A population of 10,000 in 1960 grew to 12,000 in 1970. Estimate the population in 1980 and find the doubling time.

Substitute the values into the model:

$$12000 = 10000 e^{k \cdot 10}$$

Divide by 10000:

$$\frac{12000}{10000} = e^{10k}$$ $$1.2 = e^{10k}$$

Take the natural logarithm of both sides:

$$\ln(1.2) = \ln(e^{10k})$$ $$\ln(1.2) = 10k$$ $$k = \frac{\ln(1.2)}{10}$$

The time from 1960 to 1980 is $t = 1980 - 1960 = 20$ years.
Substitute $k$ and $t=20$ into the model:

$$P(20) = 10000 e^{\left(\frac{\ln(1.2)}{10}\right) \cdot 20}$$ $$P(20) = 10000 e^{2 \ln(1.2)}$$

Using the logarithm property $a \ln b = \ln b^a$:

$$P(20) = 10000 e^{\ln((1.2)^2)}$$

Using the property $e^{\ln x} = x$:

$$P(20) = 10000 \times (1.2)^2$$ $$P(20) = 10000 \times 1.44$$ $$P(20) = 14400$$

The estimated population in 1980 is $\boxed{\text{14400}}$.

Take the natural logarithm of both sides:

$$\ln 2 = \ln(e^{kT_d})$$ $$\ln 2 = kT_d$$ $$T_d = \frac{\ln 2}{k}$$

From part (a), we found $k = \frac{\ln(1.2)}{10}$. Substitute this value:

$$T_d = \frac{\ln 2}{\frac{\ln(1.2)}{10}}$$ $$T_d = \frac{10 \ln 2}{\ln(1.2)}$$

Using a calculator:

$$\ln 2 \approx 0.6931$$ $$\ln 1.2 \approx 0.1823$$ $$T_d \approx \frac{10 \times 0.6931}{0.1823}$$ $$T_d \approx \frac{6.931}{0.1823}$$ $$T_d \approx 38.01 \text{ years}$$

The doubling time for the town's population is approximately $\boxed{\text{38.01 years}}$.

Square both equations and add them:

$$(R \cos \theta)^2 + (R \sin \theta)^2 = 1^2 + (\sqrt{3})^2$$ $$R^2 \cos^2 \theta + R^2 \sin^2 \theta = 1 + 3$$ $$R^2 (\cos^2 \theta + \sin^2 \theta) = 4$$

Since $\cos^2 \theta + \sin^2 \theta = 1$:

$$R^2 (1) = 4$$ $$R = \sqrt{4}$$

Since $R$ represents an amplitude, it must be positive:

$$R = 2$$

Divide Equation 2 by Equation 1:

$$\frac{R \sin \theta}{R \cos \theta} = \frac{\sqrt{3}}{1}$$ $$\tan \theta = \sqrt{3}$$

Since $R \cos \theta = 1 > 0$ and $R \sin \theta = \sqrt{3} > 0$, $\theta$ is in the first quadrant.

$$\theta = \arctan(\sqrt{3})$$ $$\theta = \frac{\pi}{3} \text{ radians or } 60^\circ$$

So, $\sin x + \sqrt{3}\cos x = 2 \sin\left(x + \frac{\pi}{3}\right)$.

The values are $R=\boxed{2}$ and $\theta=\boxed{\frac{\pi}{3} \text{ radians or } 60^\circ}$.