Find the equation of the normal to the curve y = 2x2+3sinx at x = 0.

Solution :

The equation of the given curve is

y = 2x2+3sinx                   ……….(i)

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

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

Now, y = 2x2+3sinx

Differentiating with respect to x,

  dydx  = 4x + 3 cos x

  (dydx)(0,0) = 4 × 0 + 3 cos 0 = 3

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

y – 0 = -13(x – 0)  or,  x + 3y = 0


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