**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.