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Question

Find the slope normal to the curve x2/3+y2/3=2 at the point (1, 1).

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Solution

On differentiating the equation wrt x, we have
23x13+23y13dydx=0
dydx=(xy)13
This is the slope of the tangent at any point on given curve. The slope of normal =dxdy=(xy)13
Hence, slope of normal at (1,1) is 1.

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