Calculate the gradient of the normal line to the curve y = \frac{6}{x} at the point where … — JAMB UTME

Calculate the gradient of the normal line to the curve $y = \frac{6}{x}$ at the point where $x = 3$ .

Answer: 1.5

The gradient of the normal line is 1.5.

Differentiating

y = 6x^{-1}

yields

\frac{dy}{dx} = -\frac{6}{x^2}

. Evaluating this derivative at

x = 3

gives the tangent gradient

m_t = -\frac{6}{9} = -\frac{2}{3}

. Because the normal line is perpendicular to the tangent line, its gradient is the negative reciprocal

m_n = -\frac{1}{m_t} = \frac{3}{2} = 1.5

Differentiate the function

y = 6x^{-1}

with respect to

x

\frac{dy}{dx} = -6x^{-2} = -\frac{6}{x^2}

The first derivative represents the formula for the tangent gradient to the curve at any given point.

Evaluate the derivative at

x = 3

to find the tangent slope (

m_t

)

m_t = -\frac{6}{3^2} = -\frac{6}{9} = -\frac{2}{3}

Substituting the given x-coordinate into the derivative gives the exact slope of the tangent at that point.

Calculate the normal slope (

m_n

) as the negative reciprocal of

m_t

m_n = -\frac{1}{m_t} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2} = 1.5

The normal line is perpendicular to the tangent line, meaning

m_t \cdot m_n = -1

The gradient of the normal to a curve

y = f(x)

x = a

is the negative reciprocal of the derivative evaluated at that point:

m_n = -\frac{1}{f'(a)}

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