The acute angle between the curves y=3x^2-2x- | vedclass

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Question

(A) To find the point of intersection, set the equations equal: $3x^2-2x-1 = x^3-1$. $x^3-3x^2+2x = 0$. $x(x^2-3x+2) = 0$. $x(x-1)(x-2) = 0$. So, $x=0

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$\operatorname{Tan}^{-1}\left(\frac{2}{121}\right)$

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(A) To find the point of intersection, set the equations equal: $3x^2-2x-1 = x^3-1$. $x^3-3x^2+2x = 0$. $x(x^2-3x+2) = 0$. $x(x-1)(x-2) = 0$. So, $x=0, 1, 2$. For $x=0$, $y=-1$ (not in the first quadrant). For $x=1$, $y=0$ (on the axis, not strictly in the first quadrant). For $x=2$, $y=3(2)^2-2(2)-1 = 12-4-1 = 7$. The point of intersection in the first quadrant is $(2, 7)$. Now, find the slopes of the tangents at $(2, 7)$: For $y=3x^2-2x-1$, $dy/dx = 6x-2$. At $x=2$, $m_1 = 6(2)-2 = 10$. For $y=x^3-1$, $dy/dx = 3x^2$. At $x=2$, $m_2 = 3(2)^2 = 12$. The angle $	heta$ between the curves is given by $	an 	heta = |(m_1-m_2)/(1+m_1m_2)|$. $	an 	heta = |(10-12)/(1+10 	imes 12)| = |-2/121| = 2/121$. Thus, $	heta = \operatorname{Tan}^{-1}(2/121)$.

The acute angle between the curves $y=3x^2-2x-1$ and $y=x^3-1$ at their point of intersection which lies in the first quadrant is

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