MA 138 - Calculus 2 for the Life SciencesFINAL EXAM

Spring 20134/30/2013

Name:

Sect. #:

Answer all of the following questions. Use the backs of the question papers for scratch paper.No books or notes may be used. You may use a calculator. You may not use a calculator thathas symbolic manipulation capabilities. When answering these questions, please be sure to:

• check answers when possible,

• clearly indicate your answer and the reasoning used to arrive at that answer(unsupported answers may receive NO credit).

QUESTION SCORE TOTAL

1. 10

2. 10

3. 10

4. 10

5. 10

6. 10

7. 10

8. 10

9. 10

10. 10

Bonus. 10

TOTAL 100

Please make sure to list the correct section number on the front page of your exam. In case youforgot your section number, consult the following table:

Sections # Lecturer Time/Location

002-003 Kate Ponto MWF 10:00 am - 10:50 am, FPAT 257

004-005 Alberto Corso MWF 01:00 pm - 01:50 pm, SRB 303

Section # Recitation Instructor Time/Location

002 Laura Graham TR 12:30 pm - 01:20 pm, CB 241

003 Laura Graham TR 02:00 pm - 02:50 pm, CP 111

004 Jonathan Thompson TR 12:30 pm - 01:20 pm, CP 367

005 Jonathan Thompson TR 02:00 pm - 02:50 pm, BS 109

1. Find the area of the region bounded by the graphs of the functions

y = 2x− x2 and y = x− 2.

x

y

pts: /10

2. (a) Use the substitution method to evaluate the definite integral∫ ln 7

ln 4

ex

(ex − 3)2dx

(b) Let f be a twice differentiable function of x. Use the integration by parts methodand the Fundamental Theorem of Calculus to evaluate the integral∫ 3

1

xf ′′(x) dx,

given that f(1) = −2 f(3) = 5 f ′(1) = −2 f ′(3) = 4.

pts: /10

3. The logistic model for growth of a population is described by the differential equationdN

dt= rN

(1− N

K

), where r is the growth rate and K is the carrying capacity. Both r

and K are positive constants.

For some species a very small population limits reproduction since there is a lack ofsuitable mates. We can model this using the differential equation

dN

dt= rN (N − a)

(1− N

K

),

where 0 < a < K.

(a) Find the equilibria N̂ for the latter differential equation.

(b) Use the analytic (eingenvalues) method to classify these points.(Hint:

dN

dt= g(N), where g(N) = − r

KN3 + r

(1 +

a

K

)N2 − raN .

)

pts: /10

4. Suppose we have two strains of bacteria with population sizes denoted by a = a(t) andb = b(t), where t represents time. If the growth rates of the bacteria are proportional tothe size of the respective population, we have differential equations

da

dt= µ a and

db

dt= λ b,

where µ and λ are positive constants. We also assume that µ ̸= λ.

The fraction of the population of type a is given by the expression

p =a

a+ b.

It can be shown that this fraction p = p(t) satisfies the following differential equation

dp

dt= (µ− λ) · p · (1− p).

(i) Find the equilibria for this differential equation and use the graphical method in orderto classify these points.

case µ > λ:

p

dp

dt

case µ < λ:

p

dp

dt

(ii) What is the biological meaning of your findings in part (i) in terms of the two strainsof bacteria?

In the case µ > λ we have:

In the case µ < λ we have:

pts: /10

5. Suppose a scientist has four colonies of the same bacteria. She wants to estimate thegrowth rate of this species of bacteria. To do this, she measures the population of eachcolony at time t = 0 and one day later at time t = 1. The data from her measurementsare listed below:

Colony P0 P1

A 9 17B 2 4.5C 4 9D 5 7

Find the least squares approximation of the form

P1 = mP0 + b

that expresses the population P1 at time t = 1 as a function of the population P0 at t = 0.

pts: /10

6. Solve the following system of linear equationsx + 2y − z = 2x + 4y + 3z = 83x + 8y + z = 12

,

by writing the corresponding augmented matrix and then by row reducing.

How many solutions does the system have?

Which of the two pictures below illustrates the geometric situation described by thegiven system of linear equations?

pts: /10

7. Match each of the following functions

f(x, y) = 3−x+y g(x, y) = xy(1−x−y) h(x, y) = xye−x2−y2 k(x, y) =√4− x2 − y2

with its graph (labeled A.-D.) and its level curves (labeled I.-IV.).

A.

B.

C.

D.

f(x, y) corresponds to

graph and

level curves

g(x, y) corresponds to

graph and

level curves

h(x, y) corresponds to

graph and

level curves

k(x, y) corresponds to

graph and

level curves

I.

II.

III.

IV.

pts: /10

8. Consider the function

f(x, y) = 4−√

x2 + y2 + 2

whose graph is given in the picture on the right.

(a) Find the z-coordinate z0 of the point P on the graphof the function f(x, y) with x-coordinate x0 = 1 andy-coordinate y0 = 1.

(b) Write the equation of the tangent plane to the graph of the function f(x, y) at thepoint P , as above, with coordinates x0 = 1 and y0 = 1.

(c) Write the linear approximation, L(x, y), of the function f at the point with x0 = 1and y0 = 1, as above, and use it to approximate f(1.1, 0.9).Compare this approximate value to the exact value f(1.1, 0.9).

pts: /10

9. If a large block of ice is placed in a room we can describe how the temperature of theblock of ice and room are changing using the system of differential equations

dI

dt= α(R− I)

dR

dt= β(I −R),

where I is the temperature of the block of ice, R is the temperature of the room and α andβ are positive constants that determine the relationship between the rates of change of thetemperatures of the block of ice and room and the difference between these temperatures.(All temperatures are measured in degrees Fahrenheit (◦F).)

(a) Find the solution of the given system of linear differential equations in the case thatα = 0.5, β = 0.25, I(0) = 32, and R(0) = 74. That is

d

dt

I

R

=

−0.5 0.5

0.25 −0.25

I

R

I(0)

R(0)

=

32

74

(b) Describe the long term behavior of your solution. In particular what happens to thetemperature of the room and to the block of ice?

(c) Which of the pictures below describes the behavior of the two temperatures?

pts: /10

10. Suppose a habitat is divided up into patches and each patch can be occupied by at mostone individual. If two species A and B live in this habitat, the growth of the popula-tion of A is controlled by the internal dynamics of the population growth of A and theinteractions between A and B. The situation for B is similar.

Suppose the members of species A are able to outcompete members of species B, that is,the members of A are able to invade patches that are occupied by species B and displacethe resident. If p1 is the fraction of the sites occupied by A and p2 is the fraction of thesites occupied by B, this situation is described by the differential equations

dp1dt

= c1p1(1− p1)−m1p1dp2dt

= c2p2(1− p1 − p2)−m2p2 − c1p1p2

where c1, c2,m1, and m2 are the colonization and mortality rates of species A and B, respec-tively.

Suppose c1 = 2, c2 = 10, m1 = 1, and m2 = 2. Hence, after some algebra, the abovesystem of nonlinear differential equations can be written as

dp1dt

= p1 (1− 2p1)

dp2dt

= p2 (8− 12p1 − 10p2)

(a) There is only one equilibrium point (p̂1, p̂2) where both species are present.Identify that point and linearize the given system of differential equations at thatpoint. Classify the type of equilibrium.

p1

p2

Plot here the nullclines andthe nontrivial equilibrium point

(b) Choose the direction field that describes the system of nonlinear differential equa-tions considered in (a).

� Direction Field A

� Direction Field B

pts: /10

Bonus. There are several ways that two species can interact in accordance with the Lotka-Volterramodel. The direction fields below correspond to four different behaviors.

For each direction field choose the type of interaction1,2 between the two species anddescribe the long term behavior of the two populations.

Type of interaction:

� Coexistence

� Founder control

� Competitive exclusion

� Predator-prey

Long term behavior:

Type of interaction:

� Coexistence

� Founder control

� Competitive exclusion

� Predator-prey

Long term behavior:

1Founder control: No species can invade the other. The competition outcome depends on the initial densities.2Competitive exclusion: One species, the strong one, outcompetes the other, the weak one.

Type of interaction:

� Coexistence

� Founder control

� Competitive exclusion

� Predator-prey

Long term behavior:

Type of interaction:

� Coexistence

� Founder control

� Competitive exclusion

� Predator-prey

Long term behavior:

pts: /10