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. \(\frac{1}{2}\)
• D. \(-\frac{1}{2}\)

Hint:

The slope of normal is the negative reciprocal of the slope of tangent. If tangent slope is \(-1\), then normal slope is \(1\).

Answer 1

The Correct Option is B

We are given the curve: \[ x^{2/3}+y^{2/3}=2. \] We need the slope of the normal at the point: \[ (1,1). \] First differentiate the equation implicitly with respect to \(x\). Differentiate \(x^{2/3}\): \[ \frac{d}{dx}(x^{2/3})=\frac{2}{3}x^{-1/3}. \] Differentiate \(y^{2/3}\): \[ \frac{d}{dx}(y^{2/3})=\frac{2}{3}y^{-1/3}\frac{dy}{dx}. \] So, \[ \frac{2}{3}x^{-1/3}+\frac{2}{3}y^{-1/3}\frac{dy}{dx}=0. \] Multiply by \(\frac{3}{2}\): \[ x^{-1/3}+y^{-1/3}\frac{dy}{dx}=0. \] Thus, \[ y^{-1/3}\frac{dy}{dx}=-x^{-1/3}. \] \[ \frac{dy}{dx}=-\frac{x^{-1/3}}{y^{-1/3}}. \] At \((1,1)\): \[ \frac{dy}{dx}=-\frac{1^{-1/3}}{1^{-1/3}}. \] \[ \frac{dy}{dx}=-1. \] So the slope of the tangent is \[ -1. \] The slope of normal is the negative reciprocal of slope of tangent: \[ m_n=-\frac{1}{m_t}. \] \[ m_n=-\frac{1}{-1}=1. \] Hence, the slope of the normal is \[ 1. \]