To find how fast $R$ is changing with respect to $T$, we need to compute the derivative of $R$ with respect to $T$. This involves applying the chain rule.

We are given two relationships:

1. $R(x) = \frac{0.0048}{x^2}$
2. $x(T) = 0.1991 + 0.000003T$

To find $\frac{dR}{dT}$, we use the chain rule: $\frac{dR}{dT} = \frac{dR}{dx} \cdot \frac{dx}{dT}$

Step 1: Find $\frac{dR}{dx}$

The formula for $R$ is $R(x) = \frac{0.0048}{x^2}$. Differentiate this with respect to $x$: $\frac{dR}{dx} = -\frac{2 \times 0.0048}{x^3} = -\frac{0.0096}{x^3}$

Step 2: Find $\frac{dx}{dT}$

We are given that $x(T) = 0.1991 + 0.000003T$, so the derivative of $x$ with respect to $T$ is: $\frac{dx}{dT} = 0.000003$

Step 3: Apply the chain rule

Now, apply the chain rule: $\frac{dR}{dT} = \frac{dR}{dx} \cdot \frac{dx}{dT} = -\frac{0.0096}{x^3} \cdot 0.000003$

Step 4: Evaluate at $T = 320$

First, find $x$ at $T = 320$: $x(320) = 0.1991 + 0.000003 \times 320 = 0.1991 + 0.00096 = 0.20006$

Now substitute $x = 0.20006$ into the expression for $\frac{dR}{dT}$: $\frac{dR}{dT} = -\frac{0.0096}{(0.20006)^3} \times 0.000003$

Step 5: Simplify

First, calculate $(0.20006)^3$: $(0.20006)^3 \approx 0.0080024$

Now substitute this value into the expression: [ \frac{dR}{dT} = -\frac{0.0096