Question

Find the equations of the tangent and normal to the curve \( x^{2/3} + y^{2/3} = 2 \) at the point (1, 1).

Hint:

Implicit differentiation is a powerful tool for finding derivatives when y is not explicitly defined as a function of x. Remember to apply the chain rule whenever you differentiate a term involving y.

Answer 1

Step 1: Understanding the Concept:
To find the equation of a tangent line to a curve at a given point, we first need to find the slope of the tangent, which is the value of the derivative \( \frac{dy}{dx} \) at that point. The slope of the normal line is the negative reciprocal of the tangent's slope.
Step 2: Key Formula or Approach:
1. Differentiate the curve's equation implicitly with respect to x to find \( \frac{dy}{dx} \).
2. Evaluate \( \frac{dy}{dx} \) at the point (1, 1) to get the slope of the tangent, \( m_t \).
3. Calculate the slope of the normal: \( m_n = -1/m_t \).
4. Use the point-slope form of a line, \( y - y_1 = m(x - x_1) \), to find the equations.
Step 3: Detailed Explanation or Calculation:
The equation of the curve is \( x^{2/3} + y^{2/3} = 2 \).
Differentiating 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 by \( \frac{2}{3} \):
\[ x^{-1/3} + y^{-1/3} \frac{dy}{dx} = 0 \] Solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -\left(\frac{y}{x}\right)^{1/3} \] Now, find the slope of the tangent at the point (1, 1):
\[ m_t = \frac{dy}{dx}\bigg|_{(1,1)} = -\left(\frac{1}{1}\right)^{1/3} = -1 \] The slope of the normal is:
\[ m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1 \] Equation of the Tangent:
Using the point-slope form with point (1, 1) and slope \( m_t = -1 \):
\[ y - 1 = -1(x - 1) → y - 1 = -x + 1 → x + y = 2 \] Equation of the Normal:
Using the point-slope form with point (1, 1) and slope \( m_n = 1 \):
\[ y - 1 = 1(x - 1) → y - 1 = x - 1 → y = x \text{ or } x - y = 0 \] Step 4: Final Answer:
The equation of the tangent is \( x + y = 2 \).
The equation of the normal is \( y = x \).