Question

Find the slope of the normal to the curve x = acos 3 θ, y = asin 3 θ at.

Answer 1

Find the slope of the normal to the curve x = acos³θ, y = asin³θ at θ = π/4

solution :- x = acos³θ
differentiate x with respect to θ,
dx/dθ = 3acos²θ(-sinθ)
dx/dθ = -3asinθcos²θ --------(1)

similarly, y = asin³θ
differentiate y with respect to θ
dy/dθ = 3asin²θcosθ ------(2)
now, dividing equation (2) by (1),
$\frac{dy/d\theta}{dx/d\theta}=\frac{-3a sin\theta cos^2\theta}{3a cos\theta sin^2\theta}\\\\\frac{dy}{dx}|_{\theta=\frac{\pi}{4}}=-cot\theta\\\\\frac{dy}{dx}|_{\theta=\frac{\pi}{4}}=-cos\frac{\pi}{4}=-1$

hence, slope of tangent = -1
we know, slope of tangent × slope of normal = -1
so, slope of normal = -1/(slope of tangent)
= -1/-1 = 1

hence, slope of normal to the curve = 1

Answer 2

Answer: 1

Step-by-step explanation:

x = acos³θ

differentiate x with respect to θ,

dx/dθ = 3acos²θ(-sinθ)

dx/dθ = -3asinθcos²θ --------(1)

similarly, y = asin³θ

differentiate y with respect to θ

dy/dθ = 3asin²θcosθ ------(2)

now, dividing equation (2) by (1),

hence, slope of tangent = -1

we know, slope of tangent × slope of normal = -1

so, slope of normal = -1/(slope of tangent)

= -1/-1 = 1

hence, slope of normal to the curve = 1