Find the equation of the normal to the curve y = \(2x^2 + 3 sin x\) at x = 0.

Solution :

The equation of the given curve is

y = \(2x^2 + 3 sin x\) ……….(i)

Putting x = 0 in (i), we get y = 0.

So, the point of contact is (0, 0).

Now, y = \(2x^2 + 3 sin x\)

Differentiating with respect to x,

\(\implies\) \(dy\over dx\) = 4x + 3 cos x

\(\implies\) \(({dy\over dx})_{(0, 0)}\) = 4 \(\times\) 0 + 3 cos 0 = 3

y – 0 = -\(1\over 3\)(x – 0) or, x + 3y = 0