The area (in sq units) of the triangle formed | vedclass

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(C) Given the curve $x=e^{\sin y}$. Taking the natural logarithm on both sides,we get $\log x = \sin y$.Differentiating both sides with respect to $x$

Vedclass · Accepted Answer

$\frac{1}{2}$

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(C) Given the curve $x=e^{\sin y}$. Taking the natural logarithm on both sides,we get $\log x = \sin y$.Differentiating both sides with respect to $x$,we have $\frac{1}{x} = \cos y \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{1}{x \cos y}$.The slope of the tangent at $(1,0)$ is $\left(\frac{dy}{dx}ight)_{(1,0)} = \frac{1}{1 \cdot \cos 0} = \frac{1}{1 \cdot 1} = 1$.The slope of the normal at $(1,0)$ is $m = -\frac{1}{	ext{slope of tangent}} = -\frac{1}{1} = -1$.The equation of the normal at $(1,0)$ is $y - 0 = -1(x - 1)$,which simplifies to $y = -x + 1$ or $x + y = 1$.This line intersects the $x$-axis at $A(1,0)$ and the $y$-axis at $B(0,1)$.The triangle formed by the normal with the coordinate axes is $	riangle OAB$,where $O$ is the origin $(0,0)$.The area of $	riangle OAB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes OA 	imes OB = \frac{1}{2} 	imes 1 	imes 1 = \frac{1}{2} 	ext{ sq units}$.

The area (in sq units) of the triangle formed by the normal drawn at the point $(1,0)$ on the curve $x=e^{\sin y}$ with the coordinate axes is

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