How to Convert Rectangular Equations to Polar Form?

[Punchlinegirl]

If someone could check my work and make sure I'm doing these problems right, I would really appreciate it.
1.Eliminate the parameter and obtain the standard form of the rectangular equation.
Circle: $$x= h + r cos \theta , y= k + r sin \theta$$
[tex] (x-h/r)^2 + (y-k/r)^2 = 1[/tex]
2.Find the arc length of the given curve on the indicated interval.
$$x=t^2 +1, y=4t^3 + 3$$
$$0 \leq t \leq -1$$
$$S= \int \sqrt (dx/dt)^2 + (dy/dt)^2$$
dx/dt= 2t, dy/dt= 12t^2
so i intregrated from -1 to 0, $$\int \sqrt 4t^2 +144t^2 dt$$
using a u subsitution, I got $$1/432(4+144t^2)^(3/2)$$
Plugging in -1 and 0 gave me -4.15, which can't be right since it's talking about arclength..
3.Convert the rectangular equation to polar.
$$x^2 + y^2 - 2ax = 0$$
$$r^2 = 2ax$$
$$r^2 / r cos \theta = 2a (r cos \theta)/ r cos \theta$$
Solving for r gave me [tex] r= 2a cos \theta [/tex]
If these are wrong, any help would be appreciated. Thanks!

 

[Punchlinegirl]

can someone please help?

 

[Tide]

1. Your equation should be

$$\frac {(x-h)^2)}{r^2} + \frac {(y-k)^2}{r^2} = 1$$

2. Check your inequality. I am guessing it should read $0 \leq x \leq 1$.

 

[BerkMath]

Question 2 looks like it gives you an elliptical integral so you will have to numerically approximate it. 3. r=2*a*cos(theta) looks good.