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Find the equation of the normal to the curve `y = 2 sin^(2) 3x "at " x = (pi)/(6)` |
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Answer» When `x = (pi)/(6)`, we have `y = 2 sin^(2) ((3pi)/(6)) = 2` So, the point of contact is `((pi)/(6), 2)` Now, `y = 2 sin^(2) 3x rArr (dy)/(dx) = (4 sin 3x) xx 3 xx (cos 3x)` `rArr (dy)/(dx) = 12 sin 3 x cos 3 x = 6 sin 6x` `:. ((dy)/(dx)) " at" (x = (pi)/(6)) = 6 sin (6 xx (pi)/(6)) = 6 sin pi = 0` So, the tangent is parallel to the x-axis. Thus, the normal is parallel to the y-axis and passes through the point `((pi)/(6), 2)` So, the equation of the normal is `x = (pi)/(6)` |
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