Equation of tangent line (rec. form) to a polar curve

[System]

Equation of tangent line (rec. form) to a polar curve!

Homework Statement

Quesiton:

Find the rectangular form of the equation of the tangent line to the polar curve r=cos^3(theta) at the point corresponding to theta=pi/4

Homework Equations

The Attempt at a Solution

How to do that?

I mean finding it in RECTANGULAR FORM !


i know that
$$\frac{dy}{dx}=\frac{\frac{dr}{d\theta} sin(\theta)+rcos(\theta)}{\frac{dr}{d\theta} cos(\theta) - r sin(\theta)}$$

I will calculate dy/dx at theta=pi/4 , and this is easy ..

The problem here is that, how can I find the equation of the tangent line in RECATNGULAR FORM ??

The equation of the tangent line is :

y-y1=m(x-x1)

m = the slope , and this one will be calculated by using the formula ..

but what about x1 and y1?

 

[tiny-tim]

Hi System!

(have a pi: π and a theta: θ and try using the X² and X₂ tags just above the Reply box )

y-y1=m(x-x1)

m = the slope , and this one will be calculated by using the formula ..

but what about x1 and y1?

(x₁,y₁) will be (r,θ) at θ = π/4.

 

[System]

I do not think so.
so y=pi/4 ?! :S

 

[tiny-tim]

(what happened to that π i gave you? )

No, you have to convert (r,θ) to (x,y) at θ = π/4.

 

[System]

ohhh i see now
x = r cosθ = cos^4 θ
y = r sinθ = cos^3θ sinθ

I will evaluate them at θ=pi/4 to get x1 & y1
and I have m from the formula of dy/dx in polar

I will substitute x1,y1 & m in the line equation and I will be finish, right?

 

[tiny-tim]

(just got up :zzz: …)

Right!