Points, Vectors, and Functions Exercises

Answer

When θ = 0, r = sin θ = 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.

Put the previous two graphs together to find the graph for 0 ≤ θ ≤ π.

Answer

Check a couple of points. When θ = 0 we should have r = 1 + cos(0) = 2. When θ = π we should have r = 1 + cos(π) = 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 . 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, so this is a reasonable graph.

Answer

When θ = 0 we find r = 0 and when we find that r = 2. 

These points are on the graph.

When θ is between 0 and ^π⁄₂ (in other words, in the first quadrant) the quantity r = 2sin θ is positive. 

We should see points in the first quadrant. 

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 that now the graph shows points in the first quadrant like it ought to. This is a reasonable graph of the specified function.