Application of derivatives – Normal

Application of derivatives – Normal

Normal: The normal to the curve at any point P on it is the straight line which passes through P and is perpendicular to the tangent to the curve at P.

Slope of Normal: We know that normal to the curve at P (x₁, y₁) is a line perpendicular to tangent at P (x₁, y₁) and passing through P.

∴ Slope of the tangent at P = 1/ Slope of the tangent at P

⇒ Slope of normal at P (x₁, y₁) = $$-\frac{1}{{{\left( \frac{dy}{dx} \right)}_{\left( {{x}_{1}},\,{{y}_{1}} \right)}}}$$,

⇒ Slope of normal at P (x₁, y₁) = $$-{{\left( \frac{dx}{dy} \right)}_{\left( {{x}_{1}},\,{{y}_{1}} \right)}}$$.

Note: If normal is parallel to x – axis

⇒ $$-{{\left( \frac{dx}{dy} \right)}_{\left( {{x}_{1}},\,{{y}_{1}} \right)}}\,=\,0$$ or $${{\left( \frac{dx}{dy} \right)}_{\left( {{x}_{1}},\,{{y}_{1}} \right)}}\,=\,0$$,

If normal is perpendicular to x – axis (perpendicular to y – axis).

⇒ $$-{{\left( \frac{dy}{dx} \right)}_{\left( {{x}_{1}},\,{{y}_{1}} \right)}}\,=\,0$$.

Example: The slope of the normal to the curve x = 1- sin θ, y = b cos θ at θ = π/2 is

Solution: Given that,

x = 1- sin θ,

y = b cos θ

at θ = π/2

differentiation x and y w. r. t ‘θ’

dx/dθ = 0 – cos θ.

dx/dθ = – cos θ

dy/dθ = – b sinθ

$$\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{-b\sin \theta }{-\cos \theta }=\frac{b\sin \theta }{\cos \theta }$$,

dy/dx = b sinθ/ cosθ

we know that

⇒ Slope of normal at θ = $$-\frac{1}{\left( \frac{dy}{dx} \right)}$$,

Where θ = π/2

Slope of tangent dy/dx = b/0

Slope of tangent = 0.