Question 4 (3 points) Now, If the herd had continued to grow at the same rate (use the same value for ' k ' that you found in the previous problem), what would its population be in the year 2010 ? 1.) In the formula \( \mathrm{A}=\mathrm{Pe}^{\mathrm{kt}} \) which variable will you be solving for? 2.) Write the equation (set up the equation using substitution and show where the variable is that you are solving 5 Page 4 of 5 for): 3.) After solving the equation you set use in "\#2, you see that the population of caribou in the year 2010 was approximately
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To find the population of the caribou herd in 2010, you're going to solve for \( A \), which represents the future population. In the formula \( A = Pe^{kt} \), \( P \) is the initial population, \( e \) is the base of the natural logarithm, \( k \) is the growth rate, and \( t \) is the time in years since the initial population was measured. Substituting into the equation to find \( A \), you might have something like this: \( A = P \cdot e^{k(t)} \). Let's say the initial population \( P \) in the year you started tracking was 100, and you've determined \( k \) from your previous calculations. If you're looking at 2010 and started tracking in, say, 2005, then \( t \) would equal 5. So your equation would read \( A = 100 \cdot e^{k(5)} \). Now you can plug in the value for \( k \) to solve for \( A \) and find out the caribou population in 2010! After plugging in all the values and performing the calculations, you would arrive at a result which tells you approximately how many caribou there were in 2010. If \( k \) were, for instance, 0.1, the equation would simplify and lead you to find that the population of caribou in 2010 was roughly around 164 (to the nearest whole number). The crowning moment of your mathematical journey! Enjoy those caribou facts!
