# Find the acute angle between the curves at their points of intersection, y = x2, y = x3. - Mathematics and Statistics

Sum

Find the acute angle between the curves at their points of intersection, y = x2, y = x3.

#### 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 = x2   ....(1) and  y = x3         .....(2)

From (1) and (2)

x3 = x2

∴ x3 - x2 = 0

∴ x2(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 = x2, "dy"/"dx" = 2"x"

For y = x3, "dy"/"dx" = 3"x"^2

Angle at O = (0, 0)

Slope of tangent to y = x2 at O

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

∴ equation of tangent to y = x2 at O is y = 0.

Slope of tangent to y = x3 at O = ("dy"/"dx")_("at O" (0,0)) = 3 xx 0 = 0

∴ equation of tangent to y = x3 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 = x2 at P

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

∴ equation of tangent to y = x2 at P is y - 1 = 2(x - 1)

∴ y = 2x - 1

Slope of tangent to y = x2 at P =("dy"/"dx")_("at O" (1,1)) = 3xx1^2 = 3

∴ equation of tangent to y = x3 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))

Concept: Vectors and Their Types
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