Solid Obtained - Calculus Two for Biological Sciences - Solved Exam, Exams of Calculus

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7) Solve the differential equation: x y

dx

dy

= + where y(0)=1.0 (10 pts)

This is a linear first order equation. Use the integration factor technique

to solve it.

( ) 1 exp( )

exp( ) exp( ) exp( ) exp( )

exp( )

exp( )

exp( )

exp( )

exp( )

( ) exp( ( 1 ) ) exp( )

y c

y c

y x x c x

x x x x x x c

du dx v x

u x dv x

x x

x

x xdx

I x

I x gx y

I x dx x

g x x

p x

y x dx

dy

Sketch a graph of the solutions that satisfy the given initial conditions on the

slope field map. (5 pts)

8) A function y(t) satisfies the differential equation:

a) Find and plot equilibrium points (10 pts)

y y y y Equilibrium po s dt

dy ( 8 )( 5 ) 0 0 , 5 , 8 int

2 = − − = ⇒ =

b) Determine whether equilibrium points are stable, unstable (or perhaps

something else). (5 pts)

Unstable

Stable

Saddle

10) The population of aphids on a rose plant increases at a rate proportional

to the number present.

a) Write a differential equation for population of aphids at time t in days.

(10 pts)

This is an exponential growth equation

kP dt

dP

b) Find the solution to the differential equation where at t=0 there were 1000

aphids and population doubles every 10 days. (10 pts)

Solution is exponential growth

ln( 2 )

2 exp( 10 )

() exp( )

o

o o

o

P

k

P P k

Pt P kt

11) Newton’s Law of cooling states that the rate of cooling is proportional to

the temperature difference between the object and its surrounding. (10 pts)

The differential equation for temperature, T, of a coffee cup as a function of

time (t) where T A is the ambient temperature of air, T o > TA is the initial

temperature of the coffee and k is the proportionality constant is: (5 pts)

A) T’ = -kT(1- T/TA)

B) T’ = -k(T - TA)

C) T’ = k(T - T A)

D) T’ = kT/TA

E) T’ = -kT/TA

Answer is B: Since the coffee is cooling rate must be negative. For T > T A

T’ = -k(T – TA) is always negative and fulfills the description of the

Newton’s law of cooling.

The solution to the differential equation is: (5 pts)

A) T(t)= T A/(1+Aexp(-kt)), where A = (T A – To)/To

B) T(t) = TA + Toexp(-kt)

C) T(t) = TAexp(-kt) – TA + To

D) T(t) = TA + (To – TA)exp(-kt)

E) T(t) = (To/TA)exp(-kt)

Answer is D: This is a 1

st

order linear equation which you can solve using

the integration factor method. You can also arrive at this answer by

eliminating bad choices.

A) is an answer to Logistics Equation which this is not.

B and E) can be eliminated since at t=0, T = To. If we set t=0 we get T A + To

in B) and T o/TA in E)

C doesn’t work since for large t, T(t) must approach TA.

Appendix:

Identities:

( 1 cos( 2 )) 2

sin ( )

cos ( ) sin ( ) 1

sec ( ) tan ( ) 1

2

2 2

2 2

x x

x x

x x

tan( )sec( )

sec( )

sec ( )

tan( ) 2

d

d

d

d

Logistics Equation:

0

1 exp( )

P

K P

A

A kt

K

Pt

kPK P dt

dP

ln(AB)=ln(A) + ln(B)

ln(2)=0.

ln(3)= 1.

ln(4)=1.

ln(5)=1.