Find the point on the curve y = x3 where the Slope of the tangent is e

1.	Find the point on the curve y = x3 where the Slope of the tangent is equal to x – coordinate of the point.
Answer» Given: The curve is y = x³ y = x³ Differentiating the above w.r.t x ⇒ \(\frac{dy}{dx}\) = 3x^{2 – 1} ⇒ \(\frac{dy}{dx}\) = 3x² ...(1) Also given the The Slope of the tangent is equal to the x – coordinate, \(\frac{dy}{dx}\) = x ...(2) From (1) & (2),we get, i.e, 3x² = x ⇒ x(3x – 1) = 0 ⇒ x = 0 or x = \(\frac{1}{3}\) Substituting x = 0 or x = \(\frac{1}{3}\)this in y = x³,we get, when x = 0 ⇒ y = 0³ ⇒ y = 0 when x = \(\frac{1}{3}\) ⇒ y = ( \(\frac{1}{3}\) )³ ⇒ y = \(\frac{1}{27}\) Thus, the required point is (0,0) & ( \(\frac{1}{3}\), \(\frac{1}{27}\))

Find the point on the curve y = x3 where the Slope of the tangent is equal to x – coordinate of the point.

Answer»

Given:

The curve is y = x³

y = x³

Differentiating the above w.r.t x

⇒ \(\frac{dy}{dx}\) = 3x^{2 – 1}

⇒ \(\frac{dy}{dx}\) = 3x² ...(1)

Also given the The Slope of the tangent is equal to the x – coordinate,

\(\frac{dy}{dx}\) = x ...(2)

From (1) & (2),we get,

i.e, 3x² = x

⇒ x(3x – 1) = 0

⇒ x = 0 or x = \(\frac{1}{3}\)

Substituting x = 0 or x = \(\frac{1}{3}\)this in y = x³,we get,

when x = 0

⇒ y = 0³

⇒ y = 0

when x = \(\frac{1}{3}\)

⇒ y = ( \(\frac{1}{3}\) )³

⇒ y = \(\frac{1}{27}\)

Thus, the required point is (0,0) & ( \(\frac{1}{3}\), \(\frac{1}{27}\))