Answer :
(a) To determine which differential equation is satisfied by [tex]\( y \)[/tex], we consider the situation where a quantity decreases over time. The differential equation that models an exponential decay process, where the rate at which a substance decreases is proportional to the amount present, is generally expressed as:
[tex]\[
\frac{d y}{d t} = k y
\][/tex]
This equation is highlighted in the options provided. So, the correct differential equation is:
[tex]\[
\frac{d y}{d t} = k y
\][/tex]
(b) Now, let's solve for the amount of [tex]\(\delta\)[/tex]-glucono-lactone remaining after 15 hours, knowing it started from 100 grams and decreased to 66.9 grams in one hour.
We'll use the formula for exponential decay:
[tex]\[
y(t) = y_0 \cdot e^{-kt}
\][/tex]
- [tex]\( y_0 = 100 \)[/tex] grams: the initial amount.
- [tex]\( y(1) = 66.9 \)[/tex] grams.
- We need to find the decay constant [tex]\( k \)[/tex] first.
First, calculate [tex]\( k \)[/tex]:
[tex]\[
66.9 = 100 \cdot e^{-k \cdot 1}
\][/tex]
Rearrange to solve for [tex]\( k \)[/tex]:
[tex]\[
e^{-k} = \frac{66.9}{100}
\][/tex]
[tex]\[
-k = \ln\left(\frac{66.9}{100}\right)
\][/tex]
[tex]\[
k = -\ln\left(\frac{66.9}{100}\right)
\][/tex]
With this calculated [tex]\( k \)[/tex], now find the amount remaining after 15 hours:
[tex]\[
y(15) = 100 \cdot e^{-k \cdot 15}
\][/tex]
Compute this using the previously determined [tex]\( k \)[/tex]:
The amount remaining after 15 hours is approximately [tex]\( 0.2407 \)[/tex] grams when rounded to four decimal places.
So, after 15 hours, about 0.2407 grams of [tex]\(\delta\)[/tex]-glucono-lactone will remain.
[tex]\[
\frac{d y}{d t} = k y
\][/tex]
This equation is highlighted in the options provided. So, the correct differential equation is:
[tex]\[
\frac{d y}{d t} = k y
\][/tex]
(b) Now, let's solve for the amount of [tex]\(\delta\)[/tex]-glucono-lactone remaining after 15 hours, knowing it started from 100 grams and decreased to 66.9 grams in one hour.
We'll use the formula for exponential decay:
[tex]\[
y(t) = y_0 \cdot e^{-kt}
\][/tex]
- [tex]\( y_0 = 100 \)[/tex] grams: the initial amount.
- [tex]\( y(1) = 66.9 \)[/tex] grams.
- We need to find the decay constant [tex]\( k \)[/tex] first.
First, calculate [tex]\( k \)[/tex]:
[tex]\[
66.9 = 100 \cdot e^{-k \cdot 1}
\][/tex]
Rearrange to solve for [tex]\( k \)[/tex]:
[tex]\[
e^{-k} = \frac{66.9}{100}
\][/tex]
[tex]\[
-k = \ln\left(\frac{66.9}{100}\right)
\][/tex]
[tex]\[
k = -\ln\left(\frac{66.9}{100}\right)
\][/tex]
With this calculated [tex]\( k \)[/tex], now find the amount remaining after 15 hours:
[tex]\[
y(15) = 100 \cdot e^{-k \cdot 15}
\][/tex]
Compute this using the previously determined [tex]\( k \)[/tex]:
The amount remaining after 15 hours is approximately [tex]\( 0.2407 \)[/tex] grams when rounded to four decimal places.
So, after 15 hours, about 0.2407 grams of [tex]\(\delta\)[/tex]-glucono-lactone will remain.
