Find the equation of the normal to the curve y = 2\sin x - \cos x at the point where x = 0… — JAMB UTME

Find the equation of the normal to the curve $y = 2\sin x - \cos x$ at the point where $x = 0$ .

x + 2y + 2 = 0

x = 0

, the

y

-coordinate is

2\sin(0) - \cos(0) = -1

. Evaluating the derivative

\frac{dy}{dx} = 2\cos x + \sin x

x = 0

yields a tangent slope of

2

. Since the normal is perpendicular to the tangent, its gradient is

-\frac{1}{2}

. Substituting the point

(0, -1)

and slope

-\frac{1}{2}

into the line formula yields

x + 2y + 2 = 0

Find the

y

-coordinate of the point of contact

x = 0

y = 2\sin(0) - \cos(0) = 0 - 1 = -1

. The point is

(0, -1)

The line equation requires a point

(x_1, y_1)

on the curve.

Differentiate the curve to find

\frac{dy}{dx}

\frac{dy}{dx} = 2\cos x - (-\sin x) = 2\cos x + \sin x

The derivative gives the gradient function of the curve.

Calculate the gradient of the tangent and normal at

x = 0

Tangent gradient

m_t = 2\cos(0) + \sin(0) = 2(1) + 0 = 2

. Normal gradient

m_n = -\frac{1}{m_t} = -\frac{1}{2}

The normal line is perpendicular to the tangent line.

Form the equation of the normal line

y - (-1) = -\frac{1}{2}(x - 0) \implies y + 1 = -\frac{1}{2}x \implies 2y + 2 = -x \implies x + 2y + 2 = 0

Apply the point-slope form

y - y_1 = m(x - x_1)

Equation of Normal to a Curve

Tahmini Süre:1m 30s

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