Question

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

Hint:

When finding the equation of the normal, remember that the slope of the normal is the negative reciprocal of the slope of the tangent.

Answer 1

Step 1: Differentiate the equation implicitly.
The given equation is: \[ x^{2/3} + y^{2/3} = 2 \] Differentiate both sides with respect to \( x \): \[ \frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0 \] Simplifying: \[ \frac{2}{3} x^{-1/3} = -\frac{2}{3} y^{-1/3} \frac{dy}{dx} \] Solving for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{y^{1/3}}{x^{1/3}} \]
Step 2: Find the slope of the tangent at the point \( (1, 1) \).
Substitute \( x = 1 \) and \( y = 1 \): \[ \frac{dy}{dx} = \frac{1^{1/3}}{1^{1/3}} = 1 \] So, the slope of the tangent at the point \( (1, 1) \) is \( 1 \).
Step 3: Find the slope of the normal.
The slope of the normal is the negative reciprocal of the slope of the tangent: \[ \text{slope of normal} = -\frac{1}{1} = -1 \]
Step 4: Use the point-slope form to find the equation of the normal.
Using the point-slope form \( y - y_1 = m(x - x_1) \), where \( m = -1 \) and the point \( (1, 1) \), we get: \[ y - 1 = -1(x - 1) \] Simplifying: \[ y - 1 = -x + 1 \] \[ y = -x + 2 \] Thus, the equation of the normal at the point \( (1, 1) \) is: \[ y = -x + 2 \]