Find the equation of the normal to the curve | vedclass

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(C) Given the curve $y = x^3$. First,find the derivative to determine the slope of the tangent: $\frac{dy}{dx} = 3x^2$. At the point $P(1, 1)$,the slo

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$x + 3y - 4 = 0$

Vedclass · Answer

(C) Given the curve $y = x^3$. First,find the derivative to determine the slope of the tangent: $\frac{dy}{dx} = 3x^2$. At the point $P(1, 1)$,the slope of the tangent is $m_t = 3(1)^2 = 3$. The slope of the normal $m_n$ is the negative reciprocal of the tangent slope: $m_n = -\frac{1}{m_t} = -\frac{1}{3}$. Using the point-slope form of the equation of a line,$y - y_1 = m_n(x - x_1)$,we get: $y - 1 = -\frac{1}{3}(x - 1)$. Multiplying both sides by $3$: $3(y - 1) = -(x - 1)$. $3y - 3 = -x + 1$. Rearranging the terms gives the equation of the normal: $x + 3y - 4 = 0$.

Find the equation of the normal to the curve $y = x^3$ at the point $P(1, 1)$.

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