Find the equations of the tangent and normal | vedclass

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Question

(A) The equation of the curve is $y=x^{3}$.On differentiating with respect to $x$,we get:$\frac{dy}{dx} = 3x^{2}$.The slope of the tangent at $(1,1)$

Vedclass · Accepted Answer

Tangent: $3x-y-2=0$,Normal: $x+3y-4=0$

Vedclass · Answer

(A) The equation of the curve is $y=x^{3}$.On differentiating with respect to $x$,we get:$\frac{dy}{dx} = 3x^{2}$.The slope of the tangent at $(1,1)$ is $\left.\frac{dy}{dx}ight|_{(1,1)} = 3(1)^{2} = 3$.The equation of the tangent at $(1,1)$ is $y-1 = 3(x-1)$,which simplifies to $3x-y-2=0$.The slope of the normal at $(1,1)$ is $-\frac{1}{	ext{slope of tangent}} = -\frac{1}{3}$.The equation of the normal at $(1,1)$ is $y-1 = -\frac{1}{3}(x-1)$,which simplifies to $3y-3 = -x+1$,or $x+3y-4=0$.

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

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