Question

Grade 12 Math:

Find the slope of the normal to the curve x = a cos³ θ, y = a sin³ θ at θ = π/4.

Don't give spam answers & don't post plagiarised answers either. I know the answer, I don't get the method though. It'll be helpful if someone solves it for me. Thanks.

Answer 1

Given x = acos³θ , y = asin³θ

Diffrentiating w.r.t. θ

dx/dθ = -3acos²θsinθ ____(1)

dy/dθ = 3asin²θcosθ ______(2)

diving (2) by (1)

dy/dθ ×dθ/dx = -( 3asin²θ cosθ /3acos²θsinθ )

dy/dx = - tanθ

dy/dx = slope of tangent = -tanθ

putting θ = π/4

∴ dy/dx =- tanπ/4 = -1

but , slope of normal = - 1/slope of tangent

∵ slope of normal = -1/-1 = 1

∴ slope of normal = 1 Answer

Answer 2

Answer:

Slope of normal = 1

Step-by-step explanation:

   $\sf x =aCos^3 \ \theta\\\\\dfrac{dy}{d \theta} = a*3Cos^2 \ \theta (-Sin \ \theta)$

         $\sf = -3aCos^2 \ \theta *Sin \ \theta$

$\sf y =aSin^3 \ \theta\\\\\dfrac{dy}{d\theta}=a*3Sin^2 \ \theta*(Cos \ \theta)\\\dfrac{dy}{d\theta}=3aSin^2 \ \theta*Cos \ \theta$

             

          $\sf Slope \ of \ Tangent = \dfrac{dy}{dx} =\dfrac{dy}{d\theta} \div \dfrac{dx}{d\theta}$

                    $\sf = \dfrac{3aSin^2 \ \theta \ Cos \ \theta}{-3aCos^2 \ \theta \ Sin \ \theta}\\\\ = \dfrac{-Sin \ \theta}{Cos \ \theta}\\\\= -tan \ \theta$

$\theta = \dfrac{\pi }{4}\\\\-tan\ \theta = -tan \ \dfrac{\pi }{4} \\\boxed{tan \ \dfrac{\pi }{4}=-1}$

Slope of tangent * slope of normal = -1

                                $\boxed{\text{Slope of normal = \dfrac{-1}{Slope \ of \ tangent} }}$

$\sf \text{Slope of normal=\dfrac{-1}{-1}}$$\sf Slope \ of \ normal =\dfrac{-1}{-1}$

                        = 1