The angle of intersection between the curves | vedclass

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(C) To find the intersection with the $x$-axis,we set $y = 0$. The integral $\int_{x^2}^{x^3} \sqrt{5 - t^2} \, dt = 0$ implies $x^2 = x^3$,so $x = 1$

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$\cot^{-1} \frac{1}{2}$

Vedclass · Answer

(C) To find the intersection with the $x$-axis,we set $y = 0$. The integral $\int_{x^2}^{x^3} \sqrt{5 - t^2} \, dt = 0$ implies $x^2 = x^3$,so $x = 1$ (since $x 
eq 0$).Using the Leibniz Rule for differentiation,the slope of the tangent is given by $y' = \frac{d}{dx} \int_{x^2}^{x^3} \sqrt{5 - t^2} \, dt = \sqrt{5 - (x^3)^2} \cdot (3x^2) - \sqrt{5 - (x^2)^2} \cdot (2x)$.At $x = 1$,the slope $m$ is $y'(1) = \sqrt{5 - 1} \cdot 3(1)^2 - \sqrt{5 - 1} \cdot 2(1) = 2 \cdot 3 - 2 \cdot 2 = 6 - 4 = 2$.The angle $	heta$ between the curve and the $x$-axis is given by $	an 	heta = |m| = 2$.Thus,$	heta = 	an^{-1} 2 = \cot^{-1} \frac{1}{2}$.

The angle of intersection between the curves $y = \int\limits_{x^2}^{x^3} \sqrt{5 - t^2} \, dt$ and the $x$-axis is (where $x \neq 0$):

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