Question

1. Armando's body metabolizes caffeine at a rate of 13.6% per hour (so the amount of caffeine in Armando's body decreases by 13.6% each hour).

a) If Armando consumes a cup of coffee with 82 mg of caffeine in it, how long will it take for Armando's body to metabolize half of the 82 mg of caffeine?

b) If Armando consumes an energy drink with 208 mg of caffeine in it, how long will it take for Armando's body to metabolize half of the 208 mg of caffeine?

c) If Armando consumes a cup of coffee with c mg of caffeine in it, how long will it take for Armando's body to metabolize half of the c mg of caffeine? (Hint: your answer will be a numerical value.)

2. At the beginning of 2000 Cesar's house was worth 230 thousand dollars and Omar's house was worth 120 thousand dollars. At the beginning of 2003, Cesar's house was worth 190 thousand dollars and Omar's house was worth 158 thousand dollars. Assume that the values of both houses vary at an exponential rate.

a) Write a function f that determines the value of Cesar's house (in thousands of dollars) in terms of the number of years t since the beginning of 2000.

b) Write a function g that determines the value of Omar's house (in thousands of dollars) in terms of the number of years t since the beginning of 2000.

c) How many years after the beginning of 2000 will Cesar's and Omar's house have the same value?

Answer 1

Solution of Q1:(a)

13.6mg is metabolised in 1 hours.

So,

50% of 82 mg, i.e.,41 mg will be metabolised in (1/13.6) ×41

=3.01 hours

(b) 13.6 mg is metabolised in 1 hour

So,

Half of 208 mg, i.e.,104 mg is metabolised in (1/13.6)×104

=7.64 hours

(c) 13.6% is metabolised in 1 hour

So,

Half of c mg, i.e.,c/2 mg is metabolised in (100/13.6c) ×c/2

=3.676 hours

Solution of Q2:

Let the function be A=A₀e^(kt)

here,A=final value

A₀=initial value

k=constant

t=time that has passed

(a)Now we find k,from the data given about Cesar's house

190=230e^(k*3)

=>3k =ln(19÷23)

=>k   =(-0.191)÷3

=>k =-0.06

So,the required function in terms of number of years is

A=A₀e^(-0.06)t

(b)Now,we find k,from the data given about Omar's house

158=120e^(k*3)

   =>3k=ln(158÷120)

=>k =0.2751÷3

=>k =0.091

So,the required function in terms of number of years is

A=A₀'e^(0.091)t

(c)Cesar's and Omar's house will have the same value,then

A₀e^(-0.06)t=A₀'e^(0.091)t

=>230e^(-0.06)t=120e^(0.091)t

=>230/190=e^(0.151)t

=>ln(23/19)=(0.151)t

=>0.65 =(0.151)t

=>t =4.308 years

Thus,the two houses will have the same value after 4.308 years approximately.