Question

Slope of the normal to the curve \(x^{2/3} + y^{2/3} = 2\) at the point (1, 1) is

• A. -1
• B. 1
• C. 1/2
• D. -1/2

Hint:

Be careful not to confuse the slope of the tangent with the slope of the normal. After finding \(\frac{dy}{dx}\), always remember to take the negative reciprocal if the question asks for the normal's slope. A common mistake is to stop after finding the tangent's slope.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the slope of the normal line to the given curve (an astroid) at the specified point (1, 1).

Step 2: Key Formula or Approach:
1. Find the derivative \(\frac{dy}{dx}\) of the curve's equation using implicit differentiation. This will give the slope of the tangent (\(m_T\)).
2. Evaluate \(\frac{dy}{dx}\) at the given point (1, 1) to find the specific slope of the tangent at that point.
3. The slope of the normal (\(m_N\)) is the negative reciprocal of the slope of the tangent: \(m_N = -\frac{1}{m_T}\).

Step 3: Detailed Explanation:
The equation of the curve is \(x^{2/3} + y^{2/3} = 2\).
Differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(x^{2/3}) + \frac{d}{dx}(y^{2/3}) = \frac{d}{dx}(2) \]
\[ \frac{2}{3}x^{(2/3 - 1)} + \frac{2}{3}y^{(2/3 - 1)} \cdot \frac{dy}{dx} = 0 \]
\[ \frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0 \]
Divide the entire equation by \(\frac{2}{3}\):
\[ x^{-1/3} + y^{-1/3} \frac{dy}{dx} = 0 \]
Solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\left(\frac{y}{x}\right)^{1/3} \]
This is the slope of the tangent, \(m_T\). Now, evaluate \(m_T\) at the point (1, 1):
\[ m_T = -\left(\frac{1}{1}\right)^{1/3} = -1 \]
The slope of the normal, \(m_N\), is the negative reciprocal of \(m_T\).
\[ m_N = -\frac{1}{m_T} = -\frac{1}{-1} = 1 \]

Step 4: Final Answer:
The slope of the normal to the curve at (1, 1) is 1.