Slope of the normal to the curve y = x^{2} - | vedclass

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(A) Given the equation of the curve: $y = x^{2} - \frac{1}{x^{2}}$.First,we find the derivative of the curve with respect to $x$ to find the slope of

Vedclass · Accepted Answer

$1/4$

Vedclass · Answer

(A) Given the equation of the curve: $y = x^{2} - \frac{1}{x^{2}}$.First,we find the derivative of the curve with respect to $x$ to find the slope of the tangent $(m_{1})$:$y = x^{2} - x^{-2}$$\frac{dy}{dx} = 2x - (-2)x^{-3} = 2x + \frac{2}{x^{3}}$.Now,evaluate the slope of the tangent at the point $(-1, 0)$:$m_{1} = \left. \frac{dy}{dx} ight|_{(-1, 0)} = 2(-1) + \frac{2}{(-1)^{3}} = -2 + \frac{2}{-1} = -2 - 2 = -4$.The slope of the normal $(m_{2})$ is the negative reciprocal of the slope of the tangent:$m_{1} 	imes m_{2} = -1$$-4 	imes m_{2} = -1$$m_{2} = \frac{-1}{-4} = \frac{1}{4}$.Therefore,the slope of the normal to the curve at $(-1, 0)$ is $1/4$.

Slope of the normal to the curve $y = x^{2} - \frac{1}{x^{2}}$ at the point $(-1, 0)$ is:

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