Revising past exam papers and I am completely lost

It would helpful if you could state what exactly you are struggling with.

a) can be done using just the definition of polar coordinates.
(a) You know that **x=r·cos(phi)** and **y=r·sin(phi)**, the standard polar coordinates. You have **r** as a function of **phi**, which we can write as **r=f(phi)**. So you simply get **x=r·cos(phi)=f(phi)cos(phi)** and **y=r·sin(phi)=f(phi)sin(phi)**.

(b) From the standard polar coordinates, you get **x\^2+y\^2=r\^2**. So you can mulitply the given polar equation by **r**, to get **r\^2=r+b·r·cos(phi)**, which in Cartesian coordinates is **x\^2+y\^2=sqrt(x\^2+y\^2)+bx**. I'm sure you can work out the rest to get the equation needed.

(c) The area of this is twice the area of its upper half. Luckily, the symmetry axis is the x-axis, so you need only look for the part above the x-axis. Those parts of the curve, it's easily calculated, correspond to **phi ∈ $0,2pi/3$** (the "bigger" one) and **phi ∈ $pi,4pi/3$** (the "smaller" one). The formula for area bounded in polar coordinates for **phi ∈ $alpha, beta$** is **A=1/2\*int$alpha,beta$ r(phi)\^2 dphi**. To find this *half,* you compute this formula for **phi ∈ $0,2pi/3$**, and then *subtract* the one for **phi ∈ $pi,4pi/3$**. What you got is area of half the figure, so you **multiply it by 2.**

(d) It says that **r=0**, so the point of intersection is **(0,0)**. So, you must find two different values **phi\_1** and **phi\_2** such that **x(phi\_1)=y(phi\_1)=0** and **x=(phi\_2)=y(phi\_2)=0**. But, from part (c), we already know those are **phi=2pi/3** and **phi=4pi/3**. So, you find the tangent vector (which in general case is simply **(x'(phi),y'(phi))** ) in that point, separately for the values **phi\_1** and **phi\_2**. Those tangent vectors, you can note as **v\_1** and **v\_2**. To find the angle between them, use the formula **cos(tau)=(v\_1\*v\_2)/(|v\_1||v\_2|)**, where **tau** is said angle.