} - 1) \times 100 \approx 12.2\%

Part d: Expected Average Cost in 2020

To find the expected cost in 2020, we use t=40t = 40 (since 2020 is 40 years after 1980) in our exponential model: C(40)=140e0.115×40C(40) = 140 \cdot e^{0.115 \times 40} Calculating the exponent: 0.115×40=4.60.115 \times 40 = 4.6 Thus: C(40)=140e4.614099.4813927C(40) = 140 \cdot e^{4.6} \approx 140 \cdot 99.48 \approx 13927

The expected average cost of physics and chemistry periodicals in 2020 is approximately $13,927.


Do you want further details on any part of the solution?

Related Questions:

  1. How would the model change if we used different data points to calculate kk?
  2. How does the choice of initial data point affect the accuracy of an exponential model?
  3. What would the model predict for a year like 2030?
  4. How can we use logarithmic transformation to linearize an exponential model?
  5. What are some limitations of using an exponential model for predicting long-term trends?

Tip: When calculating exponential growth, using the natural logarithm helps solve for growth rates easily due to the properties of ee.

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Math Problem Analysis

Mathematical Concepts

Exponential Growth
Modeling with Functions
Logarithms

Formulas

Exponential model formula: C(t) = C_0 * e^(kt)
Growth rate calculation: Growth Rate = (e^k - 1) * 100

Theorems

Properties of Exponential Functions

Suitable Grade Level

Grades 10-12