Determining a Function to Model Exponential Growth

The growth of atmospheric carbon dioxide over time can be modeled using a function based on the data from the table below. Now we will determine such a function from the data.

Year Carbon Dioxide (ppm)
1990 353
2000 375
2075 590
2175 1090
2275 2000
  1. Find an exponential function that gives the amount of carbon dioxide y in year x.

  2. Estimate the year when future levels of carbon dioxide will be double the preindustrial level of 280 ppm.

Solutions

  1. The data points showed exponential growth, so the equation will take the form LaTeX: y=y_{0} e^{kt} y=y0ekt.  We must find the values of LaTeX: y_{0} y0 and LaTeX: kk. The data began with the year 1990, so to simplify our work we let 1990 correspond to  x = 0, 1991 correspond to x = 1, and so on. Since LaTeX: y_0y0 is the initial amount, LaTeX: y_0y0 = 353 in 1990 when x = 0. 
    Thus the equation is LaTeX: y=353e^{kx}y=353ekx.
    From the last pair of values in the table, we know that in 2275 the carbon dioxide level is expected to be 2000 ppm. The year 2275 corresponds to 2275 – 1990 = 285.  Substitute 2000 for y and 285 for x and solve for k.LaTeX: y=353e^{kx}y=353ekx
    LaTeX: 2000=353e^{k\left(285\right)}2000=353ek(285)
    LaTeX: \frac{2000}{353}=e^{285k}2000353=e285k
    LaTeX: \ln\frac{2000}{353}=\ln e^{285k}ln2000353=lne285k
    LaTeX: \ln\frac{2000}{353}=285kln2000353=285k
    LaTeX: k=\frac{1}{285}\ln\frac{2000}{353}\approx0.00609k=1285ln20003530.00609
    A function modeling the data is LaTeX: y=353e^{0.00609x}y=353e0.00609x
  2. When the level is double 280 ppm, it will be 2(280)=560 ppm.
    LaTeX: y=353e^{0.00609x}y=353e0.00609x
    LaTeX: \text{560}=353e^{0.00609x}560=353e0.00609x
    Divide by 353.
    LaTeX: \frac{560}{353}=e^{0.00609x}560353=e0.00609x
    Take the natural logarithm of both sides.
    LaTeX: \ln\frac{560}{353}=\ln e^{0.00609x}ln560353=lne0.00609x
    Since LaTeX: \:\:\ln e^x=x,\:lnex=x,for all LaTeX: xx.
    LaTeX: \ln\frac{560}{353}=0.00609xln560353=0.00609x
    LaTeX: x=\frac{1}{0.00609}\cdot\ln\left(\frac{560}{353}\right)\approx75.8x=10.00609ln(560353)75.8
    Since x = 0 corresponds to 1990, the preindustrial carbon dioxide level will double in the 75th year after 1990, or during 2065, according to this model.