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

So, the equation of the normal at (0, 0) is

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


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