a)The population of a community is known to increase at a rate proportional to t

Question

a)The population of a community is known to increase at a rate proportional to the number of people present at time t. If an initial population P0 has doubled in 4 years, how long will it take to triple? (Round your answer to one decimal place.)

How long will it take to quadruple? (Round your answer to one decimal place.)
yr

b) Initially 100 milligrams of a radioactive substance was present. After 8 hours the mass had decreased by 7%. If the rate of decay is proportional to the amount of the substance present at time t, determine the half-life of the radioactive substance. (Round your answer to one decimal place.)
hr

c) A thermometer is removed from a room where the temperature is 70

Explanation / Answer

(a)

Assume continuous compounding:
Find the rate of growth, r by solving:
2 = e^(4r)
r = ln2 / 4

To find tripling time, solve for t:
3 = e^(tln2 / 4)
t = 4ln3 / ln2 = 6.34 years

Similarly, to find time taken to quadruple, solve:
4 = e^(tln2 / 4)

t = 4ln4 / ln2 = 8 years

(b)

Rate of decay proportional to amount remaining:

d n /d t = - k * n
where:
n= amount of substance remaining
k = constant
(the minus sign is because the amount of substance is decreasing)

This is a separable equation having as answer:

n = A* e ^ (-kt) (where A is yet another constant)

Initially (at t= 0) n = A so from the given data we know that A=100mg
After 8 hours (t=6) the mass decreased by 7%, so the mass left now is 93mg n= 100e^(-8k) = 93
so e^(-8k) = 93/100= 0.93

Take the natural logarithm of both sides:

-8k = ln (0.93)
k =~ 9.07 * 10 ^-3

We now have the full equation: n = 100 e ^(- 9.07*10^-3 * t)

Als, to get the half life, put n=50 (half of 100mg)

50 = 100 e ^(- 9.07*10^-3 * t)

e ^(- 9.07*10^-3 * t) = 0.5

-9.07*10^-3*t = ln(0.5)

t = 76.422 hrs

(c)

Use Newton's law of cooling:
dT/dt = k(T - Ta) .......... Ta is the ambient temperature
? dT/(T - Ta) = k ? dt
ln(T - Ta) = kt + c
T = Ce^(kt) + Ta
T[0] = 70, Ta = 40 ....

70=c+40

C = 30
T[0.5] = 60 ...

60=30*e^(0.5k) + 40

e^(0.5k) = 20/30

k = 2ln(2/3)

T = 60 (2/3)^t + 40
45 = 60 (2/3)^t + 40

(2/3)^t = 1/12

t = ln(1/12)/ln(2/3)

t = 6.13 min

Answer: ? 6.13 min

same equation for the 2nd problem
T = Ce^(kt) + Ta

T = 60(2/3)^t + 40

at t= 1min

T = 60*(2/3) + 40

T = 40+40 = 80 0 F

ANSWER : 80 0 F

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