Find the equation of the normal to the curve defined by 5 + x² * sqrt(y - 2) = y² - x * [tan(3x)/x as x-0] - 5x * [sin(x)/x as x-0] at the point (1, 3). (Here [.] denotes the greatest integer function.)

Question

Find the equation of the normal to the curve defined by 5 + x² * sqrt(y - 2) = y² - x * [tan(3x)/x as x->0] - 5x * [sin(x)/x as x->0] at the point (1, 3). (Here [.] denotes the greatest integer function.)

Accepted Answer

11x + 10y = 41. As x->0, tan(3x)/x -> 3 so [tan(3x)/x] = 3; sin(x)/x -> 1⁻ so [sin(x)/x] = 0. The curve simplifies to 5 + x²*sqrt(y-2) = y² - 3x. Differentiating implicitly and evaluating at (1,3) gives the slope of the tangent, then the normal is perpendicular to it.

Find the equation of the normal to the curve defined by 5 + x² * sqrt(y - 2) = y² - x * [tan(3x)/x as x->0] - 5x * [sin(x)/x as x->0] at the point (1, 3). (Here [.] denotes the greatest integer function.)

Solution

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