### All AP Calculus BC Resources

## Example Questions

### Example Question #11 : Polar Form

What is the polar form of ?

**Possible Answers:**

None of the above

**Correct answer:**

We can convert from rectangular to polar form by using the following trigonometric identities: and . Given , then:

### Example Question #21 : Parametric, Polar, And Vector Functions

What is the polar form of ?

**Possible Answers:**

**Correct answer:**

We can convert from rectangular to polar form by using the following trigonometric identities: and . Given , then:

### Example Question #22 : Parametric, Polar, And Vector Functions

What is the polar form of ?

**Possible Answers:**

**Correct answer:**

We can convert from rectangular to polar form by using the following trigonometric identities: and . Given , then:

Dividing both sides by , we get:

### Example Question #23 : Parametric, Polar, And Vector Functions

Convert the following cartesian coordinates into polar form:

**Possible Answers:**

**Correct answer:**

Cartesian coordinates have x and y, represented as (x,y). Polar coordinates have

is the hypotenuse, and is the angle.

Solution:

### Example Question #24 : Parametric, Polar, And Vector Functions

Convert the following cartesian coordinates into polar form:

**Possible Answers:**

**Correct answer:**

Cartesian coordinates have x and y, represented as (x,y). Polar coordinates have

is the hypotenuse, and is the angle.

Solution:

### Example Question #25 : Parametric, Polar, And Vector Functions

Calculate the polar form hypotenuse of the following cartesian equation:

**Possible Answers:**

**Correct answer:**

In a cartesian form, the primary parameters are and . In polar form, they are and

is the hypotenuse, and is the angle created by .

2 things to know when converting from Cartesian to polar.

You want to calculate the hypotenuse,

Solution:

### Example Question #26 : Parametric, Polar, And Vector Functions

Graph the equation where .

**Possible Answers:**

**Correct answer:**

At angle the graph as a radius of . As it approaches , the radius approaches .

As the graph approaches , the radius approaches .

Because this is a negative radius, the curve is drawn in the opposite quadrant between and .

Between and , the radius approaches from and redraws the curve in the first quadrant.

Between and , the graph redraws the curve in the fourth quadrant as the radius approaches from .

### Example Question #27 : Parametric, Polar, And Vector Functions

Draw the graph of from .

**Possible Answers:**

**Correct answer:**

Because this function has a period of , the x-intercepts of the graph happen at a reference angle of (angles halfway between the angles of the axes).

Between and the radius approaches from .

Between and , the radius approaches from and is drawn in the opposite quadrant, the third quadrant because it has a negative radius.

From to the radius approaches from , and is drawn in the fourth quadrant, the opposite quadrant.

Between and , the radius approaches from .

From and , the radius approaches from .

Between and , the radius approaches from . Because it is a negative radius, it is drawn in the opposite quadrant, the first quadrant.

Then between and the radius approaches from and is draw in the second quadrant.

Finally between and , the radius approaches from .

### Example Question #7 : Derivatives Of Polar Form

Find the derivative of the following function:

**Possible Answers:**

**Correct answer:**

The derivative of a polar function is given by the following:

First, we must find

We found the derivative using the following rules:

,

Finally, we plug in the above derivative and the original function into the above formula:

### Example Question #31 : Functions, Graphs, And Limits

What is the derivative of ?

**Possible Answers:**

**Correct answer:**

In order to find the derivative of a polar equation , we must first find the derivative of with respect to as follows:

We can then swap the given values of and into the equation of the derivative of an expression into polar form:

Using the trigonometric identity , we can deduce that . Swapping this into the denominator, we get: