When two curves intersect each other the angle at the intersecting point is called as angle of intersection between two curves.

Find the acute angle between the two curves y=2x^{2} and y=x^{2}-4x+4

Here the 2 curves are represented in the equation format as shown below
y=2x^{2} --> (1)
y=x^{2}-4x+4 --> (2)
Let us learn how to find angle of intersection between these curves using this equation.

Solving equ 1 and equ 2
2x^{2} = y
x^{2}- 4x + 4 = y
x^{2}+ 4x - 4 = 0
By factorizing the quadratic equation x^{2}+ 4x - 4 = 0
we get the x values as x = 0.8 and x = -4.8
From the x values the maximum value (0.8) is substituted in equation 1 to find y value

y = 2x(0.8)2 y = 1.3 From this values we get (0.8,1.3), which is an intersect point of curve.

Differentiate equ.1 and equ.2
Differentiation of equ 1 y=2x^{2}
dy/dx = 4x --> (3)

dy/dx(x^{2}) = 2x
Differentiation of equ 2 y=x^{2}-4x+4
dy/dx = 2x - 4 --> (4)

dy/dx(x) = 1 and dy/dx(constant) = 0

Find the slope by substituting intersect point (0.8,1.3) in equ.3 and equ.4, Equ. 3 4x = 4(0.8) = 3.2 = m1 Equ. 4 2x - 4 = 2(0.8) - 4 = -2.4 = m2

Find the Angle by substituting slope values in Formula
tan(θ) = (m1-m2)/(1+(m1.m2)) ∀ m1>m2
From formula θ = tan^{-1}[(m1-m2)/(1+(m1.m2))]
θ = tan^{-1}((3.2+2.4)/(1+(3.2*-2.4))
θ = tan^{-1}(5.6/-6.68)
θ = tan^{-1}(0.8383)
θ = 39.974 °
Therefore, the angle of intersection between the given curve is θ = 39.974 °

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