Let's go through both problems step by step.

1st Problem (Resistors in Parallel):

The formula for total resistance $R$ in parallel is given as:

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$

We need to determine how fast $R$ is changing, i.e., find $\frac{dR}{dt}$, given that:

• $R_1 = 60 \, \Omega$, $R_2 = 70 \, \Omega$,
• $\frac{dR_1}{dt} = 0.3 \, \Omega/s$, and $\frac{dR_2}{dt} = 0.2 \, \Omega/s$.

1. Differentiate the equation $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$ implicitly with respect to time $t$:

$-\frac{1}{R^2} \frac{dR}{dt} = -\frac{1}{R_1^2} \frac{dR_1}{dt} - \frac{1}{R_2^2} \frac{dR_2}{dt}$

2. Solve for $\frac{dR}{dt}$:

$\frac{dR}{dt} = R^2 \left( \frac{1}{R_1^2} \frac{dR_1}{dt} + \frac{1}{R_2^2} \frac{dR_2}{dt} \right)$

Now, calculate $R$ first using:

$\frac{1}{R} = \frac{1}{60} + \frac{1}{70}$

This gives:

$R \approx 32.308 \, \Omega$

Next, plug in the values to find $\frac{dR}{dt}$:

$\frac{dR}{dt} = (32.308)^2 \left( \frac{0.3}{60^2} + \frac{0.2}{70^2} \right)$

Let's compute this.

2nd Problem (Triangle with Law of Cosines):

The Law of Cosines states:

$c^2 = a^2 + b^2 - 2ab \cos(\theta)$

Where:

• $a = 9 \, m$,
• $b = 18 \, m$,
• $\theta = 60^\circ$, and $\frac{d\theta}{dt} = 2^\circ/\text{min}$.

We want to find how fast the length of the third side $c$ is changing, i.e., $\frac{dc}{dt}$.

1. Differentiate the Law of Cosines with respect to time:

$2c \frac{dc}{dt} = 2ab \sin(\theta) \frac{d\theta}{dt}$

Solve for $\frac{dc}{dt}$:

$\frac{dc}{dt} = \frac{ab \sin(\theta) \frac{d\theta}{dt}}{c}$

You can find $c$ using the Law of Cosines first with $\theta = 60^\circ$:

$c^2 = 9^2 + 18^2 - 2 \cdot 9 \cdot 18 \cdot \cos(60^\circ)$

Then substitute the values into the formula to compute $\frac{dc}{dt}$.

Would you like further assistance with the detailed calculations for both?