### All Calculus 2 Resources

## Example Questions

### Example Question #1 : Graphing Polar Form

Which of the following substitutions will help solve the following integral?

**Possible Answers:**

**Correct answer:**

As we can see in this integral, there is no reverse chain-rule u-substitution possible. The logical step is to use a trigonometric substitution. If one recalls that trig substitutions of the type could be solved with the substitution , then the answer is easily seen. However, we can also use a right triangle:

And thus we have:

or:

### Example Question #2 : Graphing Polar Form

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 #8 : Functions, Graphs, And Limits

Draw the graph of from .

**Possible Answers:**

**Correct answer:**

Between and , the radius approaches from .

From to the radius goes from to .

Between and , the curve is redrawn in the opposite quadrant, the first quadrant as the radius approaches .

From and , the curve is redrawn in the second quadrant as the radius approaches from .

### Example Question #11 : Polar Form

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 #1 : Parametric Form

Draw the graph of where .

**Possible Answers:**

**Correct answer:**

Because this function has a period of , the amplitude of the graph appear at a reference angle of (angles halfway between the angles of the axes).

Between and the radius approaches 1 from 0.

Between and , the radius approaches 0 from 1.

From to the radius approaches -1 from 0 and is drawn in the opposite quadrant, the fourth quadrant because it has a negative radius.

Between and , the radius approaches 0 from -1, and is also drawn in the fourth quadrant.

From and , the radius approaches 1 from 0. Between and , the radius approaches 0 from 1.

Then between and the radius approaches -1 from 0. Because it is a negative radius, it is drawn in the opposite quadrant, the second quadrant. Likewise, as the radius approaches 0 from -1. Between and , the curve is drawn in the second quadrant.

### Example Question #1 : Parametric, Polar, And Vector Functions

Graph where .

**Possible Answers:**

**Correct answer:**

Taking the graph of , we only want the areas in the positive first quadrant because the radius is squared and cannot be negative.

This leaves us with the areas from to , to , and to .

Then, when we take the square root of the radius, we get both a positive and negative answer with a maximum and minimum radius of .

To draw the graph, the radius is 1 at and traces to 0 at . As well, the negative part of the radius starts at -1 and traces to zero in the opposite quadrant, the third quadrant.

From to , the curves are traced from 0 to 1 and 0 to -1 in the fourth quadrant. Following this pattern, the graph is redrawn again from the areas included in to .

### Example Question #4 : Graphing Polar Form

Draw the curve of from .

**Possible Answers:**

**Correct answer:**

Taking the graph of , we only want the areas in the positive first quadrant because the radius is squared and cannot be negative.

This leaves us with the areas from to and to .

Then, when we take the square root of the radius, we get both a positive and negative answer with a maximum and minimum radius of .

To draw the graph, the radius is 0 at and traces to 1 at . As well, the negative part of the radius starts at 0 and traces to-1 in the opposite quadrant, the third quadrant.

From to , the curves are traced from 1 to 0 and -1 to 0 in the third quadrant.

Following this pattern, the graph is redrawn again from the areas included in to .

### Example Question #5 : Graphing Polar Form

What are the parameters by which one can describe the position of a point A in a polar coordinates plane?

**Possible Answers:**

By a position vector to a point A and an angle between horizontal axis and said vector (counter-clockwise positive) .

By a position vector to a point and an angle between vertical axis and said vector (counter-clockwise positive).

By a position vector to a point A and an angle between horizontal axis and said vector (clockwise positive).

By distance y from horizontal axis and distance x from vertical axis to point A.

By a position vector to a point and an angle between vertical axis and said vector (clockwise positive).

**Correct answer:**

By a position vector to a point A and an angle between horizontal axis and said vector (counter-clockwise positive) .

A point in polar coordinates is described by a position vector to a point A and an angle between horizontal axis and said vector. A convention for a positive angle is counter-clockwise.

Note, that in polar coordinates, position vector may also be of negative value, meaning pointing in the opposite direction.

### Example Question #1 : Graphing Polar Form

Describe the graph of .

**Possible Answers:**

Circle centered around the origin with a radius

Circle centered around the origin with a radius

Straight line passing through the origin and

Straight line passing through the origin and

**Correct answer:**

Straight line passing through the origin and

Graphing polar equations is different that plotting cartesian equations. Instead of plotting an coordinate, polar graphs consist of an coordinate where is the radial distance of a point from the origin and is the angle above the x-axis.

When the graph of an equation in the form , where is an angle, the angle of the graph is constant and independent of the radius. This creates a straight line radians above the x-axis passing through the origin.

In this problem, is a straight line radians or about the x-axis passing through the origin.

### Example Question #1 : Graphing Polar Form

Describe the graph of .

**Possible Answers:**

Cardiod centered around with a radius of .

Circle centered around the origin with a radius of .

Straight line passing through the origin and

Limacon with inner loop centered around

**Correct answer:**

Circle centered around the origin with a radius of .

Graphing polar equations is different that plotting cartesian equations. Instead of plotting an coordinate, polar graphs consist of an coordinate where is the radial distance of a point from the origin and is the angle above the x-axis.

When the graph of an equation in the form , where is a constant, the graph is a circle centered around the origin with a radius of .

In this problem, is a circle centered around the origin with a radius of .

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