- #1

- 144

- 0

Given the parametric equations, find an equation of the tangent line at the given point on the curve.

Find an equation of the tangent line at each given point on the curve:

x = 2cotΘ and y=2sin[itex]^{2}θ[/itex] at point ([itex]\frac{-2}{\sqrt{3}}[/itex],[itex]\frac{3}{2})[/itex]

dy/dx = dy/dθ/dx/dθ

[itex]\frac{-2}{\sqrt{3}} = 2cotθ[/itex] and [itex]\frac{3}{2}=2sin^{2}θ[/itex]

[itex]\frac{-1}{\sqrt{3}}=\frac{cosθ}{sinθ

}[/itex] and [itex]\frac{\sqrt{3}}{2}=sinθ[/itex]

letting θ = [itex]\frac{2\pi}{3}[/itex] for each function yields the point ([itex]\frac{-2}{\sqrt{3}}[/itex],[itex]\frac{3}{2})[/itex]

dy/dx = [itex]\frac{dy/dθ}{dx/dθ}[/itex] = [itex](\frac{2sin^{2}θ}{2cotθ})'[/itex]

= [itex]-2sin^{3}θcosθ[/itex]

at θ= [itex]\frac{2\pi}{3}[/itex]

slope of tangent = 9/8

so m = (y-y)/(x-x)

9/8 = [itex]\frac{(y-\frac{3}{2}}{x-(-\frac{-2}{\sqrt{3}}}[/itex]

y = [itex]\frac{9}{8} + 15/4[/itex]

Answer doesn't match. I must have made a mistake somewhere.

## Homework Statement

Find an equation of the tangent line at each given point on the curve:

x = 2cotΘ and y=2sin[itex]^{2}θ[/itex] at point ([itex]\frac{-2}{\sqrt{3}}[/itex],[itex]\frac{3}{2})[/itex]

## Homework Equations

dy/dx = dy/dθ/dx/dθ

## The Attempt at a Solution

[itex]\frac{-2}{\sqrt{3}} = 2cotθ[/itex] and [itex]\frac{3}{2}=2sin^{2}θ[/itex]

[itex]\frac{-1}{\sqrt{3}}=\frac{cosθ}{sinθ

}[/itex] and [itex]\frac{\sqrt{3}}{2}=sinθ[/itex]

letting θ = [itex]\frac{2\pi}{3}[/itex] for each function yields the point ([itex]\frac{-2}{\sqrt{3}}[/itex],[itex]\frac{3}{2})[/itex]

dy/dx = [itex]\frac{dy/dθ}{dx/dθ}[/itex] = [itex](\frac{2sin^{2}θ}{2cotθ})'[/itex]

= [itex]-2sin^{3}θcosθ[/itex]

at θ= [itex]\frac{2\pi}{3}[/itex]

slope of tangent = 9/8

so m = (y-y)/(x-x)

9/8 = [itex]\frac{(y-\frac{3}{2}}{x-(-\frac{-2}{\sqrt{3}}}[/itex]

y = [itex]\frac{9}{8} + 15/4[/itex]

Answer doesn't match. I must have made a mistake somewhere.

Last edited: