applications of derivatives

Question

The acute angle between the curves $y = |x^2 - 1|$ and $y = 1$ at their points of intersection in the first quadrant is:

RankGuru · Accepted Answer

1. Find intersection points in the first quadrant ($x > 0, y > 0$):
$|x^2 - 1| = 1 \implies x^2 - 1 = 1$ or $x^2 - 1 = -1$.
$x^2 = 2 \implies x = \sqrt{2}$. (Note: $x^2 = 0$ gives intersection at $x=0$, but 'first quadrant' implies $x > 0$, so we exclude $x=0$. The actual intersections are $(0,1)$ and $(\sqrt{2},1)$).

2. Find slopes at $x = \sqrt{2}$:
Curve 1: $y = x^2 - 1$ (since $x^2 > 1$ near $\sqrt{2}$).
$dy/dx = 2x$. At $x=\sqrt{2}, m_1 = 2\sqrt{2}$.
Curve 2: $y = 1$. Constant function.
$dy/dx = 0$. So $m_2 = 0$.

3. Calculate angle:
$	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}| = |\frac{2\sqrt{2} - 0}{1 + 0}| = 2\sqrt{2}$.
Thus, $	heta = 	an^{-1}(2\sqrt{2})$.

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