Answer

When θ = 0, r = sinθ = 0.

When

That gets us two points:

As θ moves from 0 to , the value r = sin θ increases from 0 to 1.

When we know r = 1. When θ = π,r = sin π = 0.

That gives us two points:

As θ moves from to π, the value of r moves from 1 down to 0.

0≤ θ ≤ π

We put together the graphs we found in (a) and (b) to find r = sin(θ)

Answer

Check a couple of points. When θ = 0 we should have r = 1 + cos0 = 2. When θ = π we should have r = 1 + cosπ = 0. These points are on the graph. However, for 0 ≤ θ ≤ π the value r = 1 + cos θ is positive. We should be seeing points in the first and second quadrants.  

 Since the graph we were given shows points in the third and fourth quadrants, it's the wrong graph.

Answer

 When θ = 0 we find r = cos 0 = 1 and when θ = π we find r = cos \frac{π}{2} = 0. These points are on the graph. When θ is in between 0 and π, the value of r should be positive. This makes sense with what we see on the graph, this is a reasonable graph.

Answer

\item When θ = 0 we find r = 0 and when $θ = \frac{π}{2}$ we find that r = 2. These points are on the graph. PICTURE polar funcs 19 When θ is between 0 and $\frac{π}{2}$ (in other words, in the first quadrant) the quantity r = 2sinθ is positive. We should see points in the first quadrant. PICTURE polar funcs 20 Since the graph we're given shows points in the second quadrant, it's the wrong graph.

Answer

• This is much like the previous problem, except thatnow the graph shows points in the first quadrant like it ought to. This is a reasonable graph of the specified function.