The problem consists of three independent sub-problems: (2.2) Find the equation of a curve that passes through the point $(0, 5)$ and has a slope equal to twice the y-coordinate at every point. (2.3) A bottle of soda cools from $22^{\circ}C$ to $16^{\circ}C$ in half an hour when placed in a refrigerator at $7^{\circ}C$. We need to find its temperature after another half hour. (2.4) A parchment has 74% of the radioactive carbon 14C compared to plant material today. The half-life of 14C is 5730 years. What is the age of the parchment?

1. Problem Description

The problem consists of three independent sub-problems:

(2.2) Find the equation of a curve that passes through the point $(0, 5)$ and has a slope equal to twice the y-coordinate at every point.

(2.3) A bottle of soda cools from $22^{\circ}C$ to $16^{\circ}C$ in half an hour when placed in a refrigerator at $7^{\circ}C$. We need to find its temperature after another half hour.

(2.4) A parchment has 74% of the radioactive carbon 14C compared to plant material today. The half-life of 14C is 5730 years. What is the age of the parchment?

2. Solution Steps

(2.2) The problem states that the slope of the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = 2y$. This is a separable differential equation. Separating the variables gives $\frac{dy}{y} = 2dx$. Integrating both sides, we get $\int \frac{dy}{y} = \int 2dx$, which gives $\ln|y| = 2x + C$, where C is the constant of integration. Taking the exponential of both sides, we have $|y| = e^{2x+C} = e^C e^{2x}$. We can write $y = Ae^{2x}$ where $A = \pm e^C$.

Since the curve passes through $(0, 5)$, we substitute $x = 0$ and $y = 5$ into the equation $y = Ae^{2x}$: $5 = Ae^{2(0)} = Ae^0 = A$. Thus, $A = 5$.

Therefore, the equation of the curve is $y = 5e^{2x}$.

(2.3) We use Newton's Law of Cooling: $\frac{dT}{dt} = k(T - T_a)$, where $T$ is the temperature of the object, $T_a$ is the ambient temperature, $t$ is time, and $k$ is a constant.

In this case, $T_a = 7^{\circ}C$. Separating variables, we have $\frac{dT}{T - 7} = k dt$. Integrating, we get $\ln|T - 7| = kt + C$, or $T - 7 = Ae^{kt}$, where $A = e^C$. So, $T(t) = 7 + Ae^{kt}$.

At $t = 0$, $T(0) = 22^{\circ}C$, so $22 = 7 + A$, which gives $A = 15$.

Thus, $T(t) = 7 + 15e^{kt}$.

At $t = 0.5$ hours, $T(0.5) = 16^{\circ}C$, so $16 = 7 + 15e^{0.5k}$. Then, $9 = 15e^{0.5k}$, so $e^{0.5k} = \frac{9}{15} = \frac{3}{5} = 0.6$. Thus, $0.5k = \ln(0.6)$, so $k = 2\ln(0.6)$.

Therefore, $T(t) = 7 + 15e^{2\ln(0.6)t}$.

We want to find $T(1)$. $T(1) = 7 + 15e^{2\ln(0.6)(1)} = 7 + 15e^{\ln(0.6^2)} = 7 + 15(0.6^2) = 7 + 15(0.36) = 7 + 5.4 = 12.4$.

So the temperature of the soda after another half hour is $12.4^{\circ}C$.

(2.4) We use the formula for radioactive decay: $N(t) = N_0 e^{-\lambda t}$, where $N(t)$ is the amount of the substance at time $t$, $N_0$ is the initial amount, $\lambda$ is the decay constant, and $t$ is time. The half-life $T_{1/2}$ is related to $\lambda$ by $T_{1/2} = \frac{\ln 2}{\lambda}$, so $\lambda = \frac{\ln 2}{T_{1/2}}$. In this case, $T_{1/2} = 5730$ years.

We are given that $N(t) = 0.74 N_0$. So $0.74 N_0 = N_0 e^{-\lambda t}$, which implies $0.74 = e^{-\lambda t}$. Taking the natural logarithm of both sides, we get $\ln(0.74) = -\lambda t$. Thus, $t = \frac{-\ln(0.74)}{\lambda} = \frac{-\ln(0.74)}{\ln 2 / 5730} = \frac{-\ln(0.74)}{\ln 2} \cdot 5730$.

Calculating this gives $t = \frac{-(-0.301105)}{0.693147} \cdot 5730 = \frac{0.301105}{0.693147} \cdot 5730 \approx 0.43445 \cdot 5730 \approx 2489.37$.

Thus, the age of the parchment is approximately 2489 years.

3. Final Answer

(2.2) $y = 5e^{2x}$

(2.3) $12.4^{\circ}C$

(2.4) Approximately 2489 years