Find the points on the curve y=x^{3} at which | vedclass

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

Vedclass · Accepted Answer

$(0,0)$ and $(3,27)$

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(A) The equation of the given curve is $y=x^{3}$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = 3x^{2}$.The slope of the tangent at any point $(x, y)$ on the curve is given by $\frac{dy}{dx} = 3x^{2}$.According to the problem,the slope of the tangent is equal to the $y$-coordinate of the point,so we have $3x^{2} = y$.Since the point $(x, y)$ lies on the curve,we have $y = x^{3}$.Equating the two expressions for $y$,we get $x^{3} = 3x^{2}$.$x^{3} - 3x^{2} = 0$.$x^{2}(x - 3) = 0$.This gives $x = 0$ or $x = 3$.If $x = 0$,then $y = (0)^{3} = 0$. So,the point is $(0, 0)$.If $x = 3$,then $y = (3)^{3} = 27$. So,the point is $(3, 27)$.Thus,the required points are $(0, 0)$ and $(3, 27)$.

Find the points on the curve $y=x^{3}$ at which the slope of the tangent is equal to the $y$-coordinate of the point.

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