# What is the equation (in terms of x and y) of the tangent line to the curve r=5sin4θ at θ=π/3?

y= ____________

### 1 Answer

- cidyahLv 71 decade agoFavorite Answer
r= 5 sin 4θ

dr/dθ = 5 cos 4θ (4) = 20 cos 4θ

dy/dx = (dr/dθ sin θ + r cos θ) / (dr/dθ cps θ + r sin θ)

dy/dx = ( (20 cos 4θ) (sin θ) + (5 sin 4θ) ( cos θ) ) / ( (20 cos 4θ) (cps θ) + (5 sin 4θ) ( sin θ) )

Substitute θ = π/3

cos π/3 = 1/2

sin π/3 = √3/2

cos 4π/3 = -1/2

sin 4π/3 = -√3/2

dy/dx = [ (20) (-1/2)(√3/2) + (5) ( -√3/2)(1/2) ] / [ (20) ( -1/2)(1/2) + (5) ( -√3/2)(√3/2)

dy/dx = [ -5√3 -(5/4) √3 ] / [ -5 - 15/4]

dy/dx = -25√3 /4 -35/4 = (-25√3-35) / 4

When θ=π/3, r = 5 sin 4π/3 = -5√3/2

x= r cos θ = (-5√3/2)( 1/2 ) = -5√3/4

y = r sin θ = (-5√3/2)(√3/2) = -15/4

Equation of tangent line:

y - y1 = dy/dx ( x-x1)

y - ( -15/4) = (-25√3-35) / 4 ( x - (-5√3/4) )

y + 15/4 = (-25√3-35) / 4 ( x+ -5√3/4)

y = (-15/4) - (25√3+35) / 4) ( x+ -5√3/4)

Please check all of my calculation.