The slope of the normal to the curve y=2x^{2} | vedclass

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(D) The equation of the given curve is $y=2x^{2}+3\sin x$. First,we find the slope of the tangent to the curve by differentiating $y$ with respect to

Vedclass · Accepted Answer

$-\frac{1}{3}$

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(D) The equation of the given curve is $y=2x^{2}+3\sin x$. First,we find the slope of the tangent to the curve by differentiating $y$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(2x^{2}+3\sin x) = 4x + 3\cos x$. The slope of the tangent at $x=0$ is: $\left(\frac{dy}{dx}ight)_{x=0} = 4(0) + 3\cos(0) = 0 + 3(1) = 3$. The slope of the normal is the negative reciprocal of the slope of the tangent: $	ext{Slope of normal} = -\frac{1}{	ext{Slope of tangent}} = -\frac{1}{3}$. Thus,the correct option is $D$.

The slope of the normal to the curve $y=2x^{2}+3\sin x$ at $x=0$ is

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