AP Calculus BC : Polar Form

Study concepts, example questions & explanations for AP Calculus BC

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Example Questions

Example Question #24 : Polar Form

What is the polar form of ?

Possible Answers:

None of the above

Correct answer:

Explanation:

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

 

Example Question #41 : Polar Form

What is the polar form of ?

Possible Answers:

Correct answer:

Explanation:

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

 

Example Question #51 : Polar Form

What is the polar form of ?

Possible Answers:

Correct answer:

Explanation:

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 #68 : Polar Form

Convert the following cartesian coordinates into polar form: 

Possible Answers:

Correct answer:

Explanation:

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

 is the hypotenuse, and  is the angle.

Solution:

Example Question #11 : Polar Form

Convert the following cartesian coordinates into polar form:

Possible Answers:

Correct answer:

Explanation:

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

 is the hypotenuse, and  is the angle.

Solution:

Example Question #11 : Polar Form

Calculate the polar form hypotenuse of the following cartesian equation:

Possible Answers:

Correct answer:

Explanation:

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 #2 : Graphing Polar Form

Graph the equation  where .

Possible Answers:

R_sinx

R_cosx

Faker_cosx

R_cosx_1

R_cos2x

Correct answer:

R_cosx

Explanation:

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 #1 : Graphing Polar Form

Draw the graph of  from .

Possible Answers:

R_sin2x

R_cos2x

R2_cos2x

R_sin2x

R_cosx_1

Correct answer:

R_cos2x

Explanation:

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 #1 : Derivatives Of Polar Form

Find the derivative of the following function:

Possible Answers:

Correct answer:

Explanation:

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 : Parametric, Polar, And Vector Functions

What is the derivative of ?

Possible Answers:

Correct answer:

Explanation:

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:

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