Mute swans were imported from Europe in the nineteenth century to grace ponds. (Rest of the question in description)

Question

Mute swans were imported from Europe in the nineteenth century to grace ponds. Now there is concern that their population is growing too rapidly, edging out native species. During the 2000 s, many states created mute swan controls. The goal was to reduce the mute swan count from 14313 in 2002 to 3000 swans by 2013.

 

a. Using a linear model, what would be the average rate of change per year needed to bring the swan population to that level?Round your answer to one decimal place.

 

b. Using an exponential model, what would be the decay factor needed?Round your answer to three decimal places.

Answer 1

a.  We have two points on a graph which, we are told, is to be considered linear:

 

     (14313, 2002) and (3000, 2013)

 

     The slope, m, of this line is (14313 - 3000)/(2002 - 2013) = (11313)/(-11) = -1028.4 swans per year.  This, then, is the number of swans which need to be "culled" if the target of 3000 swans in 2013 is to be met.  I shudder to think about how this culling is accomplished.  Could you kill a swan? I couldn't.

 

b.   Using an exponential model

 

     N_(initial) = N_(final)e-λt

 

     where λ is a positive constant called the decay constant.

 

     [N_(initial)/N_(final)] = e^-λt

 

     ln{[N(initial)]/N(final)]} = 1.5626 = (-λ)(Δt) = (-λ)(-11 years)

 

     ∴ λ = 0.142

Answer 2

For part a, a linear model has a constant rate of change.  Linear models usually take the form

y = mx + b

with m being the rate of change of our dependent variable y, as our independent variable x changes.  The value b is the initial value y has (when x = 0).  For this setup, we're dealing with swan populations (our y; dependent variable) as a function of time (our x; independent variable).  If we think of 2002 as our initial year (when x = 0) then we know b will be 41313 swans.  We want the number of swans to be 3000 (y = 3000) by year 2013 (11 years after 2002; x = 11).  Plugging everything into our linear equation, we have

 

y = m*x + b

 

3000 = (m)(11) + 14313

 

Simplifying and solving for m, we get

 

(3000 - 14313) / 11 = m

 

-1028.5 = m

 

Since m is our rate of change (how much the number of swans (y) changes per year (x)), we have our answer.

 

 

For part b, we have an exponential equation of the form

 

y = A*e^k*x

 

with A being the initial value (when x = 0) of our dependent variable y (swans), k being the decay/growth factor, and x being our independent variable (time).  Again, taking 2002 to be the initial year, (year zero; x = 0), we know A = 14313.  In 2013, 11 years after 2002 (x = 11), we want our y value to be 3000.  So let's plug and chug, and solve for our decay factor k.

 

y = A*e^k*x

 

3000 = 14313*e^k*11

 

3000 / 14313 = e^k*11

 

Now we take the natural log of both sides to get

 

ln(3000/14313) = k*11

 

ln(3000/14313) / 11 = k

 

-.1 = k

 

And there's your decay factor!  Hope this was a help.  :)

 

 - Jon

Answer 3

a) linear in this case means a constant decrease in absolute terms. We want the population do decrease 11,313 in 11 years. The rate of change is -11,313/11=-1028.5 swans/year.

 

b) exponential in this case means constant percentage change. We set up this equation 

14,313*k^11=3000

k^11=3000/14,313

ln(k^11)=ln(3000/14,313)=-1.5626

11*ln(k)=-1.5626

ln(k)=-1.5626/11=-.1421

k=exp(-.1421)=.8676

 

this means the population decreases 13.2% per year. (1-.8626=.1324)