Let y=e^{x^{2}} and y=e^{x^{2}} sin x be two | vedclass

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(A) For the intersection points, we set the two equations equal: $e^{x^{2}} = e^{x^{2}} \sin x$. Since $e^{x^{2}} 
eq 0$ for any real $x$, we divide

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$0$

Vedclass · Answer

(A) For the intersection points, we set the two equations equal: $e^{x^{2}} = e^{x^{2}} \sin x$. Since $e^{x^{2}} 
eq 0$ for any real $x$, we divide by $e^{x^{2}}$ to get $\sin x = 1$. This implies $x = \frac{\pi}{2} + 2n\pi$ for any integer $n$. Let $f(x) = e^{x^{2}}$ and $g(x) = e^{x^{2}} \sin x$. The derivative of $f(x)$ is $f'(x) = 2x e^{x^{2}}$. The derivative of $g(x)$ is $g'(x) = 2x e^{x^{2}} \sin x + e^{x^{2}} \cos x$. At the intersection point where $\sin x = 1$ and $\cos x = 0$, we have: $f'(x) = 2x e^{x^{2}}$ $g'(x) = 2x e^{x^{2}}(1) + e^{x^{2}}(0) = 2x e^{x^{2}}$. Since $f'(x) = g'(x)$ at the point of intersection, the slopes of the tangents are equal. Therefore, the angle between the tangents is $0$.

Let $y=e^{x^{2}}$ and $y=e^{x^{2}} \sin x$ be two given curves. Then, the angle between the tangents to the curves at any point of their intersection is:

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