It's time to get friendly with our graphing calculators. First, we need to understand how to graph polar functions by hand.

This is one of the many instances in calculus where it's helpful to use a calculator as a tool, but it's important to know what it's output means.

When checking the graph on the graphing calculator it can be helpful to spot-check points, especially at the boundaries of θ and simple angles like 0, , and to make sure *r* has the right values. Then look at what *r* is doing between those points to see if it makes sense.

**Calculator Tip:** If the calculator graph looks like jagged lines instead of looking curvy, try making the θ step size smaller (this may be called Δ θ on the calculator). This affects how carefully the calculator draws the graph.

## Practice:

- Without using a calculator, graph the polar functionr = cosθ
for
| |

Start by finding *r* at the endpoints. When θ = 0, r = cos0 = 1.
When ,
As θ moves from 0 to , the value of r = cos θ decreases from 1 to 0.
When , r = 0. When θ = π, r = cosπ = -1. Remembering how we graph polar coordinates with negative *r*, we find these points: When θ is in the second quadrant and *r* is negative, the point (r,θ) is in the fourth quadrant. As θ moves closer to π, the points (r,θ) move closer to the positive *x*-axis. As θ moves from to π, the value r = cosθ goes from 0 to -1. The points (r,θ) move further from the origin as θ gets closer to π. To graph r = cosθ for 0≤ θ≤ π we put the graphs from (a) and (b) together: | |

Determine whether the following is a correct graph of
r = cos θ
for
| |

When we should have
and we do:
When θ = 0 we should have r = cos 0 = 1,
and we do:
For θ in between and 0 the value of *r* should be positive, and it is:
This appears to be a correct graph of r = cosθ.
| |

Determine whether the following is a correct graph of
r = sin θ
for
PICTURE: polar funcs 13 | |

When $θ = -\frac{π}{2}$ we should have
r = sin -\frac{π}{2} = -1.
But the point
(r,θ) = (-1,-\frac{π}{2})
isn't on the graph we were given.
PICTURE polar funcs 14
This must not be the right graph.
| |

Without using a calculator, graph the function r = sin θ for

0 ≤ θ ≤ π

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)

Exercise. Determine if each graph is a reasonable graph of the given equation.

- r = 1 + cos θ for 0≤ θ ≤ π

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.

Exercise. Determine if each graph is a reasonable graph of the given equation.

- \item $r = cos \frac{θ}{2}$ for $0≤ θ ≤ π$
PICTURE graph this (the correct 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.

Exercise. Determine if each graph is a reasonable graph of the given equation.

r = 2sinθ for 0≤ θ ≤ frac{π/2

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.

Exercise. Determine if each graph is a reasonable graph of the given equation.

r = 2sinθ for 0≤ θ ≤ π/2

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.