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for each scenario, write an exponential model in function notation. cho…

Question

for each scenario, write an exponential model in function notation. choose variables that would make sense for the problem. (there are no \correct\ variables, but try to have them fit.)

  1. a car that is worth $25,000, decreases in value by 15% per year.
  2. mr. brusts iq is currently 173, but it is decaying at a rate of 4.5% every year.
  3. a plague of mice has hit australia again! starting with only 30 mice, they can increase by 650% every month.
  4. there are 2,300 counts of bacteria in a petri dish. its total count increases by 168% per day.
  5. during a cleveland browns game, sullys blood pressure rises 7.8% each quarter. at kickoff, his systolic blood pressure is 120.
  6. mr. beans yard is getting overrun with weeds. the first year he bought his home, there was 1800 square feet of grass. it is decreasing by 16.1% per year.

Explanation:

Step1: Recall decay/growth formula

Exponential decay: $f(t) = P(1-r)^t$; Exponential growth: $f(t) = P(1+r)^t$, where $P$ = initial value, $r$ = rate, $t$ = time.

Step2: Model problem 32 (car value decay)

Initial value $P=25000$, decay rate $r=0.15$, time $t$ (years), value $V(t)$.
$V(t) = 25000(1-0.15)^t = 25000(0.85)^t$

Step3: Model problem 33 (IQ decay)

Initial value $P=173$, decay rate $r=0.045$, time $t$ (years), IQ $I(t)$.
$I(t) = 173(1-0.045)^t = 173(0.955)^t$

Step4: Model problem 34 (mouse growth)

Initial value $P=30$, growth rate $r=6.5$, time $t$ (months), mouse count $M(t)$.
$M(t) = 30(1+6.5)^t = 30(7.5)^t$

Step5: Model problem 35 (bacteria growth)

Initial value $P=2300$, growth rate $r=1.68$, time $t$ (days), bacteria count $B(t)$.
$B(t) = 2300(1+1.68)^t = 2300(2.68)^t$

Step6: Model problem 36 (blood pressure growth)

Initial value $P=120$, growth rate $r=0.078$, time $t$ (quarters), blood pressure $P(t)$.
$P(t) = 120(1+0.078)^t = 120(1.078)^t$

Step7: Model problem 37 (grass decay)

Initial value $P=1800$, decay rate $r=0.161$, time $t$ (years), grass area $G(t)$.
$G(t) = 1800(1-0.161)^t = 1800(0.839)^t$

Answer:

  1. $V(t) = 25000(0.85)^t$, where $V(t)$ = car value after $t$ years
  2. $I(t) = 173(0.955)^t$, where $I(t)$ = IQ after $t$ years
  3. $M(t) = 30(7.5)^t$, where $M(t)$ = mouse count after $t$ months
  4. $B(t) = 2300(2.68)^t$, where $B(t)$ = bacteria count after $t$ days
  5. $P(t) = 120(1.078)^t$, where $P(t)$ = blood pressure after $t$ quarters
  6. $G(t) = 1800(0.839)^t$, where $G(t)$ = grass area after $t$ years