Question

the mass of a substance, which follows a continuous exponential growth model, is being studied in a lab. the doubling time for this substance was observed to be 9 hours. there were 1.8 mg of the substance present at the beginning of the study.
(a) let ( t ) be the time (in hours) since the beginning of the study, and let ( y ) be the amount of the substance at time ( t ).
write a formula relating ( y ) to ( t ).
use exact expressions to fill in the missing parts of the formula.
do not use approximations.
( y = square e^{square t} )
(b) how much will be present in 6 hours?
do not round any intermediate computations, and round your answer to the nearest tenth.
(square) mg

Explanation:

Part (a)

Step 1: Recall the continuous exponential growth formula

The general form of continuous exponential growth is \( y = y_0 e^{kt} \), where \( y_0 \) is the initial amount, \( k \) is the growth rate, and \( t \) is time. We know the doubling time \( T = 9 \) hours. For doubling time, when \( t = T \), \( y = 2y_0 \). Substituting into the formula: \( 2y_0 = y_0 e^{k \cdot 9} \).

Step 2: Solve for \( k \)

Divide both sides by \( y_0 \): \( 2 = e^{9k} \). Take the natural logarithm of both sides: \( \ln(2) = 9k \), so \( k=\frac{\ln(2)}{9} \).

Step 3: Substitute \( y_0 \) and \( k \) into the formula

The initial amount \( y_0 = 1.8 \) mg. So the formula is \( y = 1.8 e^{\frac{\ln(2)}{9}t} \).

Step 1: Substitute \( t = 6 \) into the formula from part (a)

We have \( y = 1.8 e^{\frac{\ln(2)}{9} \cdot 6} \).

Step 2: Simplify the exponent

Simplify \( \frac{\ln(2)}{9} \cdot 6=\frac{2\ln(2)}{3}=\ln(2^{\frac{2}{3}})=\ln(\sqrt[3]{4}) \). So \( y = 1.8 e^{\ln(\sqrt[3]{4})} \).

Step 3: Use the property \( e^{\ln(x)} = x \)

Since \( e^{\ln(\sqrt[3]{4})}=\sqrt[3]{4} \), then \( y = 1.8 \times \sqrt[3]{4} \). Calculate \( \sqrt[3]{4}\approx1.5874 \), then \( y\approx1.8\times1.5874\approx2.8573 \). Round to the nearest tenth: \( y\approx2.9 \).

Answer:

\( y = 1.8 e^{\frac{\ln(2)}{9}t} \)

Part (b)