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kondor19780726 [428]
2 years ago
11

The concentration of dichlorodiphenyltrichloroethane (DDT - an infamous pesticide) in Lake Michigan has been declining exponenti

ally over many years. In 1971, the DDT concentration was 13.00 ppm (parts per million). In 1984, the DDT concentration was 2.22 ppm.
Using these data points, determine the exponential decay function to model the DDT concentration decline in terms of years, tt, where t=0t=0 represents the year 1971.


A) Write your function as y = Ce^{rt}y=Ce^rt. Round your value of rr to four decimal places.


B) Another way to write the function down is known as the half-life formula, written as y = C(0.5)^{t/H}y=C(0.5)^t/H, where HH is the time required for the concentration to get cut in half. Using the initial two data points, find the value of HH, rounded to 2 decimal places. Show all Algebraic work.


C) The goal was to reduce the DDT concentration to 0.98 ppm by 1995. Was this possible according to your model?
Mathematics
1 answer:
mojhsa [17]2 years ago
8 0

Answer: A) y=13e^{-0.1359t}

              B) H = 5.10

              C) Yes

Step-by-step explanation: <u>Exponential</u> <u>Decay</u> <u>function</u> is a model that describes the reducing of an amount by a constant rate over time. Generally, it is written in the form: y(t)=Ce^{rt}

A) C is initial quantity, in this case, the initial concentration of DDT. To determine r, using the data given:

y(t)=Ce^{rt}

2.22=13e^{13r}

e^{13r}=0.1708

Using a natural logarithm property called <em>power rule:</em>

13r=ln(0.1708)

r=\frac{ln(0.1708)}{13}

r=-0.1359

The decay function for concentration of DDT through the years is y(t)=13e^{-0.1359t}

B) The value of H is calculated by y=C(0.5)^{\frac{t}{H} }

2.22=13(0.5)^{\frac{13}{H} }

(0.5)^{\frac{13}{H} }=0.1708

Again, using power rule for logarithm:

\frac{13}{H} log(0.5)=log(0.1708)

\frac{13}{H} =\frac{log(0.1708)}{log(0.5)}

\frac{13}{H} =2.55

H = 5.10

Constant H in the half-life formula is H=5.10

C) Using model y(t)=13e^{-0.1359t} to determine concentration of DDT in 1995:

y(24)=13e^{-0.1359.24}

y(24) = 0.5

By 1995, the concentration of DDT is 0.5 ppm, so using this model is possible to reduce such amount and more of DDT.

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