I have no clue where to begin on this problem. Can some one help me please.
One demographer believes that the population growth of a certain country is best modeled by the function P (t) =15 e^.08t, while a second demographer believes that the population growth of that same country is best modeled by the function P (t) =15 +3t. In each case, t is the number of years from the present and P (t) is given in millions of people. For what values of t do these two models give the same population? In how many years is the population predicted by the exponential model twice as large as the population predicted by the linear model?
Just get past the words and see what they're really telling you.
There are two functions:
p(t) = 15e^.08t
q(t) = 15 + 3t
when are they equal?
15e^.08t = 15+3t
p(0) = 15
q(0) = 15
p(1) = 16.249
q(1) = 18
They start out the same, but you know that the exponential will eventually grow much faster. Look at a graph.
p(t) = q(t) when t = 20.235 or so
when does
p(t) = 2q(t)
again, from the graph, t=34.501
To find the values of t for which the two models give the same population, we need to equate the two functions and solve for t.
Equating the functions:
15e^(0.08t) = 15 + 3t
Now, let's solve this equation step by step.
Step 1: Divide both sides of the equation by 15.
e^(0.08t) = 1 + 0.2t
Step 2: Subtract 1 from both sides of the equation.
e^(0.08t) - 1 = 0.2t
Step 3: Multiply both sides of the equation by 5.
5(e^(0.08t) - 1) = t
At this point, it is difficult to solve for t algebraically, so we can use numerical methods or approximation techniques to find the approximate values of t.
One way to do this is by plotting the graphs of the two functions on a graphing calculator or software and finding the points of intersection. We can also use trial and error by substituting different values of t into the equations until we find values that satisfy the equation.
For the second part of the question, we need to find the number of years in which the population predicted by the exponential model (P(t) = 15e^(0.08t)) is twice as large as the population predicted by the linear model (P(t) = 15 + 3t).
Let's set up the equation and solve for t:
15e^(0.08t) = 2(15 + 3t)
Start by dividing both sides of the equation by 15.
e^(0.08t) = 2 + 0.4t
Similarly to the previous equation, it is difficult to solve for t algebraically. Therefore, we can use numerical methods or approximation techniques to find the approximate value of t at which the exponential model is twice as large as the linear model.