Question

Find the equation of the normal to the curve y = 3x² + 2sinx + 4cos x + 10 at x = 0.

Answer 1

when x = 0, y = 14

Hence, the point is (0, 14)

Now, dy/dx = 6x + 2cos x – 4sinx

Thus, m = ( dy/dx)_x=0 = 2

Hence, the equation of the normal is

y – 14 = – 1/2(x – 0)

⇒ 2y – 28 = –x

⇒ x + 2y = 28.