Step 1: Understanding the Concept:
This problem requires finding the slope of the normal to a curve at a given point. This involves two main steps: first, find the slope of the tangent to the curve at that point by implicit differentiation. Second, find the slope of the normal, which is the negative reciprocal of the tangent's slope.
Step 2: Key Formula or Approach:
1. Differentiate the curve's equation implicitly to find an expression for $\frac{dy}{dx}$.
2. Evaluate $\frac{dy}{dx}$ at the given point $(1,1)$ to find the slope of the tangent, $m_{\text{tangent}}$.
3. The slope of the normal is $m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}$.
Step 3: Detailed Explanation:
The equation of the curve is $x^{2/3} + y^{2/3} = 2$.
Differentiate both sides with respect to x:
\[ \frac{d}{dx}(x^{2/3}) + \frac{d}{dx}(y^{2/3}) = \frac{d}{dx}(2) \]
\[ \frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0 \]
Divide the entire equation by $\frac{2}{3}$:
\[ x^{-1/3} + y^{-1/3}\frac{dy}{dx} = 0 \]
Solve for $\frac{dy}{dx}$:
\[ y^{-1/3}\frac{dy}{dx} = -x^{-1/3} \]
\[ \frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\left(\frac{y}{x}\right)^{1/3} \]
Now, find the slope of the tangent at the point $(1, 1)$ by substituting $x=1$ and $y=1$:
\[ m_{\text{tangent}} = -\left(\frac{1}{1}\right)^{1/3} = -1 \]
The slope of the normal is the negative reciprocal of the slope of the tangent:
\[ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{-1} = 1 \]
Step 4: Final Answer:
The slope of the normal to the curve at the point (1, 1) is 1. Therefore, option (B) is correct.