The equation of the normal to the curve y^3 + | vedclass

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Question

(B) Given the curve equation: $y^3 + 2xy + x^3 = (x - 1)^3$. Differentiating both sides with respect to $x$: $3y^2 \frac{dy}{dx} + 2y + 2x \frac{dy}{d

Vedclass · Accepted Answer

$5x - y = 6$

Vedclass · Answer

(B) Given the curve equation: $y^3 + 2xy + x^3 = (x - 1)^3$. Differentiating both sides with respect to $x$: $3y^2 \frac{dy}{dx} + 2y + 2x \frac{dy}{dx} + 3x^2 = 3(x - 1)^2$. At the point $(1, -1)$,substitute $x = 1$ and $y = -1$: $3(-1)^2 \frac{dy}{dx} + 2(-1) + 2(1) \frac{dy}{dx} + 3(1)^2 = 3(1 - 1)^2$. $3 \frac{dy}{dx} - 2 + 2 \frac{dy}{dx} + 3 = 0$. $5 \frac{dy}{dx} + 1 = 0 \implies \frac{dy}{dx} = -\frac{1}{5}$. The slope of the tangent at $(1, -1)$ is $m = -\frac{1}{5}$. The slope of the normal is $m_{\perp} = -\frac{1}{m} = 5$. The equation of the normal at $(1, -1)$ is: $y - (-1) = 5(x - 1)$. $y + 1 = 5x - 5$. $5x - y = 6$.

The equation of the normal to the curve $y^3 + 2xy + x^3 = (x - 1)^3$ at the point $(1, -1)$ is:

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