The problem describes a scenario where 12% of the air in the lungs is replaced with each breath. The initial amount of air is 500 mls. We need to: (a) Write an exponential decay model in the form $y = ae^{kx}$ for the amount of original air left in the lungs after $x$ breaths. (b) Calculate the amount of original air remaining after 24 breaths. (c) Describe the behavior of the air content as the number of breaths increases.

1. Problem Description

The problem describes a scenario where 12% of the air in the lungs is replaced with each breath. The initial amount of air is 500 mls. We need to:

(a) Write an exponential decay model in the form $y = ae^{kx}$ for the amount of original air left in the lungs after $x$ breaths.

(b) Calculate the amount of original air remaining after 24 breaths.

(c) Describe the behavior of the air content as the number of breaths increases.

2. Solution Steps

(a) Writing the exponential decay model.

Since 12% of the air is replaced with each breath, 88% remains. Therefore, the decay factor is 0.

8

8. The initial amount of air, $a$, is 500 mls.

Since the model is in the form $y = ae^{kx}$, we can express the decay as $y = a(0.88)^x$.

To rewrite the base (0.88) in terms of $e$, we can use the following:

$0.88 = e^{\ln(0.88)}$

$y = 500(e^{\ln(0.88)})^x$

$y = 500e^{\ln(0.88)x}$

Thus, $k = \ln(0.88)$

(b) Amount of original air after 24 breaths.

Now we substitute $x = 24$ into the equation $y = 500e^{\ln(0.88)x}$.

$y = 500e^{\ln(0.88)(24)}$

$y = 500e^{24\ln(0.88)}$

$y \approx 500e^{-3.058}$

$y \approx 500(0.047)$

$y \approx 23.5$

(c) Behavior of the air content as the number of breaths increases.

As the number of breaths increases, the amount of original air remaining in the lungs approaches zero. The exponential decay model indicates that the amount of air decreases exponentially as the number of breaths increases. The term $e^{kx}$ approaches 0 as $x$ approaches infinity since $k = \ln(0.88)$ is negative.

3. Final Answer

(a) The exponential decay model is $y = 500e^{\ln(0.88)x}$.

(b) After 24 breaths, approximately 23.5 mls of the original air remains.

(c) As the number of breaths increases, the amount of original air in the lungs approaches zero.