Question: User: A certain substance decomposes according to a continuous exponential decay model. To begin an experiment, the initial amount of the substance is $5800$ kg.
After $18$ hours, $1856$ kg of the substance is left.
(a) Let $t$ be the time (in hours) since the beginning of the experiment,
and let $y$ be the amount of the substance (in kg) 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 of the substance is left $22$ hours after the beginning of
the experiment?
Do not round any intermediate computations, and round your
answer to the nearest whole number.
\square kg
Graph/maths placement: the main text block and formula boxes are in the left-center area, the small tool palette is top-right, and the answer box for part (b) is bottom-left.
1. **State the problem:**
We have a substance undergoing continuous exponential decay. The initial amount is $5800$ kg, and after $18$ hours, $1856$ kg remains. We want to find the formula relating the amount $y$ to time $t$, and then find the amount left after $22$ hours.
2. **Formula for continuous exponential decay:**
The general formula is:
$$y = y_0 e^{kt}$$
where:
- $y_0$ is the initial amount,
- $k$ is the decay constant (negative for decay),
- $t$ is time,
- $y$ is the amount at time $t$.
3. **Find the decay constant $k$ using given data:**
Given:
- $y_0 = 5800$
- $y(18) = 1856$
Substitute into the formula:
$$1856 = 5800 e^{18k}$$
Divide both sides by $5800$:
$$\frac{1856}{5800} = e^{18k}$$
Write the fraction explicitly:
$$\frac{1856}{5800} = e^{18k}$$
Take the natural logarithm of both sides:
$$\ln\left(\frac{1856}{5800}\right) = 18k$$
Solve for $k$:
$$k = \frac{\ln\left(\frac{1856}{5800}\right)}{18}$$
4. **Write the exact formula for $y$:**
$$y = 5800 e^{\left(\frac{\ln\left(\frac{1856}{5800}\right)}{18}\right) t}$$
This is the exact expression without approximations.
5. **Calculate the amount after $22$ hours:**
Substitute $t=22$:
$$y(22) = 5800 e^{\left(\frac{\ln\left(\frac{1856}{5800}\right)}{18}\right) \times 22}$$
Simplify the exponent:
$$y(22) = 5800 e^{\ln\left(\frac{1856}{5800}\right) \times \frac{22}{18}}$$
Using the property $e^{\ln(a) b} = a^b$:
$$y(22) = 5800 \left(\frac{1856}{5800}\right)^{\frac{22}{18}}$$
6. **Compute the numerical value:**
Calculate the fraction:
$$\frac{1856}{5800} \approx 0.32$$
Calculate the exponent:
$$\frac{22}{18} = \frac{11}{9} \approx 1.2222$$
Calculate:
$$0.32^{1.2222} \approx 0.263$$
Multiply by $5800$:
$$5800 \times 0.263 \approx 1525.4$$
Rounded to the nearest whole number:
$$1525$$
**Final answers:**
(a) $$y = 5800 e^{\left(\frac{\ln\left(\frac{1856}{5800}\right)}{18}\right) t}$$
(b) Amount left after $22$ hours is approximately $1525$ kg.