Find the equations of the tangent and normal | vedclass

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Question

(A) Given the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=2$. Differentiating with respect to $x$,we get: $\frac{2}{3}x^{-\frac{1}{3}} + \frac{2}{3}y^{-\fr

Vedclass · Accepted Answer

Tangent: $x+y-2=0$,Normal: $y-x=0$

Vedclass · Answer

(A) Given the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=2$. Differentiating with respect to $x$,we get: $\frac{2}{3}x^{-\frac{1}{3}} + \frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx} = 0$. This simplifies to $\frac{dy}{dx} = -\left(\frac{y}{x}\right)^{\frac{1}{3}}$. At the point $(1,1)$,the slope of the tangent $m_t = -\left(\frac{1}{1}\right)^{\frac{1}{3}} = -1$. The equation of the tangent is $y - 1 = -1(x - 1)$,which simplifies to $x + y - 2 = 0$. The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$. The equation of the normal is $y - 1 = 1(x - 1)$,which simplifies to $y - x = 0$.

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

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