MATH SOLVE

3 months ago

Q:
# In a study by the Texas A & M University, found there were 23,000 prairie dogs in Texas in 2000. By 2010, the number has increased to 40,000. The exponential growth function D=2.3e^kt describes the prairie dog population, D in ten thousands, t years after 2000. A) solve for kB)by this model how many prairie dogs existed last year?C)in which year will the population double?

Accepted Solution

A:

The population function is given as [tex]D=2.3e^{kt}[/tex].(A) [tex][tex]D=2.3e^{kt}\\\frac{D}{2.3}=e^{kt}\\\ln\frac{D}{2.3}=\ln e^{kt}\\\ln\frac{D}{2.3}=kt\\k=\frac{\ln\frac{D}{2.3}}{t}\\k = \frac{\ln\frac{4}{2.3}}{10}=0.0553[/tex][/tex]The value of k is 0.0553.(I used D=4 for t=10 to determine the value of k.)(B) With the above, we have a fully determined growth function:[tex]D(t)=2.3e^{0.0553t}[/tex]and can answer this question now. Last year was 2018, so t=18:[tex][tex]D(18)=2.3e^{0.0553\cdot 18 }=6.22[/tex][/tex]Last year (2018), there existed approximately 62200 prairie dogs.(C) We are looking for t* for which D(t*) will be twice the population as of last year (Note: this is my interpretation of the vague question wording - it is not 100% clear with respect to which population we are considering the doubling):[tex]D(t^*)=2D(18)=12.45=2.3e^{0.0553t^*}\\\frac{12.45}{2.3}=e^{0.0553t^*}\\\ln\frac{12.45}{2.3}=\ln e^{0.0553t^*}\\\ln\frac{12.45}{2.3}=0.0553t^*\\\frac{\ln\frac{12.45}{2.3}}{0.0553}=t^*\\t^*=30.54[/tex]The doubling will occur approximately in 2030 (mid-year).