# Homework Help: Equation of tangent line (rec. form) to a polar curve

1. Jun 4, 2010

### System

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

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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?

2. Jun 4, 2010

### 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 )
(x1,y1) will be (r,θ) at θ = π/4.

3. Jun 4, 2010

### System

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

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

4. Jun 4, 2010

### tiny-tim

(what happened to that π i gave you? )

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

5. Jun 4, 2010

### System

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

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?

6. Jun 5, 2010

### tiny-tim

(just got up :zzz: …)

Right!