Question

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

• A. $0$ (zero)
• B. $\pi$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is A $0$ (zero)

At the point of intersection,
$e^{x^2}=e^{x^2}\sin x$
$\Rightarrow \sin x=1$

$y=e^{x^2}$
The slope of the above curve is,
$\frac{dy}{dx}=2x{e}^{x^2}$
Again, $y=e^{x^2}\sin x$
Differentiate the above equation,
$\frac{dy}{dx}=2x{e}^{x^2}\sin x+e^{x^2}\cos x$
When $\sin x=1$, $\cos x=0$
$\Rightarrow \frac{dy}{dx}=2x{e}^{x^2}$

Thus, slopes of tangents are equal at the point of intersection.
Therefore, the angle between them is zero.