Find the acute angle between the curves at their points of intersection, y = x^{2}, y = x^{3}.

#### Solution

The angle between the curves is the same as the angle between their tangents at the points of intersection.

We find the points of intersection of y = x^{2} ....(1) and y = x^{3} .....(2)

From (1) and (2)

x^{3} = x^{2}

∴ x^{3} - x^{2} = 0

∴ x^{2}(x - 1) = 0

∴ x = 0 or x = 1

When x = 0, y = 0.

When x = 1, y = 1.

∴ the points of intersection are

O = (0, 0) and P = (1, 1)

For y = x^{2}, `"dy"/"dx" = 2"x"`

For y = x^{3}, `"dy"/"dx" = 3"x"^2`

**Angle at O = (0, 0)**

Slope of tangent to y = x^{2} at O

`= ("dy"/"dx")_("at O" (0,0)) = 2xx0 = 0`

∴ equation of tangent to y = x^{2} at O is y = 0.

Slope of tangent to y = x^{3} at O = `("dy"/"dx")_("at O" (0,0)) = 3 xx 0 = 0`

∴ equation of tangent to y = x^{3} at P is y = 0.

∴ the tangents to both curves at (0, 0) are y = 0

∴ angle between them is 0.

**Angle at P = (1, 1)**

Slope of tangent to y = x^{2} at P

`= ("dy"/"dx")_("at O" (1,1)) = 2xx1 = 2`

∴ equation of tangent to y = x^{2} at P is y - 1 = 2(x - 1)

∴ y = 2x - 1

Slope of tangent to y = x^{2} at P =`("dy"/"dx")_("at O" (1,1)) = 3xx1^2 = 3`

∴ equation of tangent to y = x^{3} at P is y - 1 = 3(x - 1)

∴ y = 3x - 2

We have to find angle between y = 2x - 1 and y = 3x - 2

Lines through origin parallel to these tagents are

y = 2x and y = 3x

∴ `"x"/1 = "y"/2 and "x"/1 = "y"/3`

These lines lie in XY-plane.

∴ the direction ratios of these lines are 1, 2, 0 and 1, 3, 0.

The angle θ between them is given by

cos θ = `((1)(1) + (2)(3) + (0)(0))/(sqrt(1^2 + 2^2 + 0^2)sqrt(1^2 + 3^2 + 0^2))`

`= (1 + 6 + 0)/(sqrt5 sqrt10)`

`= 7/sqrt50 = 7/(5sqrt2)`

∴ θ = `cos^-1(7/(5sqrt2))`

Hence, the required angles are 0 and `cos^-1(7/(5sqrt2))`