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

Slope of normal = 1

Explanation:

`\sf x =aCos^3 \ \theta\\\\(dy)/(d \theta) = a*3Cos^2 \ \theta (-Sin \ \theta)\\`

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

`\sf y =aSin^3 \ \theta\\\\(dy)/(d\theta)=a*3Sin^2 \ \theta*(Cos \ \theta)\\(dy)/(d\theta)=3aSin^2 \ \theta*Cos \ \theta`

`\sf Slope \ of \ Tangent = (dy)/(dx) =(dy)/(d\theta) / (dx)/(d\theta)`

`\sf = (3aSin^2 \ \theta \ Cos \ \theta)/(-3aCos^2 \ \theta \ Sin \ \theta)\\\\ = (-Sin \ \theta)/(Cos \ \theta)\\\\= -tan \ \theta`

`\theta = (\pi )/(4)\\\\-tan\ \theta = -tan \ (\pi )/(4) \\\boxed{tan \ (\pi )/(4)=-1}`

Slope of tangent * slope of normal = -1

`\boxed{\text{Slope of normal =$(-1)/(Slope \ of \ tangent)$}}`

`\sf \text{Slope of normal=(-1)/(-1)}`
`\sf Slope \ of \ normal =(-1)/(-1)`

= 1

Answer 2

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