# AP Calculus BC : Parametric Form

## Example Questions

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### Example Question #1 : Parametric Form

Rewrite as a Cartesian equation:

Explanation:

So

or

We are restricting  to values on , so  is nonnegative; we choose

.

Also,

So

or

We are restricting  to values on , so  is nonpositive; we choose

or equivalently,

to make  nonpositive.

Then,

and

### Example Question #2 : Parametric Form

Rewrite as a Cartesian equation:

Explanation:

, so

This makes the Cartesian equation

.

### Example Question #3 : Parametric Form

If  and , what is  in terms of  (rectangular form)?

Explanation:

Given  and  , we can find  in terms of  by isolating  in both equations:

Since both of these transformations equal , we can set them equal to each other:

### Example Question #4 : Parametric Form

Given  and , what is the arc length between ?

Explanation:

In order to find the arc length, we must use the arc length formula for parametric curves:

.

Given   and , we can use using the Power Rule

for all , to derive

and

.

Plugging these values and our boundary values for  into the arc length equation, we get:

Now, using the Power Rule for Integrals

for all ,

we can determine that:

### Example Question #5 : Parametric Form

Given  and , what is the length of the arc from ?

Explanation:

In order to find the arc length, we must use the arc length formula for parametric curves:

.

Given  and , we can use using the Power Rule

for all  , to derive

and

.

Plugging these values and our boundary values for into the arc length equation, we get:

Now, using the Power Rule for Integrals

for all ,

we can determine that:

### Example Question #6 : Parametric Form

Find the length of the following parametric curve

,   ,   .

Explanation:

The length of a curve is found using the equation

We use the product rule,

, when  and  are functions of ,

the trigonometric rule,

and

and exponential rule,

to find  and

In this case

,

The length of this curve is

Using the identity

Using the identity

Using the trigonometric identity  where  is a constant and

Using the exponential rule,

Using the exponential rule, , gives us the final solution

### Find dy/dx at the point corresponding to the given value of the parameter without eliminating the parameter:

Explanation:

The formula for dy/dx for parametric equations is given as:

From the problem statement:

If we plug these into the above equation we end up with:

If we plug in our given value for t, we end up with:

This is one of the answer choices.

### Example Question #8 : Parametric Form

Draw the graph of  from .

Explanation:

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 #7 : Parametric Form

Draw the graph of  where .

Explanation:

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 #10 : Parametric Form

Graph  where .

Explanation:

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 .

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