Migratory birds fly north past a check-point at 300-300cos(pi(t)/6) birds per month. t= number of months since the beginning of the year.

Question

A. Find the total number of birds for the year.

B. When are the birds migrating the fastest.

C. When is the migratory rate increasing the fastest?

Answer 1

Hello Jasmine,

 

Let the rate of migration be R(t) 

R(t) = 300(1 - cos(πt/6))

A.

The total number of birds for the year is the integral of R(t) from t=1 to t = 12

       N(t) =∫ R(t) dt

             =∫ R(t) dt

             = 300∫(1 - cos(πt/6))dt

             = 300(t - (6/π)sin(πt/6)) +C

 

Total number of birds =N(12) - N(1)

                                = 300(12 -(6/π)sin(2π))+ C - 300(1-(6/π)sin(π/6)) - C

                                = 300 (12 - 0) - 300(1-3/π)

                                =3586 birds

B.

The maximum of N(t) occurs when R(t) = 0

R(t) = 300(1 - cos(πt/6))

   0  =300(1 - cos(πt/6))

cos(πt/6) = 1

       πt/6 = cos^-1(1) = nπ, n =0, 1, 2, ...

            t = 6n, n =0, 1, 2, 

The birds are migrating the fastest in  beginning of the last month(n = 2) t=12

January (n = 0), and  June (n = 1) are not migration months.

C.

To find when R(t) is increasing the fasting 

R'(t) = 300( (π/6)sin (πt/6))

       = 50π sin(πt/6), 

For maximum R(t), R'(t) = 0

sin((πt/6))= 0

 πt/6 = sin^-1(0)= nπ, n = 0, 1, 2, ...

t = 6n,   n = 0, 1, 2, ...

R(t) = 300(1 - cos(nπ)) = 300 for n = 2

The migratory rate is increasing the fastest when t = 12 , or December.